Wiredfaculty Logo

For Daily Free Study Material Join wiredfaculty Whatsapp Group | Download Android App | Ncert Book Download

Physics Part I Chapter 2 Units And Measurement

    • Sponsor Area

    NCERT Solution For Class 11 Physics Physics Part I

    Units And Measurement Here is the CBSE Physics Chapter 2 for Class 11 students. Summary and detailed explanation of the lesson, including the definitions of difficult words. All of the exercises and questions and answers from the lesson's back end have been completed. NCERT Solutions for Class 11 Physics Units And Measurement Chapter 2 NCERT Solutions for Class 11 Physics Units And Measurement Chapter 2 The following is a summary in Hindi and English for the academic year 2021-2022. You can save these solutions to your computer or use the Class 11 Physics.

    Question 1
    CBSEENPH110024915
    Wired Faculty App

    Fill up 3.0 m/s2 = ___km/hr2
    Solution

    Here we have to convert m/s2 into km/h2

    1 km = 1000 m

    1 m = 10-3 km

    1 hour = 60 min = 60 × 60 sec
    or 1 sec =  wiredfaculty.com  h

    Substituting the above values

    3 m s-2 = 3 × 10-3 km ×  wiredfaculty.com  h-2

    = 38880 km-h-2h

    = 3.88 × 104 km h-2

    Question 2
    CBSEENPH110024927
    Wired Faculty App

    E, m, L, G denote energy, mass, angular momentum & gravitation constant respectively. The dimensions of EL EL2M5G2
    Solution

    dimension of energy  = ML2T-2m⇒ML→ML2T-1G→M-1L3T-2dimension of EL2M5G2⇒M0L0T0dimension less means angle

    Question 3
    CBSEENPH11016190
    Wired Faculty App

    What is measurement?

    Solution
    Measurement is the process of attaching a numeric value to an aspect of a natural phenomenon. Measurement is making the quantitative knowledge of a physical quantity. 
    Question 4
    CBSEENPH11016191
    Wired Faculty App

    What is the need of measurement?

    Solution
    To get the complete knowledge of any physical quantity, measurement is needed. Measurement of any physical quantity involves comparison with a certain basic, arbitrarily chosen, internationally accepted reference standard called unit. Knowledge without measurement is incomplete and unsatisfactory. 
    Question 5
    CBSEENPH11016192
    Wired Faculty App

    Define physical quantity.

     
    Solution
    A measurable quantity in terms of which laws of Physics can be expressed is called physical quantity.

    The result of a measurement of a physical quantity is expressed by a number accompanied by a unit.
    Question 6
    CBSEENPH11016194
    Wired Faculty App

    Give four examples of physical quantities.

    Solution
    Examples of Physical quantitie are force, mass, density and power.
    Question 7
    CBSEENPH11016195
    Wired Faculty App

    Define unit.

    Solution

    The quantity used as a standard of measurement is called unit. 

    Example: Unit of time is second. 

    Question 8
    CBSEENPH11016196
    Wired Faculty App

    What are the two main required characteristics of a unit?

    Solution
    Characteristics of unit are:

    1. Invariability
    2. Avaialbility
    Question 9
    CBSEENPH11016197
    Wired Faculty App

    Can we use same units for different physical quantities?

    Solution
    No, different units are assigned for different physical quantities. Hence, we need different units for different physical quantities. 

    But, we can use different units for the same physical quantity. 
    Question 10
    CBSEENPH11016198
    Wired Faculty App

    Can we use different units to measure same physical quantity?

    Solution
    Yes, different units can be used to measure the same physical quantity. 
    For example, second, minute, hour etc. are the different units used to measure the physical quantity time. 
    Question 11
    CBSEENPH11016200
    Wired Faculty App

    To measure a physical quantity X, the unit U is repeatedly used for n times. What is the magnitude of physical quantity?

    Solution
    Here, 
    Physical quantity is X
    Unit is U
    Therefore, magnitude of physical quantity is X = nU.
    Question 12
    CBSEENPH11016201
    Wired Faculty App

    Does the numerical factor in the quantitative measurement depend on the system of unit used?

    Solution
    Yes. the numerical factor  in the quantitative measurement depend on the system of unit used.

    For example, let us consider a length. 
    1 km = 1000 m

    Here, we can see that the magnitude of the length has changed when the unit has changed from kilometre to metre.
    Question 13
    CBSEENPH11016202
    Wired Faculty App

    Is one unit sufficient to measure a physical quantity?

    Solution
    A physical quantity can be measured using a number of units. More than one unit is required to measure a  physical quantity. 
    For eg. time can be measured using the units second, minute, hour. 
    Question 14
    CBSEENPH11016212
    Wired Faculty App

    A physical quantity X is measured in two units i.e. U1 and U2 The unit U1 needs to be used for n1 time to measure X. How many times the unit U2 will be required to measure X?

    Solution
    Measure of any quantity is given by,

    Measure = physical quantityunit

              n1 = XU1

    Therefore, the number of times unit U2 will be required to measure X is given by, 

                                   n2 = n1U1U2
    Question 15
    CBSEENPH11016213
    Wired Faculty App

    Can a physical quantity have two or more units? Give example.

    Solution
    Yes. A physical quantity can have more than two or more units.
    For e.g., The units of length are Å, cm, m, parsec, light year. Length is measurable in all these units. 
    Question 16
    CBSEENPH11016214
    Wired Faculty App

    What are fundamental quantities?

    Solution
    Fundamental units are the physical quantities, which do not depend on other quantities.
    Examples of fundamental units are mass, length, temperature etc. 
    Fundamental quantities have fundamental units which are not derived from other units. 
    Quantity
     Unit
    Mass
     kilogram
    Temperature
     kelvin
    Question 17
    CBSEENPH11016215
    Wired Faculty App

    What are derived quantities?

    Solution
    Physical quantities that are derived from the base or fundamental quantities is known as derived quantities.

    e.g. force , velocity etc.

    Velocity = Displacementtime

    Here, velocity is derived from displacement and time and hence it is a derived quantity. 

    Sponsor Area

    Question 18
    CBSEENPH11016217
    Wired Faculty App

    How many fundamental quantities are there in mechanics?

    Solution
    There are three fundamental quantities in mechanics.
    Length, time and mass forms the bases of mechanics. 
    Question 19
    CBSEENPH11016218
    Wired Faculty App

    How many fundamental quantities are there in Physics?

    Solution
    There are seven fundamental quantities in Physics which are length, mass, time, electric current, thermodynamic temperature, amount of substance, luminous intensity. 
    Question 20
    CBSEENPH11016219
    Wired Faculty App

    Name the fundamental quantities of mechanics.

    Solution
    The fundamental quantities of mechanics are mass, length and time.
    Question 21
    CBSEENPH11016220
    Wired Faculty App

    Name the different system of units in mechanics that are internationally accepted.

    Solution
    By international convention there are three system of units that are internationally accepted. 

    The three systems are:
    i) mks (metre-kilogram-second) system
    ii) cgs (centimetre-gram-second) system
    iii) fps (foot-pound-second)system
    Question 22
    CBSEENPH11016221
    Wired Faculty App

    Which of the system of units are decimally related with each other?

    Solution

    mks and cgs system of units are decimally related with each other. 

     

    Question 24
    CBSEENPH11016223
    Wired Faculty App

    What is cgs system of units?

    Solution
    The centimetre–gram–second system of units (abbreviated CGS or cgs) is a variant of the metric system based on the centimetre as the unit of length, the gram as the unit of mass, and the second as the unit of time.
    Question 25
    CBSEENPH11016224
    Wired Faculty App

    What is mks system of units?

    Solution
    MKS is a system of measurement based on measuring lengths in meters, mass in kilograms, and time in seconds.
    Question 26
    CBSEENPH11016225
    Wired Faculty App

    Why mks system of units is most commonly used for scientific measurement?

    Solution
    MKS system of units is most commonly used for scientific measurement because magnitude of units in this system is of moderate size i.e. neither very small nor very large. These system of units are appropraite for measurement. 
    Question 27
    CBSEENPH11016226
    Wired Faculty App

    Can we measure all the physical quantities of heat and thermodynamic in mks system?

    Solution
    No. All the physical quantities of heat and thermodynamics cannot be measured in mks system. These quantities are measured using the Standard International system.
    Question 28
    CBSEENPH11016227
    Wired Faculty App

    Name the system of units in which all the physical quantities can be measured.
    Solution
    All the physical quantities can be measured using the Standard International (SI) system of units.
    Question 29
    CBSEENPH11016228
    Wired Faculty App

    Name the physical quantities which are quantitatively similar but have different units.

    Solution
    Mechanical energy, heat energy, electrical energy are similar types of quantity but their units are joule, calorie and watt hour respectively. Even though all of them are energies, their units are different. 
    Question 30
    CBSEENPH11016229
    Wired Faculty App

    What do you mean by rational system of units?

    Solution
    The system in which all similar types of physical quantities are expressed in one unit is known as rational system of units. Rational system is also known as the metric system of units and measurement. 
    Question 31
    CBSEENPH11016230
    Wired Faculty App

    Which system of units is rational system of units?

    Solution
    Standard International (SI) system of units is rational system of units.
    Question 32
    CBSEENPH11016231
    Wired Faculty App

    What is coherent system of units?

    Solution
    Coherent system of units are driven by equations and derived from fundamental units. In this system of units, all the derived units can be obtained by multiplying or dividing the fundamental units without introducing the numerical factor other than 1.
    The International System of Units (SI) is a coherent system. 
    Question 33
    CBSEENPH11016232
    Wired Faculty App

    What is prefix?

    Solution
    Prefix is multiple or sub-multiple that increases or decreases the size of unit. 
    For example, the prefix mega to the SI unit Joule gives us megajoule. It increases the size of the unit. 
    Question 34
    CBSEENPH11016233
    Wired Faculty App

    What are the multiple values of prefixes giga and tera?

    Solution
    The multiple values of prefixes giga and tera are 109 and 1012 respectively. 
    Question 35
    CBSEENPH11016234
    Wired Faculty App

    What is the multiple values of prefixes kilo and mega?

    Solution
    The multiple values of prefixes kilo and mega are 103 and 106respectively.
    Question 36
    CBSEENPH11016235
    Wired Faculty App

    What is atto?

    Solution
    Atto is used in units of measurement whose multiple value is 10–18. 
    Question 37
    CBSEENPH11016237
    Wired Faculty App

    Name the prefixes used for multiple values 1015 and 10–12.

    Solution
    The prefix used for given multiple values are:
    10-15 - pico
    10-12 - fermi

    Question 38
    CBSEENPH11016238
    Wired Faculty App

    1 pico second =..........second

    Solution
    10-12
    Question 39
    CBSEENPH11016239
    Wired Faculty App

    What are the multiple values of prefixes nano and micro?

    Solution
    Multiple values of prefixes nano and micro are 10–9 and 10–6 respectively.

    Sponsor Area

    Question 40
    CBSEENPH11016241
    Wired Faculty App

    What is the main advantage of prefix?

    Solution
    Prefix enables us to increase or decrease the size of unit without changing the basic unit of a physical quantity.
    Question 41
    CBSEENPH11016242
    Wired Faculty App

    Define 1 kilogram.

    Solution

    The kilogram is equal to the mass of the international prototype of the kilogram (a platinum-iridium alloy cylinder) kept at International Bureau of Weights and Measures, at Sevres, near Paris, France. 

    Question 42
    CBSEENPH11016244
    Wired Faculty App

    Define atomic mass unit.

    Solution

    The amount of substance of a system, which contains substance as many elementary entities as there are atoms in 0.012 kilogram of carbon -12.

    1 u = 1.66 x 10-27kg 

    Question 43
    CBSEENPH11016245
    Wired Faculty App

    Define kilogram in terms of mass of carbon -12.

    Solution
    One kilogram is the mass of 5.0188x1025 atoms of C–12 isotope. 
    Question 44
    CBSEENPH11016246
    Wired Faculty App

    Name the unit that is used to measure the atomic masses.

    Solution
    The unit u is used to measure atomic masses.

    The atomic mass unit which is abbreviated as amu or u is roughly equal to the mass of 1 proton or 1 neutron.
    Question 45
    CBSEENPH11016247
    Wired Faculty App

    How many kg are there in one slug?

    Solution
    The slug is the unit of mass in the US common system of units.

    1 slug = 14.57 kg.
    Question 46
    CBSEENPH11016248
    Wired Faculty App

    How many kilograms are there in one pound?

    Solution
    Pound is a unit of weight in the UK system.

    1 Pound = 0.453592 kg
    Question 47
    CBSEENPH11016249
    Wired Faculty App

    Define Chandra Shekher Limit (CSL).

    Solution
    It is the largest practical unit of mass and is equal to 1.4 times the solar mass.

    The Chandrashekhar limit is the maximum mass of a stable white dwarf star. 
    Question 48
    CBSEENPH11016250
    Wired Faculty App

    What is the order of mass of neutron and proton?

    Solution
    Mass of neutron and proton is of the order of 10-27 kg. Mass of neutron and proton is measured in the order of atomic mass unit (amu). 
    Question 49
    CBSEENPH11016251
    Wired Faculty App

    What is inertial mass?
    Solution

    The mass measured by studying the effect of external force on the body other than gravity is called inertial mass.

    Question 50
    CBSEENPH11016252
    Wired Faculty App

    What is gravitational mass?
    Solution
    Gravitational mass is determined by the strength of the gravitational force experienced by the body when in the gravitational field 'g'. 
    Question 51
    CBSEENPH11016253
    Wired Faculty App

    Which of the two masses is greater-inertial mass or gravitational mass?

    Solution
    Inertial mass and gravitational mass are precisely equal to each other. The equivalence of these masses makes it possible for all the objects to fall at the same rate.
    Question 52
    CBSEENPH11016254
    Wired Faculty App

    How do the inertial and gravitational masses differ - quantitatively or qualitatively.

    Solution
    The inertial mass and gravitational mass differ qualitatively.

    Inertial mass is found out using Newton's law. On applying a force of F Newton to an object, a being the acceleration we will get the inertial mass as m= Fa.

    Graviational mass is defined by the gravitational force which states that there is gravitational force between any two pair of objects. 
    F = Gm1m2r2
    where, m1 and m2 are the masses of the objects. 
    Question 53
    CBSEENPH11016256
    Wired Faculty App

    What docs the common balance compare or measure?

    Solution
    The common balance measures or compares the gravitational masses.
    The arms of the balance move up and down, leaving the only position for equilibrium as exactly straight, for equal masses. 
    Question 54
    CBSEENPH11016257
    Wired Faculty App

    What does the inertial balance compare or measure?

    Solution
    An inertial balance is used to measure the value of the inertial mass. It can also be used to compare the gravitational masses. 

    Inertial balance is basically a spring device that vibrates the sample which is being measured.
    Question 55
    CBSEENPH11016258
    Wired Faculty App

    What is the working principle of common balance?

    Solution
    Common balance is the balance which has each arm suspended.The unknown mass is placed in one arm and the known mass in another until they both become equal. 
    Therefore this balance works on the principle of moment of weights. When the weights become balanced, equilibrium is attained. 
    Question 56
    CBSEENPH11016259
    Wired Faculty App

    What is the working principle of inertial balance?

    Solution
    Inertial balance works on the principle that the time period of vibration of a cantilever is directly proportional to the square root of mass of the load placed at the free end of cantilever. 
    Question 57
    CBSEENPH11016260
    Wired Faculty App

    What is length of an object?

    Solution
    Length of an object is defined as the distance between two points at the extreme ends of the object. 
    Question 58
    CBSEENPH11016261
    Wired Faculty App

    What is unit of length in MKS system?

    Solution
    Meter is the unit of length in MKS system. 
    Question 59
    CBSEENPH11016262
    Wired Faculty App

    What is unit of length in FPS system?

    Solution
    Foot is the unit of length in FPS system. 
    Question 60
    CBSEENPH11016263
    Wired Faculty App

    Define the unit meter in terms of size of the earth.

    Solution
    One metre in terms of the size of the earth is defined as one-ten millionth of the length of meridian line from north pole of the earth to the equator, passing through Paris. 
    Question 61
    CBSEENPH11016264
    Wired Faculty App

    Define meter.

    Solution
    The metre is the length of the path travelled by light in vacuum during a time interval of 1/299,792,458 of a second. 
    Question 62
    CBSEENPH11016265
    Wired Faculty App

    Why particular alloy of platinum-iridium is chosen to make the rod used to define the unit of mass?

    Solution
    The platinum-iridium alloy is highly chemically stable. The relatively inert nature of the material minimises surface contamination and enhances the mass stability. 
    Platinum and its alloys are reasonably easy to machine, enabling a good surface finish to be achieved on the artifact that again reduces the effect of surface contamination. The addition of 10% iridium to the platinum greatly increases its hardness and so reduces wear. 

    Question 63
    CBSEENPH11016266
    Wired Faculty App

    Define the unit meter on atomic scale. 

    Solution
    Unit metre on atomic scale is defined as the distance occupied by 1650763.73 waves (in vacuum) emitted due to electronic transition from 5d5 to 2p10 by individual atom of Krypton-86 isotope. 
    Question 65
    CBSEENPH11016268
    Wired Faculty App

    What is the relation between meter, cm and foot?

    Solution
    1 foot = 30.48 cm = 0.3048 m. 
    Question 66
    CBSEENPH11016269
    Wired Faculty App

    Define cubit. Is it a definite quantity?

    Solution
    Cubit is a measur of length, approximately equal to the length of a forearm.
    It measures the length of an arm from elbow to the tip of middle finger. Cubit is not a definite quantity. It varies from person to person.
    Question 67
    CBSEENPH11016270
    Wired Faculty App

    What is millimeter?

    Solution
    Millimetre is unit of length and is equal to one-thousandth part of meter. 
    Question 68
    CBSEENPH11016271
    Wired Faculty App

    Define atomic unit

    Solution
    Atomic unit is the mean distance from the centre of the sun to the centre of earth and is numerically equal to 1.496x1011 m. 
    Question 69
    CBSEENPH11016272
    Wired Faculty App

    Define a light year.

    Solution
    Light year is the distance travelled by light in one year in vacuum. 
    1 light-year = 9.5x 1015 m. 
    Question 70
    CBSEENPH11016273
    Wired Faculty App

    1 Parallactic second =........m

    Solution
    1 Parallactic second = 3.08x1016 m.
    Question 71
    CBSEENPH11016274
    Wired Faculty App

    1 parsec  is how many light years? 

    Solution
    1 parsec is equal to 3.26 light years.
    Question 72
    CBSEENPH11016275
    Wired Faculty App

    How many miles are there in one kilometer?

    Solution
    1 kilometer = 0.6213 mile 
    Question 73
    CBSEENPH11016276
    Wired Faculty App

    Express mile in yards.

    Solution
    1 mile = 1760 yards 
    Question 74
    CBSEENPH11016277
    Wired Faculty App

    What is one micron?

    Solution
    Micron which is a unit of length is equal to 10–6 m. 
    Question 75
    CBSEENPH11016278
    Wired Faculty App

    Define the unit Angstrom.

    Solution
    Angstrom is a unit of length and is equal to 10-10 m.
    Question 76
    CBSEENPH11016279
    Wired Faculty App

    What is the smallest practical unit of length?

    Solution
    Fermi is the smallest practical unit of length and is equal to 10-15 m.
    Question 77
    CBSEENPH11016281
    Wired Faculty App

    Express furlong in yards and feet.

    Solution
    A furlong is a measure of distance in imperial units.

    1 furlong = 220 yard = 660 feet.
    Question 78
    CBSEENPH11016283
    Wired Faculty App

    How many nanometers are there in one micron?

    Solution
    There are 1000 nm in one micron.
    Question 79
    CBSEENPH11016284
    Wired Faculty App

    What does a fermi measure?

    Solution

    Fermi is a unit of length used to measure the size of nucleus. 

    1 fermi = 10–15m

    Sponsor Area

    Question 80
    CBSEENPH11016286
    Wired Faculty App

    When the earth is at apogee, then its distance from the sun is greater or less than 1 Astronomical unit?

    Solution
    At apogee, the earth is at a distance greater than 1astronomical unit.

    [Note: Apogee, or more formally apsis, is the point, in an elliptical Earth orbit, of greatest distance from the Earth.]
    Question 81
    CBSEENPH11016287
    Wired Faculty App

    Which of the following is the shortest unit of length : foot, furlong, yard.

    Solution
    Foot is the smallest unit of length.

    1 foot = 0.3048 m
    Question 82
    CBSEENPH11016289
    Wired Faculty App

    Arrange the following units in ascending order of their magnitude: 
    Light year, mile, parsec, foot, kilometer, furlong, yard and astronomical unit

    Solution
    Units in ascending order is as given below:

    Foot, yard, furlong, kilometer, mile, astronomical unit, light year, parsec. 
    Question 83
    CBSEENPH11016290
    Wired Faculty App

    Which unit would you like to measure the distance between the sun and the earth: Fermi or light year

    Solution
    Light year is the unit used to measure the distance between the Sun and the Earth.
    Question 84
    CBSEENPH11016291
    Wired Faculty App

    Do the notations Å and AU stand for same unit? 

    Solution
    No. Å and AU stand for different units.  Å is Angstrom and a.u is atomic unit.
    Question 85
    CBSEENPH11016293
    Wired Faculty App

    Comment on the statement: There is no way of defining a length standard in terms of a time standard.

    Solution
    The given statement is wrong.

    For instance the unit of length, light-year is defined in terms of unit of time. 
    Question 86
    CBSEENPH11016295
    Wired Faculty App

    How would you measure the height of an inaccessible tower?

    Solution
    The height of an inaccessible tower is triangulation method. 
    Triangulation is a way of determining something's location using the locations of other things. 
    Question 88
    CBSEENPH11016298
    Wired Faculty App

    How can we measure the diameter of the moon?

    Solution
    Angular method is used to measure the diameter of the moon. 

    Note: Angular diameter of the moon is the angle subtended by two diametrically opposite ends of the moon at a point on the earth. 
    Question 89
    CBSEENPH11016300
    Wired Faculty App

    What are the full forms of RADAR and SONAR?

    Solution

    RADAR stands for Radio Detection and Ranging.

    SONAR stands for Sound Navigation and Ranging.

    Question 90
    CBSEENPH11016301
    Wired Faculty App

    What is echo method of measuring the large distances?

    Solution
    Echo method: an echo is the phenomenon of repetition of sound on reflection from an obstacle. 
    In echo method, a wave (sound or electromagnetic) is sent towards the object and then reflected wave is again received back at the same position. Knowing the speed of wave and time after which reflected wave is received, the distance from which wave is reflected can be calculated. 
    Question 91
    CBSEENPH11016303
    Wired Faculty App

    Name the techniques that use echo method to measure the large distances?

    Solution
    RADAR and SONAR use echo method to measure laerge distances. 
    Question 92
    CBSEENPH11016305
    Wired Faculty App

    Can we use echo method to find very small distances of the order of nanometer?

    Solution
    Yes. Echo method can be used to measure small distances of the order of nanometer.
    The techniques used are interference and diffraction. 
    Question 93
    CBSEENPH11016306
    Wired Faculty App

    Which method is used to measure the distance of inferior planets from the sun?

    Solution
    Copernicus method can be used to measure the distance of inferior planets from the Sun.
    This method is basically used to find the distance between those planets which are near to sun rather than that of earth. The assumption used in this method is, orbits in which the planets are moving is circular. 
    Question 94
    CBSEENPH11016307
    Wired Faculty App

    Name the law that is used to measure the distance of superior planets from the sun?

    Solution
    Kepler's third law is used to measure the distance of the superior planets from the sun. 

    Kepler's third law states, the square of the orbital period of a planet is proportional to the cube of the semi major axis of its orbit.
    Question 95
    CBSEENPH11016308
    Wired Faculty App

    What is Avogadro’s hypothesis of the volume occupied by atom?

    Solution
    According to Avogadro's hypothesis, the actual volume occupied by atoms in a given mass of substance is just two third the total volume occupied by same mass of substance. 
    Question 96
    CBSEENPH11016309
    Wired Faculty App

    What are the proper units of length for measuring the terrestrial distances? 
    Solution
    The proper units for measuring terrestrial distances are light year, parsec and astronomical unit.
    Question 97
    CBSEENPH11016310
    Wired Faculty App

    What is the SI and CGS unit of time?

    Solution
    Unit of time in both SI and CGS system is seconds.
    Question 98
    CBSEENPH11016311
    Wired Faculty App

    Define the unit of time in terms of mean solar day.

    Solution

    One second is equal to 1/86400 th part of a mean solar day.

    The time between two successive passages of the mean sun across the meridian at noon. It is equal to24 hours 3 minutes and 56.555 seconds of mean sidereal time.


    Question 99
    CBSEENPH11016313
    Wired Faculty App

    What is the duration of one mean solar day?

    Solution
    One mean solar day equals 86400 second. 
    Question 100
    CBSEENPH11016315
    Wired Faculty App

    Define 1 second on atomic scale. 

    Solution
    One second on atomic scale is the duration of 9,192,631,770 periods of radiation corresponding to the transition between the two hyperfine levels of the ground state of cesium –133 atom. (1967)
    Question 101
    CBSEENPH11016319
    Wired Faculty App

    Define Solar day. 

    Solution
    Solar day is the time between successive meridian transits of the sun at a particular place. It is the time taken by the earth to complete one rotation about its axis with respect to the sun averaged over solar year.
    Question 102
    CBSEENPH11016321
    Wired Faculty App

    Express solar year in terms of solar day and siderial day

    Solution
    One solar year is equal to 366.25 siderial day.
    And, 1 siderial day is equal to 365.25 average solar day. 
    Question 103
    CBSEENPH11016322
    Wired Faculty App

    Define tropical year.

    Solution
    The time taken by the earth to make one revolution around the sun.
    Tropical year also known as solar year, is the time that the Sun takes to return to the same position in the cycle of seasons, as seen from Earth.
    Question 104
    CBSEENPH11016325
    Wired Faculty App

    What is the difference between tropical year and light year?

    Solution
    Tropical year is the unit of time while the light year is the unit of distance.
    Question 105
    CBSEENPH11016328
    Wired Faculty App

    Define one second in terms of tropical year.

    Solution
    One second in terms of a tropical year is expressed as, 
    1 sec = 3.1689 ×10-8 tropical year. 

    Question 106
    CBSEENPH11016330
    Wired Faculty App

    Define Lunar month.

    Solution
    It is the time taken by the moon to complete one revolution around the earth in its orbit. Lunar month is the average time between successive new or full 
    moons, equal to 29 days, 12 hours, 44 minutes.
    Question 107
    CBSEENPH11016331
    Wired Faculty App

    What does the Chronometer measure?

    Solution
    Chronometer measures Time.
    Question 108
    CBSEENPH11016332
    Wired Faculty App

    Define ampere.

    Solution
    Ampere is that constant current which, if maintained in two straight paralle conductors of infinite length, of negligible circular cross-section, and placed 1 metre apart in vacuum, would produce between these conductors a force equal to 2×10-7 newtom per metre of length.
    Question 109
    CBSEENPH11016333
    Wired Faculty App

    Define candela.

    Solution
    The candela is the luminous intensity, in a given direction, of a source that emits monochromatic radiation of frequency 540×1012 hertz and that has a radiant intensity in that direction of 1683 watt per steradian.
    Question 110
    CBSEENPH11016334
    Wired Faculty App

    Define kelvin.

    Solution
    The kelvin, is the fraction 1/273.16 of the thermodynamic temperature of the triple point of water.
    Question 111
    CBSEENPH11016335
    Wired Faculty App

    Define mole.

    Solution
    The mole is the amount of substance of a system, which contains as many elementray entities as there are atoms in 0.012 kilograms of carbon-12. 
    Question 112
    CBSEENPH11016336
    Wired Faculty App

    What are supplementary units?

    Solution
    Supplementary units are the dimensionless units that are used along with the base units to form derived units in the International system. The class of supplementary contains only two purely geometrical units, that is the radian and the steradian.
    The supplementary unit of plane angle is radian and that of solid angle is steradian. 
    Question 113
    CBSEENPH11016338
    Wired Faculty App

    Define radian.

    Solution
    Radian is an SI unit of plane angle. Radian is the angle subtended, at the center of a circle, by an arc whose length is equal to radius of the circle. One radian is equivalent to 57.296 degrees. 
    Question 114
    CBSEENPH11016339
    Wired Faculty App

    Define steradian.

    Solution
    Steradian is a unit of measurement for the solid angles.
    Steradian is the angle subtended, at the center of a sphere, by a surface whose magnitude of area is equal to square of the radius of the sphere.

    The solid angle of a sphere at it's centre is 4π steradians. 
    Question 115
    CBSEENPH11016340
    Wired Faculty App

    How does the unit radian depend on the unit of length used to measure length of arc and radius?

    Solution
    The unit radian is independent of the unit of length used to measure length of arc and radius.
    Question 116
    CBSEENPH11016341
    Wired Faculty App

    How many bars are there in one atmospheric pressure?

    Solution
    1 Atmospheric pressure = 1 bar.
    Question 117
    CBSEENPH11016342
    Wired Faculty App

    What is the density of water in SI and cgs systems?

    Solution
    Density of water in SI system is 1000 kg/m3 and in cgs system it is 1 gram/c.c. 
    Question 118
    CBSEENPH11016343
    Wired Faculty App

    What is solid angle subtended by a sphere at the center of the sphere? 

    Solution
    Sphere subtends an angle of 4π steradian at the center of the sphere. 
    Question 119
    CBSEENPH11016344
    Wired Faculty App

    What is the density of air at Normal Temperature and Pressure (N.T.P.)?

    Solution
    Density of air at NTP is 1.293 kg/m3. 

    Sponsor Area

    Question 120
    CBSEENPH11016345
    Wired Faculty App

    Give different units of energy.

    Solution
    Different units of energy are Joule, erg, eV, calories, kilo-watt hour. 
    Question 121
    CBSEENPH11016346
    Wired Faculty App

    Express eV in joules.

    Solution
    1 eV = 1.6 x 10–19 J
    Question 122
    CBSEENPH11016347
    Wired Faculty App

    In which unit nuclear cross-section is measured?

    Solution
    Nuclear cross-section is measured in barn and one barn is equal to 10–28m2.
    Question 123
    CBSEENPH11016352
    Wired Faculty App

    Is measurement and counting necessary? Explain giving one or two examples. 

    Solution

    Measurement or counting is necessary as the knowledge without counting is inadequate and unsatisfactory. Measurement will make the understanding of the object easier. 

    For instance, suppose you go to a tailor to get trousers stitched. If the tailor stitches them without taking your measurement it would be ill-fitting and look shabby. Similarly, if you order a door to be made by a carpenter, without measurement, the door may not fit in the opening allowed for it. Taking the measurement or counting is always advised for an easier understanding. 

    Question 124
    CBSEENPH11016354
    Wired Faculty App

    What is the difference between measurement and counting?

    Solution

     

    Measurement  Counting
    1. Measurement needs unit. 1. Counting does not need unit.
    2. Measurement is done for continuous data. 2. Counting is done for discrete data.
    3. Measurement can result in any real value outcome. 3. The outcome of counting is always a natural number.
    Question 125
    CBSEENPH11016357
    Wired Faculty App

    Define a physical quantity. Give examples of physical and non-physical quantities.

    Solution

    Physical quantity is quantity in terms of which law of physics can be expressed and which can be measured directly or indirectly. 

    Examples of physical quantities are Work, power, force, energy, solar constant, etc. 

    Examples of non-physical quantities: love, sympathy. 

    Question 126
    CBSEENPH11016359
    Wired Faculty App

    What information is required to make the measurement of a physical quantity?

    Solution

    The unit used for a certain quantity is required to completely measure a physical quantity. The number of times the unit is contained in the physical quantity is to be known. 
    Given a physical quantity X, the unit used is u and is required fo n times.
    Then, 
                                       X = nu


    Question 127
    CBSEENPH11016360
    Wired Faculty App

    Can a physical quantity have two or more units? Does the outcome of measurement depend on the system of unit used to make measurement? Is there any relation between the measures of a physical quantity using different units? If yes write the relation. 
    Solution
    Yes. A Physical quantity cannot have more than two units. The outcome of the measurement does not depend on the system of unit used to make the measurement. That is, the magnitude of a physical quantity remains same whatever the system of unit is used.

     Let two units u1 and u2 be used to measure a physical quantity X. If n1 and n2 are the numerical factors of measure, then

                             X = n1u1 = n2u2
    Question 128
    CBSEENPH11016364
    Wired Faculty App

    Does the numerical factor in the measurement change with change of unit?

    Solution
    The numerical factor in the measurement change with the change of unit. Smaller is the unit used to make the measurement, greater is the numerical factor and vice versa.

     
    Question 129
    CBSEENPH11016367
    Wired Faculty App

    Define a unit. What are its required characteristics?

    Solution
    The unit is a standard of measurement that contains definite amount of an invariable and accessible quantity. Other physical quantities can be compared w.r.t unit of a physical system.

    The required characteristics of unit are:

    (i) It should not change with time, place and physical conditions.

    (ii) It should be easily reproducible.

    (iii) It should neither be too small nor too large.

    (iv) It should be independent of all other units. 
    Question 130
    CBSEENPH11016378
    Wired Faculty App

    What are fundamental units? 

    Solution
    A fundamental unit a unit of measurement for a measurable physical property from which every other unit for that quantity can be derived. Fundamental units are assigned to fundamental quantities. 
    The fundamental unit for a particular measurable quantity is chosen by convention. 

    Units for mass, length, time, electric current are taken as fundamental units.
    Question 131
    CBSEENPH11016380
    Wired Faculty App

    What is SI system of units?

    Solution

    The SI system of units is the system consisting of seven fundamental quantities. The seven fundamental quantities and their respective units are given below in the table. 

    Physical quantity

    1. Mass

    2. Length

    3. Time

    4. Electric current

    5. Luminous intensity

    6. Temperature

    7. Amount of substance

    Fundamental Unit

    Kilogram

    Metre

    Second

    Ampere

    Candela

    Kelvin

    Mole
    Question 132
    CBSEENPH11016387
    Wired Faculty App

    What are the advantages of SI system of units?

    Solution

    Advantages of SI system of units are:

    (i)    SI is coherent system of units, i.e., a system based on certain set of fundamental units. 

    (ii)   SI is rational system of units. i.e., it assigns only one unit to a particular physical quantity. 

    (iii)  SI is metric system, i.e., multiples and sub-multiples of the system can be expressed as a power of 10. 

    (iv)  SI is internationally accepted system of units. 

    Question 133
    CBSEENPH11016396
    Wired Faculty App

    What is prefix? Name the different metric prefixes and their multiple values.

    Solution

    The prefix is multiple or sub-multiple that increases or decreases the size of unit. 

    Below is given a list of prefixes that is used. 

    prefix

    atto

    fermi

    pico

    nano

    micro

    milli

    centi

    deci

    value

    10-18

    10–15

    10–12

    10–9

    10–6

    10–3

    10–2

    10–1

    abb.

    a

    f

    P

    n

    H

    m

    c

    d

    prefix

    exa

    peta

    tera

    giga

    mega

    kilo

    hecta

    deca

    value

    10–18

    10–15

    10–12

    10–9

    10–6

    10–3

    10–2

    10–1

    abb.

    E

    P

    T

    G

    M

    K

    H

    D

     
    Question 134
    CBSEENPH11016399
    Wired Faculty App

    What is the significance of defining prefixes in units?

    Solution
    The physical quantities have wide range of variation as we come from microscopic to macroscopic scale. A single unit would not be sufficient to measure a physical quantity over a wide range. 
    Usually one unit is defined for a physical quantity and the size of the unit is increased or decreased by using prefix with that single unit. 
    For example, let us consider the thickness of a paper of the order 0.0001 m. MKS unit of length is metre. Thus, to measure the thickness of one page, we use the unit millimeter or micrometer. Milli and micro are the prefixes that decrease the unit meter by factor 103 and 106 respectively. 
    Question 135
    CBSEENPH11016401
    Wired Faculty App

    Why we exercise different methods to measure the same physical quantity of different magnitude? Explain by giving one example.

    Solution

    Different methods are employed to measure the same physical quantities of different magnitude. This is because, the magnitude varies over a wide range from one measurement to another. 
    Precision and accuracy of the measurement also determines the method employed. 

    For example, a physical balance can measure the weight of the commodity items, but the same method cannot be used to measure the mass of the earth or other heavenly bodies.
    If we have to buy 100gm of sugar, we use ordinary physical balance.
    The ordinary balances are not very accurate and may have an error of ±5% or even greater.
    But, if we have to buy gold or platinum jewelary and use the same physical balance as used by grocer, then error of ±5% makes a big difference to customer and jeweller. Here we need very accurate and precise measurement. 

    Question 136
    CBSEENPH11016404
    Wired Faculty App

    Comment upon the statement that mass of the earth is very large

    Solution
    The statement 'mass of the earth is very large' is right when we compare it with the mass of small objects. But, it becomes meaningless when it is compared with the mass of other objects like Sun, bigger planets etc. 

    For example, the mass of the earth is very large as compared to mass of man, but very small as compared to the mass of the sun or stars.

    Question 137
    CBSEENPH11016432
    Wired Faculty App

    What is the mass of electron, proton, moon, earth and Universe.

    Solution

    The mass of entities is given by,

    Entity

    Electron

    Proton

    Moon

    Earth

    Universe

    Mass

    9.1x1031 kg

    1.67x10–27 kg

    7.33x1022 kg

    5.98x1024 kg

    2.23x1038 kg
     
    Question 138
    CBSEENPH11016434
    Wired Faculty App

    Define the unit of length m in terms of size of the earth, meter bar and wavelength of light.

    What are the advantages of defining the unit of length in terms of wavelength of light? 

    Solution

    Unit of length in terms of size of earth: One meter is one ten millionth of the distance from equator to the north pole along the meridian line through Paris.

    Unit of length in terms of meter bar: One meter is the distance between two fine lines engraved on gold plugs near the platinum-iridium rod kept at temperature of 273.16K, preserved at International Bureau of weights and measures near Paris.

    Unit of length in terms of wavelength: It is the distance occupied by 1650763.73 wave (in vacuum) of wave emitted due to electronic transition from 5d5 to 2p10 by individual atom of Kr-86 isotope. 

    The advantage of defining the unit of length in terms of wavelength of light is that this standard does not change with time, physical conditions and is easily reproducible.

    Question 139
    CBSEENPH11016437
    Wired Faculty App

    A photosensitive metal is exposed by electromagnetic radiation of frequency greater than threshold frequency. It is observed that after 5 seconds mass of metal decrease by 0.091 pg. How many photoelectrons are emitted from the surface?

    Solution
    The mass of electron is 9.1 x 10–31 kg.

    Decrease in the mass of metal = 0.091 pg = 0.091×10-12 g

    Thus, emitting one electron shall decrease the mass of metal by a mass equal to mass of electron.

    Therefore, the number of electrons emitted is,


    n = Δmm=0.091x10-129.1x10-28=1014 
    Question 140
    CBSEENPH11016440
    Wired Faculty App

    Can we measure the mass of the sun using simple physical balance? If not, discuss one method to measure its mass. Estimate it.

    Solution
    We cannot use simple physical balance to measure the mass of the sun. 
    The indirect method to measure the mass of the sun is Kepler's third law of planetary motion. 

    The Sun’s gravity is the force that acts on the Earth to keep it moving in a circle,
    straight F subscript straight g space equals space fraction numerator GM subscript sun straight m subscript Earth over denominator straight r squared end fraction space space space space space space space space space space space space space... space left parenthesis straight i right parenthesis straight r space is space the space radius space of space the space earth apostrophe straight s space orbit space in space kms. Acceleration space of space the space Earth space in space orbit space is space given space by comma straight a space equals space straight v squared over straight r where comma space straight v space equals space Earth apostrophe straight s space orbital space speed. straight F space equals space straight m subscript Earth space end subscript straight a space equals space straight m subscript Earth space end subscript straight v squared over straight r space space space space... space left parenthesis ii right parenthesis On space comparing space and space solving space equations space left parenthesis straight i right parenthesis space and space left parenthesis ii right parenthesis comma space we space get straight M subscript sun space equals space fraction numerator straight v squared straight r over denominator straight G end fraction Earth apostrophe straight s space orbital space speed space left parenthesis straight v right parenthesis thin space can space be space expressed space as the space circumference space of space Earth apostrophe straight s space orbit space divided space by space it apostrophe straight s space orbital space period. So comma space straight v space equals space fraction numerator 2 πr over denominator straight P end fraction Hence comma space mass space of space the space sun space is space given space by comma straight M subscript sun space equals space fraction numerator 4 straight pi squared straight r cubed over denominator GP squared end fraction

     
    Question 141
    CBSEENPH11016441
    Wired Faculty App

    Discuss the triangulation method to measure the height of accessible object.

    Solution
    Triangulation method uses the geometry of the triangle and is used for measuring large distances such as height of a mountins, distances of moon, planet and starts from the earth etc. 

    The height of a distant but an accessible object like BC can be determined by measuring the angle θ subtended at A. 

    That is , 

    Angle of elevation of top at A = θ 

    The distance from the foot of the object is x. 



    Now, from the figure, 

                        tan θ = hx 

         ⇒ h = x tan θ, is the required height of the accessible object. 
    Question 142
    CBSEENPH11016449
    Wired Faculty App

    Discuss the triangulation method to measure the height of inaccessible object.

    Solution
    An object whose distance cannot be measured from the observation point, is called inaccessible object. 
    To measure the height of such an object BC, measure the two angles of elevation of the top of object C from two different points  A1 and A2. 
    . 
    Let space space straight theta subscript 1 space and space straight theta subscript 2 space end subscript be the angles of elevation of top
    C at A1 and A2 respectively.
    From figure, 
    cot space straight theta subscript 1 space equals space straight x subscript 1 over straight h space and space cotθ subscript 2 space equals space straight x subscript 2 over straight h
    rightwards double arrow      <pre>uncaught exception: <b>mkdir(): Permission denied (errno: 2) in /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/util/sys/Store.class.php at line #56mkdir(): Permission denied</b><br /><br />in file: /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/util/sys/Store.class.php line 56<br />#0 [internal function]: _hx_error_handler(2, 'mkdir(): Permis...', '/home/config_ad...', 56, Array) #1 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/util/sys/Store.class.php(56): mkdir('/home/config_ad...', 493) #2 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/plugin/impl/FolderTreeStorageAndCache.class.php(110): com_wiris_util_sys_Store->mkdirs() #3 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/plugin/impl/RenderImpl.class.php(231): com_wiris_plugin_impl_FolderTreeStorageAndCache->codeDigest('mml=<math xmlns...') #4 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/plugin/impl/TextServiceImpl.class.php(59): com_wiris_plugin_impl_RenderImpl->computeDigest(NULL, Array) #5 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/service.php(19): com_wiris_plugin_impl_TextServiceImpl->service('mathml2accessib...', Array) #6 {main}</pre>
    rightwards double arrow     straight x subscript 2 minus straight x subscript 1 equals straight h left parenthesis cot space straight theta subscript 2 minus space cotθ subscript 1 right parenthesis
    rightwards double arrow     straight h space equals space fraction numerator straight x subscript 2 minus straight x subscript 1 over denominator cotθ subscript 2 minus cotθ subscript 1 end fraction , is the height of the inaccessible object. 
    Question 143
    CBSEENPH11016450
    Wired Faculty App

    What is parallax?

    Solution
    Parallax is the difference or displacement in the apparent position of the object viewed along two different lines of sight. Parallax is measured by the angle or semi-angle of inclination between these two lines.

    Question 144
    CBSEENPH11016455
    Wired Faculty App

    What is echo? Discuss the echo method to measure the large distances. 

    Solution
    The reception of wave after reflection is called 'echo'.
    When echo method is used to measure large distances, we send the wave (sound or light) towards the object and the reflected wave is again received.
    Let 't' be the time after which echo is observed.
    Obviously in 't' time the wave travels the distance twice the distance between source of wave and object.
    If c is the velocity of wave then, 
                   ct = 2x
    rightwards double arrow           x = ct/2
    Question 145
    CBSEENPH11016457
    Wired Faculty App

    State Kepler's third law.What are its uses in measurement?

    Solution

    Kepler's third law states the time period, T of revolution of planet around the sun is directly proportional to cube of the semi major axis a of the orbit of planet.
    i.e.               straight T squared space straight alpha space straight a cubed
                     space straight T squared space equals fraction numerator 4 straight pi squared over denominator GM end fraction straight a cubed

    Kepler's law is used to measure the mass of the sun and to compare the radius of orbits of different planets.

    Question 146
    CBSEENPH11016458
    Wired Faculty App

    Define dimensions of a physical quantity.
    Solution
    The dimensions of a physical quantity refer to the type of units that must be used in order to obtain the measure of that quantity. They are expressed in terms of the base quantities and derived quantities. 
    Question 147
    CBSEENPH11016459
    Wired Faculty App

    Name the two quantities that have no dimensions but variable.

    Solution
    Two quantities that have no dimensions but variable are:
    (i) Coefficient of friction
    (ii) Strain 
    Question 148
    CBSEENPH11016460
    Wired Faculty App

    Can constant quantity have dimensions? Give examples.

    Solution
    Yes. Gravitational constant and Planck's constant have dimensions. 
    Question 149
    CBSEENPH11016461
    Wired Faculty App

    Does the dimensional formula of a physical quantity represent the quantitative part of measurement?

    Solution
    No, the dimensional formula of a physical quantity represent the quantitative part of measurement. The magnitude is not considered. And, it is the quality of the physical quantity that enters. 
    Question 150
    CBSEENPH11016462
    Wired Faculty App

    What are dimensionless constants? Give example.

    Solution
    The constants which do not possess any dimension are known as dimensionless quantities.  
    Examples π, e.
    Question 151
    CBSEENPH11016463
    Wired Faculty App

    State the principle of homogeneity of dimensional equation.

    Solution
    This principle of homogeneity of dimensional equation states that if a physical relation is true, then dimensions of each term on both the sides of the equation are same. 
    Question 152
    CBSEENPH11016464
    Wired Faculty App

    Can a quantity have dimensions but no units?

    Solution
    No, all the quantities which have dimensions possess units. 
    Question 153
    CBSEENPH11016465
    Wired Faculty App

    Can a quantity have units but still be dimensionless? Give examples.

    Solution
    Yes, there are certain quantities which have units but they are dimensionless. 
    For example, angle, loudness, solid angle. 
    Question 154
    CBSEENPH11016466
    Wired Faculty App

    Does the ratio of same quantities depend upon system of unit used?

    Solution
    No, the ratio of same quantities does not depend upon the system of unit used. 
    Question 155
    CBSEENPH11016467
    Wired Faculty App

    Mass x Pressure over Density is represented by which physical quantity? 

    Solution

    Work or energy. 

    Question 156
    CBSEENPH11016468
    Wired Faculty App

    If one says that the area of disc of radius r and diameter d is πr2d, how would you refute?

    Solution
    The dimensional formula of area is [M°L2T°]
    Dimensional formula of πr2d is [M°L3T°]
    Hence the statement is False. 
    Question 157
    CBSEENPH11016470
    Wired Faculty App

    Are the dimensions of coefficient of viscosity and coefficient of friction same?

    Solution
    No, the dimensions are different.
     Dimensional formula of coefficient of viscosity is [M
    1L–1T-1]
    Coefficient of friction is dimensionless.
    Question 158
    CBSEENPH11016471
    Wired Faculty App

    What is the dimensional formula of mass per unit length?

    Solution
    [M1L–1T0]
    Question 159
    CBSEENPH11016472
    Wired Faculty App

    Name the physical quantities which have same dimensions as that of [M1L–1T–2 ].

    Solution
    Pressure, Young's modulus, Bulk modulus and stress have the dimensional formula [M1L–1T–2 ]. 
    Question 160
    CBSEENPH11016473
    Wired Faculty App

    What is the dimensional formula of rate of flow of liquid?

    Solution

    The volume of the liquid flowing through pipe per second is the rate of flow of liquid. 

    So, dimensional formula is [M0L3T–1].

    Question 161
    CBSEENPH11016474
    Wired Faculty App

    Name the pair that has same dimensional formula?

    Solution
    Angle and strain have dimensional formula [MoLoTo]. 
    Question 162
    CBSEENPH11016475
    Wired Faculty App

    What is the dimensional formula of surface tension?

    Solution
    Surface tension is defined as force per unit length.
    Therefore, the dimensional formula is [M–1L0T2].
    Question 163
    CBSEENPH11016476
    Wired Faculty App

    What is the dimensional formula of specific gravity?

    Solution
    Specific gravity is the ratio of density of substance to density of water.
    Dimensional formula of specific gravity is [M0L0T0]
    Question 164
    CBSEENPH11016477
    Wired Faculty App

    What is the dimensional formula of impulse?

    Solution
    Impulse is forcecross times time space.
    So, dimensional formula is [M1L1T–1].
    Question 165
    CBSEENPH11016478
    Wired Faculty App

    What is the dimensional formula of intensity?

    Solution
    The dimensonal formula os intensity is [M1L0T–3]
    Question 166
    CBSEENPH11016479
    Wired Faculty App

    What is the dimensional formula of angular acceleration?

    Solution
    Angular acceleration is the change in angular accelearation with respect to time. 
    The dimensional fornula [M0L0T–2].
    Question 167
    CBSEENPH11016480
    Wired Faculty App

    What is the dimensional formula of spring constant?

    Solution
    The dimensional formula of spring constant is [M1L0T–2] .
    Question 168
    CBSEENPH11016481
    Wired Faculty App

    What is the dimensional formula of torque per unit twist?

    Solution
    The dimensional formula os torque per unit twist is [M1L2T–2].
    Question 169
    CBSEENPH11016482
    Wired Faculty App

    What is the dimensional formula of gravitational constant?

    Solution
    The dimensional formula of gravitaional constant is [M-1L3T-2].
    Question 170
    CBSEENPH11016483
    Wired Faculty App

    What is the dimensional formula of velocity graident?

    Solution
    The dimensional formula of velocity gradient is [M0L0T–1].
    Question 171
    CBSEENPH11016484
    Wired Faculty App

    Do the physical quantities - angular velocity and frequency have same dimensions?

    Solution
    Yes, angular velocity and frequency have the same dimension which is [MoLoT-1]. 
    Question 172
    CBSEENPH11016485
    Wired Faculty App

    What is the dimensional formula of angular momentum?

    Solution
    The dimensional formula of angular momentum is [M1L2T–1].
    Question 173
    CBSEENPH11016486
    Wired Faculty App

    What are the units and dimensional formula of intensity.

    Solution

    Dimensional formulaof Intensity is  [M1L0T–3]
    SI unit of intensity is  W/m2.
    In CGS units it is erg/s.cm2. 

    Question 174
    CBSEENPH11016488
    Wired Faculty App

    If all the terms in an equation have same units, is it necessary that they have same dimensions?

    Solution
    Yes, terms having same units will have the same dimension. 
    Question 175
    CBSEENPH11016489
    Wired Faculty App

    Find the dimensions of linear momentum.

    Solution
    Linear momentum is given by, 
                            p = mv
    So, the dimensional formula is [M1L1T–1]
    Question 176
    CBSEENPH11016490
    Wired Faculty App

    What is the dimensional formula of surface energy?

    Solution
    The dimensional formula of surface energy is [M1L0T-2].
    Question 177
    CBSEENPH11016491
    Wired Faculty App

    Name the physical quantity that has same dimensional formula as that of Planck's constant.

    Solution
    Angular momentum has the same dimensional formula as that of Planck's constant. 
    Question 178
    CBSEENPH11016492
    Wired Faculty App

    What is the dimensional formula of specific heat?

    Solution
    Specific heat is fraction numerator Energy over denominator mass cross times temperature end fraction
    Therefore, the dimensional formula is [M0L2T–2K–1].
    Question 179
    CBSEENPH11016493
    Wired Faculty App

    What is the dimensional formula of latent heat?

    Solution
    Latent heat is given by energy per unit mass.
    The dimensional formula is [M0L2T–2K0].
    Question 180
    CBSEENPH11016494
    Wired Faculty App

    What is the dimensional formula of mechanical equivalent of heat?

    Solution
    Mechanical equivalent of heat is dimensionless.
    That is, it has the dimensional formula [M0L0T0].
    Question 181
    CBSEENPH11016495
    Wired Faculty App

    Name two physical quantities which are dimensionless.

    Solution
    Coefficient of friction and refractive index have no units. Hence, they are dimensionless. 
    Question 182
    CBSEENPH11016496
    Wired Faculty App

    Name a quantity that has no unit and no dimensions.

    Solution
    Coefficient of friction has no units and hence no dimensions. 
    Question 184
    CBSEENPH11016498
    Wired Faculty App

    What is the dimensional formula of universal gas constant?

    Solution
    The dimensional formula of universal gas constant is [M1 L2 T–2 K–1].
    Question 185
    CBSEENPH11016499
    Wired Faculty App

    Name the quantity that has unit and no dimensions.

    Solution
    Loudness and angle have units but has no dimension. 
    Question 186
    CBSEENPH11016501
    Wired Faculty App

    If force F, velocity v and momentum p are taken as fundamental quantities, then what is the dimensional formula of time?

    Solution

    Let time, T = K Favbpc 
    So, [MoLoT1] = [MLT -2]a [ LT -1]b [ M1L1T -1]c 
                        = Ma+c L a+b+c T -2a-b-c 
    That is, 
    a+c = 0
    a+b+c = 0
    -2a-b-c = 1 
    On solving the above three equations, we get
    a = -1 ; b= 0 and c= 1
    Therefore, the dimensional formula of time is [F-1v0p1]

    Question 187
    CBSEENPH11016506
    Wired Faculty App

    What do you understand by dimensional formula and dimensional equation?

    Solution

    Dimensional formula of a physical quantity is an expression which tells us:
     
    i) the fundamental units on which the quantity depends on, and 
    ii) the nature of dependence. 

    For example, dimensional formula of area is [M0L2T0].

    This expression tells us that area does not depend upon mass and time but unit of length is required twice. 

    Dimensional equation: Physical quantity whebn equated with dimensional formula becomes dimensional equation. 

    i.e. dimensional equation of area will be given as,
                              A = [M0L2T0]

    Question 188
    CBSEENPH11016507
    Wired Faculty App

    Name two sets of three quantities that have same dimensions.

    Solution

    Quantities having same dimensions are, 
    (i) Work, energy, torque.

    (ii) Angle, strain, coefficient of friction.

    Question 189
    CBSEENPH11016508
    Wired Faculty App

    Comment on the statement: Dimensional formula deals with the qualitative nature of physical quantity.

    Solution

    For a physical quantity, the measure value is given by,
                                          nu,
    where, n is the quantitative part and u is qualitative part.
    The dimensional formula deals with u part only.
    For example, if dimensional formula of physical quantity is [MaLbTc],
    the value of u in SI system is kgambsc

    and in cgs system it is gmacmbse. 

    Question 190
    CBSEENPH11016509
    Wired Faculty App

    If all the terms in an equation have same dimensions, is it necessary that they have same units? Explain your answer by giving one example.

    Solution

    For quantities having same dimensions, it is not necessary that they should have the same units. 
    For example, in the equation W = Ttheta space plus m v squared

    where, W is work,
    T is torque,
    m is mass, and
    v is velocity.

    Dimensional formula of each term in the equation is [M1L2T–2 ] but the unit of work is Joule and that of torque is Newton-meter. That is, the units are different. 

    Question 191
    CBSEENPH11016510
    Wired Faculty App

    What are the applications of dimensional analysis?

    Solution

    Applications of dimensioanl analysis are:
    (i) Used to check the correctness of the physical relation.

    (ii) It is used to convert one system of units into another.

    (iii) Used to derive the new physical relations.

    Question 192
    CBSEENPH11016511
    Wired Faculty App

    What are the limitations of dimensional analysis?

    Solution

    The limitations of dimensional analysis are:
    (i) We cannot derive the formulae involving trigonometric functions, exponential functions, log functions etc., which have no dimension. 

    (ii) It does not give us any information about the dimensional constants in the formula. 

    (iv) We cannot find the exact sign of plus or minus connecting two or more terms in relation. 

    (v) This method fails to derive the relation of mechanics involving more than three physical quantities.

    Question 193
    CBSEENPH11016513
    Wired Faculty App

    A dimensionally correct equation is also numerically correct. Comment on the statement and explain.

    Solution

    This statement is wrong. A dimensionally correct equation may or may not be numerically correct.
    For example, the equation for the angle subtended by an arc of length l of circle of radius r at the center is written as θ = Error converting from MathML to accessible text.

    Dimensional formula of θ = [M0L0T0]
    Also dimensional formula of (Error converting from MathML to accessible text.) = [M0L0T0]

    Therefore the equation is dimensionally correct.
    The angle subtended by an arc of length l, circle of radius r, at the center is given by t/r.
    Thus, we can say that formula θ = r/l is dimensionally correct but numerically wrong. 

    Question 194
    CBSEENPH11016520
    Wired Faculty App

    What are the dimensions of Planck’s constant and gravitational constant?

    Solution
    Energy of a photon is given by, E = hv
    where 'h' is the planck's constant. 
    space space therefore space space space space space space space space space space space space straight h space equals space straight E over straight v equals fraction numerator open square brackets straight M to the power of 1 straight L squared straight T to the power of negative 2 end exponent close square brackets over denominator straight M to the power of 0 straight L to the power of 0 straight T to the power of 1 end fraction equals open square brackets straight M to the power of 1 straight L squared straight T to the power of negative 1 end exponent close square brackets
    Gravitational constant, 
    We know, italic space italic space F italic equals G fraction numerator M m over denominator r to the power of italic 2 end fraction 
    space space space space space thereforespace space space space space space space space space space space space space straight G space equals space Fr squared over Mm equals fraction numerator open square brackets straight M to the power of 1 straight L to the power of 1 straight T to the power of negative 2 end exponent close square brackets open square brackets straight L squared close square brackets over denominator open square brackets straight M to the power of 1 close square brackets open square brackets straight M to the power of 1 close square brackets end fraction
                    space space space space equals space open square brackets straight M to the power of negative 1 end exponent straight L cubed straight T to the power of negative 2 end exponent close square brackets        
    Question 195
    CBSEENPH11016525
    Wired Faculty App

    What are the dimensions of Vander Waal’s constant? 

    Solution
     Vander Waal's gas equation is
    space space space space space space space open square brackets straight P plus straight a over straight v squared close square brackets open parentheses straight v minus straight b close parentheses equals nRT
    Now, according go to the principle of homogeneity of  dimensions, straight P space and space straight a over straight r squared have same dimensions.
    space space therefore    <pre>uncaught exception: <b>mkdir(): Permission denied (errno: 2) in /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/util/sys/Store.class.php at line #56mkdir(): Permission denied</b><br /><br />in file: /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/util/sys/Store.class.php line 56<br />#0 [internal function]: _hx_error_handler(2, 'mkdir(): Permis...', '/home/config_ad...', 56, Array) #1 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/util/sys/Store.class.php(56): mkdir('/home/config_ad...', 493) #2 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/plugin/impl/FolderTreeStorageAndCache.class.php(110): com_wiris_util_sys_Store->mkdirs() #3 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/plugin/impl/RenderImpl.class.php(231): com_wiris_plugin_impl_FolderTreeStorageAndCache->codeDigest('mml=<math xmlns...') #4 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/plugin/impl/TextServiceImpl.class.php(59): com_wiris_plugin_impl_RenderImpl->computeDigest(NULL, Array) #5 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/service.php(19): com_wiris_plugin_impl_TextServiceImpl->service('mathml2accessib...', Array) #6 {main}</pre>
       rightwards double arrow  a = PV2 
    This dimesnsional formula of a is 
    space space straight a space equals space open square brackets straight M to the power of 1 straight L to the power of negative 1 end exponent straight T to the power of negative 2 end exponent close square brackets open square brackets straight M to the power of 0 straight L to the power of 6 straight T to the power of 0 close square brackets equals open square brackets bold M to the power of bold 1 bold L to the power of bold 5 bold T to the power of bold minus bold 2 end exponent close square brackets
    Similarly b = v
    Therefore dimensional formula of b is, 
    straight b space equals space open square brackets bold M to the power of bold 0 bold L to the power of bold 3 bold T to the power of bold 0 close square brackets  
    Question 196
    CBSEENPH11016529
    Wired Faculty App

    What is dimensional formula of solar constant?

    Solution
    The amount of radiant energy received per second per unit area by the perfectly black body placed on the earth with its surface perpendicular to the direction of radiation from the sun, is called as soalr consatnt. 
    So, the dimensional formula of solar consatnt is, 
    space space therefore  <pre>uncaught exception: <b>mkdir(): Permission denied (errno: 2) in /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/util/sys/Store.class.php at line #56mkdir(): Permission denied</b><br /><br />in file: /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/util/sys/Store.class.php line 56<br />#0 [internal function]: _hx_error_handler(2, 'mkdir(): Permis...', '/home/config_ad...', 56, Array) #1 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/util/sys/Store.class.php(56): mkdir('/home/config_ad...', 493) #2 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/plugin/impl/FolderTreeStorageAndCache.class.php(110): com_wiris_util_sys_Store->mkdirs() #3 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/plugin/impl/RenderImpl.class.php(231): com_wiris_plugin_impl_FolderTreeStorageAndCache->codeDigest('mml=<math xmlns...') #4 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/plugin/impl/TextServiceImpl.class.php(59): com_wiris_plugin_impl_RenderImpl->computeDigest(NULL, Array) #5 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/service.php(19): com_wiris_plugin_impl_TextServiceImpl->service('mathml2accessib...', Array) #6 {main}</pre>
           space space space space space equals space open square brackets straight M to the power of 1 straight L to the power of 0 straight T to the power of negative 3 end exponent close square brackets
    Question 197
    CBSEENPH11016533
    Wired Faculty App

    What is the dimensional formula of quantity by <pre>uncaught exception: <b>mkdir(): Permission denied (errno: 2) in /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/util/sys/Store.class.php at line #56mkdir(): Permission denied</b><br /><br />in file: /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/util/sys/Store.class.php line 56<br />#0 [internal function]: _hx_error_handler(2, 'mkdir(): Permis...', '/home/config_ad...', 56, Array) #1 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/util/sys/Store.class.php(56): mkdir('/home/config_ad...', 493) #2 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/plugin/impl/FolderTreeStorageAndCache.class.php(110): com_wiris_util_sys_Store->mkdirs() #3 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/plugin/impl/RenderImpl.class.php(231): com_wiris_plugin_impl_FolderTreeStorageAndCache->codeDigest('mml=<math xmlns...') #4 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/plugin/impl/TextServiceImpl.class.php(59): com_wiris_plugin_impl_RenderImpl->computeDigest(NULL, Array) #5 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/service.php(19): com_wiris_plugin_impl_TextServiceImpl->service('mathml2accessib...', Array) #6 {main}</pre>is pressure, a is Vander Waal's constant, V is volume and F is force?

    Solution

    Dimensional formula of the quantity is given by, 
    space space space space straight X equals Pa over VF space space space space space space space equals fraction numerator open square brackets straight M to the power of 1 straight L to the power of negative 1 end exponent straight T to the power of negative 2 end exponent close square brackets space open square brackets straight M to the power of 1 straight L to the power of 5 straight T to the power of negative 2 end exponent close square brackets over denominator open square brackets straight M to the power of 0 straight L cubed straight T to the power of 0 close square brackets open square brackets straight M to the power of 1 straight L to the power of 1 straight T to the power of negative 2 end exponent close square brackets end fraction
       space space space equals open square brackets straight M to the power of 1 straight L to the power of 0 straight T to the power of negative 2 end exponent close square brackets

    Question 198
    CBSEENPH11016536
    Wired Faculty App

    If E, G, J and m represent the energy, gravitational constant, angular momentum and

    mass, then what is dimensional formula of quantity <pre>uncaught exception: <b>mkdir(): Permission denied (errno: 2) in /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/util/sys/Store.class.php at line #56mkdir(): Permission denied</b><br /><br />in file: /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/util/sys/Store.class.php line 56<br />#0 [internal function]: _hx_error_handler(2, 'mkdir(): Permis...', '/home/config_ad...', 56, Array) #1 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/util/sys/Store.class.php(56): mkdir('/home/config_ad...', 493) #2 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/plugin/impl/FolderTreeStorageAndCache.class.php(110): com_wiris_util_sys_Store->mkdirs() #3 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/plugin/impl/RenderImpl.class.php(231): com_wiris_plugin_impl_FolderTreeStorageAndCache->codeDigest('mml=<math xmlns...') #4 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/plugin/impl/TextServiceImpl.class.php(59): com_wiris_plugin_impl_RenderImpl->computeDigest(NULL, Array) #5 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/service.php(19): com_wiris_plugin_impl_TextServiceImpl->service('mathml2accessib...', Array) #6 {main}</pre>

    Solution

    The dimensional formula of energy,  <pre>uncaught exception: <b>mkdir(): Permission denied (errno: 2) in /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/util/sys/Store.class.php at line #56mkdir(): Permission denied</b><br /><br />in file: /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/util/sys/Store.class.php line 56<br />#0 [internal function]: _hx_error_handler(2, 'mkdir(): Permis...', '/home/config_ad...', 56, Array) #1 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/util/sys/Store.class.php(56): mkdir('/home/config_ad...', 493) #2 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/plugin/impl/FolderTreeStorageAndCache.class.php(110): com_wiris_util_sys_Store->mkdirs() #3 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/plugin/impl/RenderImpl.class.php(231): com_wiris_plugin_impl_FolderTreeStorageAndCache->codeDigest('mml=<math xmlns...') #4 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/plugin/impl/TextServiceImpl.class.php(59): com_wiris_plugin_impl_RenderImpl->computeDigest(NULL, Array) #5 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/service.php(19): com_wiris_plugin_impl_TextServiceImpl->service('mathml2accessib...', Array) #6 {main}</pre>
    The dimensional formula of gravitational constant,<pre>uncaught exception: <b>mkdir(): Permission denied (errno: 2) in /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/util/sys/Store.class.php at line #56mkdir(): Permission denied</b><br /><br />in file: /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/util/sys/Store.class.php line 56<br />#0 [internal function]: _hx_error_handler(2, 'mkdir(): Permis...', '/home/config_ad...', 56, Array) #1 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/util/sys/Store.class.php(56): mkdir('/home/config_ad...', 493) #2 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/plugin/impl/FolderTreeStorageAndCache.class.php(110): com_wiris_util_sys_Store->mkdirs() #3 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/plugin/impl/RenderImpl.class.php(231): com_wiris_plugin_impl_FolderTreeStorageAndCache->codeDigest('mml=<math xmlns...') #4 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/plugin/impl/TextServiceImpl.class.php(59): com_wiris_plugin_impl_RenderImpl->computeDigest(NULL, Array) #5 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/service.php(19): com_wiris_plugin_impl_TextServiceImpl->service('mathml2accessib...', Array) #6 {main}</pre>
    The dimensional formula of, angular momentum, <pre>uncaught exception: <b>mkdir(): Permission denied (errno: 2) in /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/util/sys/Store.class.php at line #56mkdir(): Permission denied</b><br /><br />in file: /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/util/sys/Store.class.php line 56<br />#0 [internal function]: _hx_error_handler(2, 'mkdir(): Permis...', '/home/config_ad...', 56, Array) #1 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/util/sys/Store.class.php(56): mkdir('/home/config_ad...', 493) #2 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/plugin/impl/FolderTreeStorageAndCache.class.php(110): com_wiris_util_sys_Store->mkdirs() #3 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/plugin/impl/RenderImpl.class.php(231): com_wiris_plugin_impl_FolderTreeStorageAndCache->codeDigest('mml=<math xmlns...') #4 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/plugin/impl/TextServiceImpl.class.php(59): com_wiris_plugin_impl_RenderImpl->computeDigest(NULL, Array) #5 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/service.php(19): com_wiris_plugin_impl_TextServiceImpl->service('mathml2accessib...', Array) #6 {main}</pre>
    The dimension formula of mass, m = space space space open square brackets straight M to the power of 1 straight L to the power of 0 straight T to the power of blank close square brackets
    space space therefore  The dimensional formula of space space space space space fraction numerator EJ squared over denominator straight m to the power of 5 straight G squared end fraction is, 
    fraction numerator open square brackets straight M to the power of 1 straight L squared straight T to the power of negative 2 end exponent close square brackets open square brackets straight M to the power of 1 straight L squared straight T to the power of negative 1 end exponent close square brackets squared over denominator straight M to the power of 5 open square brackets straight M to the power of negative 1 end exponent straight L cubed straight T to the power of negative 2 end exponent close square brackets squared end fraction equals open square brackets straight M to the power of 0 straight L to the power of 0 straight T to the power of 0 close square brackets
    Hence, this is a dimensionless quantity. 

    Question 199
    CBSEENPH11016539
    Wired Faculty App

    The SI unit of energy is kg m2/sec2, that of speed v is m/s and that of acceleration a is m/s2.

    Which of the formulae for kinetic energy given below can you rule out by dimensional arguments (m stands for mass of body):

    (a) KE = m2 v3 
    (b) KE = 1/2 mv2 
    (c) KE = ma
    (d) KE = 3 mv2   
    (e) KE = 1/2 mv2  + ma

    Solution

    The dimensional formula for KE is [M1L2T–2].
    According to the principle of homogeneity of dimensional analysis, the dimensions of each term on both the sides of correct formula or equation will be the same.
    The dimensions of quantity on R.H.S. of different equations are: 

    <pre>uncaught exception: <b>mkdir(): Permission denied (errno: 2) in /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/util/sys/Store.class.php at line #56mkdir(): Permission denied</b><br /><br />in file: /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/util/sys/Store.class.php line 56<br />#0 [internal function]: _hx_error_handler(2, 'mkdir(): Permis...', '/home/config_ad...', 56, Array) #1 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/util/sys/Store.class.php(56): mkdir('/home/config_ad...', 493) #2 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/plugin/impl/FolderTreeStorageAndCache.class.php(110): com_wiris_util_sys_Store->mkdirs() #3 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/plugin/impl/RenderImpl.class.php(231): com_wiris_plugin_impl_FolderTreeStorageAndCache->codeDigest('mml=<math xmlns...') #4 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/plugin/impl/TextServiceImpl.class.php(59): com_wiris_plugin_impl_RenderImpl->computeDigest(NULL, Array) #5 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/service.php(19): com_wiris_plugin_impl_TextServiceImpl->service('mathml2accessib...', Array) #6 {main}</pre> 
    Since dimensional formula of the R.H.S. of (a), (c) and both the terms of R.H.S. of (e) are not same as that of L.H.S. i.e. [M1L2T2].
    Therefore, the formulae (a), (c) and (e) are ruled out dimensionally.

    Question 200
    CBSEENPH11016541
    Wired Faculty App

    Given y = a sin (ωt– kx) where 'x' is position and 't' is time. Show that ω/k has dimensions of velocity.

    Solution
    Here, y = a sin (ωt- kx), where x is position and t is time. 
    Sine has dimensions of angle, which is dimensionless.
    therefore ωt is dimensionless. 
    And straight omega has the same dimensional formula as that of frequency. 
    So, straight omega space equals space open parentheses straight M to the power of 0 straight L to the power of 0 straight T to the power of negative 1 end exponent close parentheses
    Similarly, kx is dimensionless and dimensions of k are,               k italic space italic equals italic space open square brackets M to the power of italic 0 L to the power of italic minus italic 1 end exponent T to the power of italic 0 close square brackets
    Now dimensional formula of, 
    straight omega over straight k equals fraction numerator open square brackets straight M to the power of 0 straight L to the power of 0 straight T to the power of negative 1 end exponent close square brackets over denominator open square brackets straight M to the power of 0 straight L to the power of negative 1 end exponent straight T to the power of 0 close square brackets end fraction equals open square brackets straight M to the power of 0 straight L to the power of 1 straight T to the power of negative 1 end exponent close square brackets , which is same as the dimensions of velocity. 
    Question 201
    CBSEENPH11016544
    Wired Faculty App

    If space straight x space equals space straight a space plus space bt space plus fraction numerator straight c over denominator straight d space plus space straight t end fraction where x has dimensions of length and t has dimension of time,

    then what are the dimensions of d and d.b?

    Solution
    Since we can add only similar types of physical quantities, therefore d and t have same dimensional formula.
    space space therefore space space straight d space equals space open square brackets straight M to the power of 0 straight L to the power of 0 straight T to the power of 1 close square brackets 
    According to the principle of homogeneity, the dimensions of each term on both sides are same.
    Therefore x and bt have same dimensional formula. i.e.
               straight x equals space bt
    or,    <pre>uncaught exception: <b>mkdir(): Permission denied (errno: 2) in /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/util/sys/Store.class.php at line #56mkdir(): Permission denied</b><br /><br />in file: /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/util/sys/Store.class.php line 56<br />#0 [internal function]: _hx_error_handler(2, 'mkdir(): Permis...', '/home/config_ad...', 56, Array) #1 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/util/sys/Store.class.php(56): mkdir('/home/config_ad...', 493) #2 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/plugin/impl/FolderTreeStorageAndCache.class.php(110): com_wiris_util_sys_Store->mkdirs() #3 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/plugin/impl/RenderImpl.class.php(231): com_wiris_plugin_impl_FolderTreeStorageAndCache->codeDigest('mml=<math xmlns...') #4 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/plugin/impl/TextServiceImpl.class.php(59): com_wiris_plugin_impl_RenderImpl->computeDigest(NULL, Array) #5 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/service.php(19): com_wiris_plugin_impl_TextServiceImpl->service('mathml2accessib...', Array) #6 {main}</pre> 
    Now dimensional formula of d.b
    <pre>uncaught exception: <b>mkdir(): Permission denied (errno: 2) in /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/util/sys/Store.class.php at line #56mkdir(): Permission denied</b><br /><br />in file: /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/util/sys/Store.class.php line 56<br />#0 [internal function]: _hx_error_handler(2, 'mkdir(): Permis...', '/home/config_ad...', 56, Array) #1 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/util/sys/Store.class.php(56): mkdir('/home/config_ad...', 493) #2 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/plugin/impl/FolderTreeStorageAndCache.class.php(110): com_wiris_util_sys_Store->mkdirs() #3 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/plugin/impl/RenderImpl.class.php(231): com_wiris_plugin_impl_FolderTreeStorageAndCache->codeDigest('mml=<math xmlns...') #4 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/plugin/impl/TextServiceImpl.class.php(59): com_wiris_plugin_impl_RenderImpl->computeDigest(NULL, Array) #5 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/service.php(19): com_wiris_plugin_impl_TextServiceImpl->service('mathml2accessib...', Array) #6 {main}</pre> 
    Question 202
    CBSEENPH11016547
    Wired Faculty App

    The potential energy of the particle varies with distance ‘x’ as straight U equals space fraction numerator Ax squared over denominator straight B squared plus straight x squared end fraction Find the dimensional formula of AB.

    Solution
    Given, x is the distance. Therefore the dimensional formula for 'x' is [L2]
    Here, B2 is added in X2. 
    Therefore
    space space space space space space space space space space space space straight B squared equals open square brackets straight L close square brackets squared rightwards double arrow space space space space space space space space straight B space equals space open square brackets straight L close square brackets space space Also comma space space space straight A equals fraction numerator straight B squared plus straight x squared over denominator straight x squared end fraction straight U
    Thus the dimensional formula of AB is given by,
    equals space open square brackets straight M to the power of 1 straight L squared straight T to the power of negative 2 end exponent close square brackets open square brackets straight L close square brackets space equals space open square brackets straight M to the power of 1 straight L cubed straight T to the power of negative 2 end exponent close square brackets
     
    Question 203
    CBSEENPH11016548
    Wired Faculty App

    A famous relation in physics relates, moving mass m to the rest mass m0 of a particle in terms of its speed v and the speed of light c. A boy recalls the relation almost correctly but forgets where to put the constant c. He writes:

    <pre>uncaught exception: <b>mkdir(): Permission denied (errno: 2) in /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/util/sys/Store.class.php at line #56mkdir(): Permission denied</b><br /><br />in file: /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/util/sys/Store.class.php line 56<br />#0 [internal function]: _hx_error_handler(2, 'mkdir(): Permis...', '/home/config_ad...', 56, Array) #1 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/util/sys/Store.class.php(56): mkdir('/home/config_ad...', 493) #2 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/plugin/impl/FolderTreeStorageAndCache.class.php(110): com_wiris_util_sys_Store->mkdirs() #3 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/plugin/impl/RenderImpl.class.php(231): com_wiris_plugin_impl_FolderTreeStorageAndCache->codeDigest('mml=<math xmlns...') #4 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/plugin/impl/TextServiceImpl.class.php(59): com_wiris_plugin_impl_RenderImpl->computeDigest(NULL, Array) #5 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/service.php(19): com_wiris_plugin_impl_TextServiceImpl->service('mathml2accessib...', Array) #6 {main}</pre>
    Guess where to put the missing c.

    Solution

    As per the dimensional analysis, each term on both the sides should have the same dimensions.
    Here, m and mo have same dimensions.
    Therefore, the quantity within square root must be dimensionless.
    Since dimensional formula of v is [M0L1T–1], therefore v2 has dimensions of [M0L2T-2], which is not dimensionless.
    Since, v and c have same dimensions, therefore
    straight v over straight c and space straight c over straight v have no dimensions. 
    Thus the quantity in square root should be either
    <pre>uncaught exception: <b>mkdir(): Permission denied (errno: 2) in /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/util/sys/Store.class.php at line #56mkdir(): Permission denied</b><br /><br />in file: /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/util/sys/Store.class.php line 56<br />#0 [internal function]: _hx_error_handler(2, 'mkdir(): Permis...', '/home/config_ad...', 56, Array) #1 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/util/sys/Store.class.php(56): mkdir('/home/config_ad...', 493) #2 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/plugin/impl/FolderTreeStorageAndCache.class.php(110): com_wiris_util_sys_Store->mkdirs() #3 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/plugin/impl/RenderImpl.class.php(231): com_wiris_plugin_impl_FolderTreeStorageAndCache->codeDigest('mml=<math xmlns...') #4 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/plugin/impl/TextServiceImpl.class.php(59): com_wiris_plugin_impl_RenderImpl->computeDigest(NULL, Array) #5 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/service.php(19): com_wiris_plugin_impl_TextServiceImpl->service('mathml2accessib...', Array) #6 {main}</pre>
    Since, straight c squared over straight v squared greater than 1
    <pre>uncaught exception: <b>mkdir(): Permission denied (errno: 2) in /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/util/sys/Store.class.php at line #56mkdir(): Permission denied</b><br /><br />in file: /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/util/sys/Store.class.php line 56<br />#0 [internal function]: _hx_error_handler(2, 'mkdir(): Permis...', '/home/config_ad...', 56, Array) #1 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/util/sys/Store.class.php(56): mkdir('/home/config_ad...', 493) #2 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/plugin/impl/FolderTreeStorageAndCache.class.php(110): com_wiris_util_sys_Store->mkdirs() #3 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/plugin/impl/RenderImpl.class.php(231): com_wiris_plugin_impl_FolderTreeStorageAndCache->codeDigest('mml=<math xmlns...') #4 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/plugin/impl/TextServiceImpl.class.php(59): com_wiris_plugin_impl_RenderImpl->computeDigest(NULL, Array) #5 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/service.php(19): com_wiris_plugin_impl_TextServiceImpl->service('mathml2accessib...', Array) #6 {main}</pre>  is a complex number, therefore not possible.
    Hence the quantity within the square root should be <pre>uncaught exception: <b>mkdir(): Permission denied (errno: 2) in /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/util/sys/Store.class.php at line #56mkdir(): Permission denied</b><br /><br />in file: /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/util/sys/Store.class.php line 56<br />#0 [internal function]: _hx_error_handler(2, 'mkdir(): Permis...', '/home/config_ad...', 56, Array) #1 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/util/sys/Store.class.php(56): mkdir('/home/config_ad...', 493) #2 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/plugin/impl/FolderTreeStorageAndCache.class.php(110): com_wiris_util_sys_Store->mkdirs() #3 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/plugin/impl/RenderImpl.class.php(231): com_wiris_plugin_impl_FolderTreeStorageAndCache->codeDigest('mml=<math xmlns...') #4 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/plugin/impl/TextServiceImpl.class.php(59): com_wiris_plugin_impl_RenderImpl->computeDigest(NULL, Array) #5 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/service.php(19): com_wiris_plugin_impl_TextServiceImpl->service('mathml2accessib...', Array) #6 {main}</pre>, 
    therefore space space space space space Correct space relation space is space straight m space equals space space fraction numerator straight m subscript 0 over denominator square root of 1 minus begin display style straight v squared over straight c squared end style end root end fraction

    Question 204
    CBSEENPH11016552
    Wired Faculty App

    A book with many printing errors contains four different formulas for the displacement ‘y’ of a particle undergoing a certain periodic motion: 

    (a) y = a sin wt               (b) y = a sin vt

    left parenthesis straight c right parenthesis space space space space straight y equals space straight a over straight T sin open parentheses straight t over straight a close parentheses        left parenthesis straight d right parenthesis space space space space straight y space equals space fraction numerator straight a over denominator square root of 2 end fraction left parenthesis sin space wt space plus space cos space wt right parenthesis

    (a = maximum displacement of the particle, v = speed of the particle, T = time-period of motion).

    Rule out the wrong formulas on dimensional grounds.

    Solution

    (a) The formula is dimensionally correct.

    (b) The formula is wrong, because ‘vt’ has the dimensions of length, according to the gven equation. 

    (c) The formula is wrong because (a/T) has dimension [M0L1T–1] and (T/a) has dimensions [M1L1T-1].

    (d) The formula is dimensionally correct.

    Question 205
    CBSEENPH11016554
    Wired Faculty App

    If force is taken as fundamental quantity instead of mass, then what is the dimensional formula of mass in terms of length, time and force?

    Solution
    Dimensional formula of force is,
    straight F space equals space open square brackets straight M to the power of 1 straight L to the power of 1 straight T to the power of negative 2 end exponent close square brackets
    rightwards double arrow space straight M space equals space open square brackets straight M to the power of 1 straight L to the power of negative 1 end exponent straight T to the power of negative 2 end exponent close square brackets
    <pre>uncaught exception: <b>mkdir(): Permission denied (errno: 2) in /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/util/sys/Store.class.php at line #56mkdir(): Permission denied</b><br /><br />in file: /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/util/sys/Store.class.php line 56<br />#0 [internal function]: _hx_error_handler(2, 'mkdir(): Permis...', '/home/config_ad...', 56, Array) #1 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/util/sys/Store.class.php(56): mkdir('/home/config_ad...', 493) #2 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/plugin/impl/FolderTreeStorageAndCache.class.php(110): com_wiris_util_sys_Store->mkdirs() #3 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/plugin/impl/RenderImpl.class.php(231): com_wiris_plugin_impl_FolderTreeStorageAndCache->codeDigest('mml=<math xmlns...') #4 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/plugin/impl/TextServiceImpl.class.php(59): com_wiris_plugin_impl_RenderImpl->computeDigest(NULL, Array) #5 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/service.php(19): com_wiris_plugin_impl_TextServiceImpl->service('mathml2accessib...', Array) #6 {main}</pre>Dimensional formula of mass in terms of F, L and T is space space open square brackets straight F to the power of 1 straight L to the power of negative 1 end exponent straight T squared close square brackets
    Question 206
    CBSEENPH11016555
    Wired Faculty App

    A man walking briskly in rain with speed v must slant his umbrella forward making an angle straight theta with the vertical. A student derives the following relation between θ and v: tan θ = v and checks that the relation has a correct limit : as v →0, θ → 0 as expected. (We are assuming there is no strong wind and that the rain falls vertically for a stationary man.) Do you think this relation can be correct?  If not, guess the correct relation. 
    Solution
    The relation tan straight theta = v is wrong because the dimensions on both the sides of the equation are not equal.
    Dimension of right hand side is [MoL1T-1] 
    Dimension of Left hand side quantity is [MoLoTo] 
    Hence, the equation is wrong. 
    To make the relation dimensionally correct, R.H.S should be divided by the speed of the rainfall. 
    That is, 
    tan space theta space equals space fraction numerator v over denominator v apostrophe end fraction, is a dimensionally correct relation. 
    Question 207
    CBSEENPH11016567
    Wired Faculty App

    What is the relation between the value of gravitational constant in SI system and cgs system?

    Solution
    Dimensional formula of gravitational constant is [M–1L3T–2].
    If n1 and n2 be the numerical values of gravitational constant in SI and cgs systems, then
                               straight n subscript 1 SI equals straight n subscript 2 cgs
    rightwards double arrow space space SI over cgs equals straight n subscript 2 over straight n subscript 1 equals open parentheses fraction numerator straight M subscript 1 over denominator straight M 2 end fraction close parentheses to the power of negative 1 end exponent open parentheses straight L subscript 1 over straight L subscript 1 close parentheses cubed open parentheses straight T subscript 1 over straight T subscript 1 close parentheses to the power of negative 2 end exponent
                 equals open parentheses fraction numerator 1 kg over denominator 1 gm end fraction close parentheses to the power of negative 1 end exponent open parentheses fraction numerator 1 straight m over denominator 1 cm end fraction close parentheses cubed open parentheses fraction numerator 1 straight s over denominator 1 straight s end fraction close parentheses to the power of negative 2 end exponent
                 space space space space space space space space space space space space space space space equals open parentheses 1000 over 1 close parentheses to the power of negative 1 space end exponent open parentheses 100 over 1 close parentheses cubed space open parentheses 1 over 1 close parentheses to the power of negative 2 end exponent   
                  = 1000
    space space therefore   SI = 1000 cgs
                   
    Question 208
    CBSEENPH11016569
    Wired Faculty App

    Convert 1J into ergs.

    Solution
    SI unit of work is Joule and
    CGS unit of work is ergs. 
    Work is force acting on a unit distance. 
    Therefore,
    Dimensional formula of work =left square bracket straight M to the power of 1 straight L squared straight T to the power of negative 2 end exponent right square bracket 
    space space therefore  space space space straight n subscript 2 equals space straight n subscript 1 space space open parentheses straight M subscript 1 over straight M subscript 2 close parentheses to the power of 1 open parentheses straight L subscript 1 over straight L subscript 2 close parentheses squared open parentheses straight T subscript 1 over straight T subscript 2 close parentheses to the power of negative 2 end exponent
            space space space space space space equals space open parentheses fraction numerator 1 kg over denominator 1 gm end fraction close parentheses to the power of 1 open parentheses fraction numerator 1 straight c over denominator 1 cm end fraction close parentheses squared open parentheses fraction numerator 1 straight s over denominator 1 straight s end fraction close parentheses to the power of negative 2 end exponent
                 = 107    
    space space therefore   1J =  107  ergs 
    Question 209
    CBSEENPH11016571
    Wired Faculty App

    In new system of units, the mass is measured in grams, length is measured in kilometers and time is measured in minutes. What is the value of 1W in new system of units?

    Solution
    Watt(W) is unit of power in SI system and dimensional formula of power is given by,
                             space open square brackets straight M to the power of 1 straight L squared straight T to the power of negative 3 end exponent close square brackets
    In SI system, 
    M1 = 1 kg, L1 = 1 m, T1 = 1s
    In new system, 
    M2 = 1 kg, L1 = 1 m, T2 = 1 min
    We know, 
            straight n subscript 2 equals straight n subscript 1 space open parentheses straight M subscript 1 over straight M subscript 2 close parentheses to the power of straight a open parentheses straight L subscript 1 over straight L subscript 2 close parentheses to the power of b open parentheses straight T subscript 1 over straight T subscript 2 close parentheses to the power of straight c
    rightwards double arrow  <pre>uncaught exception: <b>mkdir(): Permission denied (errno: 2) in /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/util/sys/Store.class.php at line #56mkdir(): Permission denied</b><br /><br />in file: /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/util/sys/Store.class.php line 56<br />#0 [internal function]: _hx_error_handler(2, 'mkdir(): Permis...', '/home/config_ad...', 56, Array) #1 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/util/sys/Store.class.php(56): mkdir('/home/config_ad...', 493) #2 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/plugin/impl/FolderTreeStorageAndCache.class.php(110): com_wiris_util_sys_Store->mkdirs() #3 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/plugin/impl/RenderImpl.class.php(231): com_wiris_plugin_impl_FolderTreeStorageAndCache->codeDigest('mml=<math xmlns...') #4 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/plugin/impl/TextServiceImpl.class.php(59): com_wiris_plugin_impl_RenderImpl->computeDigest(NULL, Array) #5 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/service.php(19): com_wiris_plugin_impl_TextServiceImpl->service('mathml2accessib...', Array) #6 {main}</pre>
              <pre>uncaught exception: <b>mkdir(): Permission denied (errno: 2) in /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/util/sys/Store.class.php at line #56mkdir(): Permission denied</b><br /><br />in file: /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/util/sys/Store.class.php line 56<br />#0 [internal function]: _hx_error_handler(2, 'mkdir(): Permis...', '/home/config_ad...', 56, Array) #1 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/util/sys/Store.class.php(56): mkdir('/home/config_ad...', 493) #2 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/plugin/impl/FolderTreeStorageAndCache.class.php(110): com_wiris_util_sys_Store->mkdirs() #3 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/plugin/impl/RenderImpl.class.php(231): com_wiris_plugin_impl_FolderTreeStorageAndCache->codeDigest('mml=<math xmlns...') #4 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/plugin/impl/TextServiceImpl.class.php(59): com_wiris_plugin_impl_RenderImpl->computeDigest(NULL, Array) #5 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/service.php(19): com_wiris_plugin_impl_TextServiceImpl->service('mathml2accessib...', Array) #6 {main}</pre>
              = 216 
    <pre>uncaught exception: <b>mkdir(): Permission denied (errno: 2) in /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/util/sys/Store.class.php at line #56mkdir(): Permission denied</b><br /><br />in file: /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/util/sys/Store.class.php line 56<br />#0 [internal function]: _hx_error_handler(2, 'mkdir(): Permis...', '/home/config_ad...', 56, Array) #1 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/util/sys/Store.class.php(56): mkdir('/home/config_ad...', 493) #2 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/plugin/impl/FolderTreeStorageAndCache.class.php(110): com_wiris_util_sys_Store->mkdirs() #3 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/plugin/impl/RenderImpl.class.php(231): com_wiris_plugin_impl_FolderTreeStorageAndCache->codeDigest('mml=<math xmlns...') #4 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/plugin/impl/TextServiceImpl.class.php(59): com_wiris_plugin_impl_RenderImpl->computeDigest(NULL, Array) #5 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/service.php(19): com_wiris_plugin_impl_TextServiceImpl->service('mathml2accessib...', Array) #6 {main}</pre> 1W = 216 gm km2 / min2 
    Question 210
    CBSEENPH11016582
    Wired Faculty App

    A calorie is a unit of heat or energy and it equals about 4.2 J where 1J = 1 kg m2 sñ2. Suppose we employ a system of units in which the unit of mass equals αkg, the unit of length equals β m, the unit of time is γ s. Show that a calorie has a magnitude 4.2 straight alpha to the power of negative 1 end exponent straight beta to the power of negative 1 end exponent straight gamma squared in terms of the new units. 
    Solution
    We know that dimensional formula of heat energy is open square brackets straight M to the power of 1 straight L squared straight T to the power of negative 2 end exponent close square brackets
    It is given that,
    1 cal = 4.2J 
    In SI system,   
    M1 = 1 kg, L1 = 1 m,  T1 = 1 s 
    In new system of units, 
    straight M subscript 2 equals straight alpha space kg comma space straight L subscript 2 equals straight beta space straight m comma space straight T subscript 2 equals straight gamma space straight s
    We  know,  
    <pre>uncaught exception: <b>mkdir(): Permission denied (errno: 2) in /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/util/sys/Store.class.php at line #56mkdir(): Permission denied</b><br /><br />in file: /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/util/sys/Store.class.php line 56<br />#0 [internal function]: _hx_error_handler(2, 'mkdir(): Permis...', '/home/config_ad...', 56, Array) #1 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/util/sys/Store.class.php(56): mkdir('/home/config_ad...', 493) #2 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/plugin/impl/FolderTreeStorageAndCache.class.php(110): com_wiris_util_sys_Store->mkdirs() #3 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/plugin/impl/RenderImpl.class.php(231): com_wiris_plugin_impl_FolderTreeStorageAndCache->codeDigest('mml=<math xmlns...') #4 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/plugin/impl/TextServiceImpl.class.php(59): com_wiris_plugin_impl_RenderImpl->computeDigest(NULL, Array) #5 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/service.php(19): com_wiris_plugin_impl_TextServiceImpl->service('mathml2accessib...', Array) #6 {main}</pre> 
    space space space space straight n subscript 2 equals 4.2 open parentheses fraction numerator 1 space kg over denominator straight alpha space kg end fraction close parentheses to the power of 1 open parentheses fraction numerator 1 space straight m over denominator straight beta space straight m end fraction close parentheses squared open parentheses fraction numerator 1 space straight s over denominator straight gamma space straight s end fraction close parentheses to the power of negative 2 end exponent
          <pre>uncaught exception: <b>mkdir(): Permission denied (errno: 2) in /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/util/sys/Store.class.php at line #56mkdir(): Permission denied</b><br /><br />in file: /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/util/sys/Store.class.php line 56<br />#0 [internal function]: _hx_error_handler(2, 'mkdir(): Permis...', '/home/config_ad...', 56, Array) #1 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/util/sys/Store.class.php(56): mkdir('/home/config_ad...', 493) #2 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/plugin/impl/FolderTreeStorageAndCache.class.php(110): com_wiris_util_sys_Store->mkdirs() #3 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/plugin/impl/RenderImpl.class.php(231): com_wiris_plugin_impl_FolderTreeStorageAndCache->codeDigest('mml=<math xmlns...') #4 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/plugin/impl/TextServiceImpl.class.php(59): com_wiris_plugin_impl_RenderImpl->computeDigest(NULL, Array) #5 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/service.php(19): com_wiris_plugin_impl_TextServiceImpl->service('mathml2accessib...', Array) #6 {main}</pre>
    therefore space space space space space space space space 1 space cal space equals space equals 4.2 straight alpha to the power of negative 1 end exponent straight beta to the power of negative 2 end exponent straight gamma squared space new space unit 
    Question 211
    CBSEENPH11016592
    Wired Faculty App

    Find the value of force of 200 dyne in a system based on meter, kg and minute as the fundamental units.

    Solution

    We know,
                     n1U1 = n2U2
    Dimensional formula of force is <pre>uncaught exception: <b>mkdir(): Permission denied (errno: 2) in /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/util/sys/Store.class.php at line #56mkdir(): Permission denied</b><br /><br />in file: /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/util/sys/Store.class.php line 56<br />#0 [internal function]: _hx_error_handler(2, 'mkdir(): Permis...', '/home/config_ad...', 56, Array) #1 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/util/sys/Store.class.php(56): mkdir('/home/config_ad...', 493) #2 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/plugin/impl/FolderTreeStorageAndCache.class.php(110): com_wiris_util_sys_Store->mkdirs() #3 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/plugin/impl/RenderImpl.class.php(231): com_wiris_plugin_impl_FolderTreeStorageAndCache->codeDigest('mml=<math xmlns...') #4 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/plugin/impl/TextServiceImpl.class.php(59): com_wiris_plugin_impl_RenderImpl->computeDigest(NULL, Array) #5 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/service.php(19): com_wiris_plugin_impl_TextServiceImpl->service('mathml2accessib...', Array) #6 {main}</pre>   
    Here, n1 = 200
    space space therefore space space space straight n subscript 2 equals space 200 open parentheses fraction numerator 1 gm over denominator 1 kg end fraction close parentheses to the power of 1 open parentheses fraction numerator 1 cm over denominator 1 straight m end fraction close parentheses to the power of 1 open parentheses fraction numerator 1 straight s over denominator 1 min end fraction close parentheses to the power of negative 2 end exponent
             space equals space 200 open parentheses 1 over 1000 close parentheses to the power of 1 open parentheses 1 over 100 close parentheses to the power of 1 open parentheses 1 over 60 close parentheses to the power of negative 2 end exponent
               <pre>uncaught exception: <b>mkdir(): Permission denied (errno: 2) in /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/util/sys/Store.class.php at line #56mkdir(): Permission denied</b><br /><br />in file: /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/util/sys/Store.class.php line 56<br />#0 [internal function]: _hx_error_handler(2, 'mkdir(): Permis...', '/home/config_ad...', 56, Array) #1 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/util/sys/Store.class.php(56): mkdir('/home/config_ad...', 493) #2 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/plugin/impl/FolderTreeStorageAndCache.class.php(110): com_wiris_util_sys_Store->mkdirs() #3 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/plugin/impl/RenderImpl.class.php(231): com_wiris_plugin_impl_FolderTreeStorageAndCache->codeDigest('mml=<math xmlns...') #4 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/plugin/impl/TextServiceImpl.class.php(59): com_wiris_plugin_impl_RenderImpl->computeDigest(NULL, Array) #5 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/service.php(19): com_wiris_plugin_impl_TextServiceImpl->service('mathml2accessib...', Array) #6 {main}</pre>
    <pre>uncaught exception: <b>mkdir(): Permission denied (errno: 2) in /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/util/sys/Store.class.php at line #56mkdir(): Permission denied</b><br /><br />in file: /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/util/sys/Store.class.php line 56<br />#0 [internal function]: _hx_error_handler(2, 'mkdir(): Permis...', '/home/config_ad...', 56, Array) #1 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/util/sys/Store.class.php(56): mkdir('/home/config_ad...', 493) #2 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/plugin/impl/FolderTreeStorageAndCache.class.php(110): com_wiris_util_sys_Store->mkdirs() #3 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/plugin/impl/RenderImpl.class.php(231): com_wiris_plugin_impl_FolderTreeStorageAndCache->codeDigest('mml=<math xmlns...') #4 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/plugin/impl/TextServiceImpl.class.php(59): com_wiris_plugin_impl_RenderImpl->computeDigest(NULL, Array) #5 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/service.php(19): com_wiris_plugin_impl_TextServiceImpl->service('mathml2accessib...', Array) #6 {main}</pre> 

    Question 212
    CBSEENPH11016597
    Wired Faculty App

    The surface tension of a liquid is 40 dyne/cm. Express it in N/m.

    Solution

    We know,
    1 dyne = 10-5 N
    and
        1cm = 10-2m
    Therefore, 
    Surface  tension in terms of N/m is given by, 
     40 dyne over cm equals 40 fraction numerator 10 to the power of negative 5 end exponent straight N over denominator 10 to the power of negative 2 end exponent straight m end fraction 
                   = 40 x 10-3 N/m
                   = 0.040 N/m

    Question 213
    CBSEENPH11016599
    Wired Faculty App

    If unit of velocity is increased by factor four times, then by what factor the unit of momentum and kinetic energy will increase?

    Solution

    Momentum of a particle is given by, p = mv
    Kinetic energy is given by K.E = 1/2 mv2 
    i.e. straight p space straight alpha space straight v space space space space and space space space straight K space space straight alpha space straight v squared
    Therefore, if unit of velocity becomes four times, the unit of momentum will also increase by the factor four. 
    Since kinetic energy is directly proportional to the square of velocity, therefore, unit of kinetic energy increases by factor sixteen. 

    Question 214
    CBSEENPH11016601
    Wired Faculty App

    Convert 10N weight into gm-cm/min2.

    Solution

    We know,
         straight I space straight N space equals space 1 fraction numerator kg space straight m over denominator straight S squared end fraction 
    Therefore,   
    10 straight N equals 10 fraction numerator kg space space straight m over denominator straight s squared end fraction space space space space space space space space equals 10 cross times fraction numerator 1000 gm cross times 100 cm over denominator open parentheses 1 divided by 60 space min close parentheses squared end fraction
       space space space space equals 10 cross times 1000 cross times 100 cross times 3600 fraction numerator gm. cm over denominator min squared end fraction      

        space space space equals 3.6 cross times 10 to the power of 9 fraction numerator gm. cm over denominator min squared end fraction 
           = 3.6cross times109 gm-cm/min2

    Question 215
    CBSEENPH11016602
    Wired Faculty App

    If the units of mass, length and time are 1000 lb , 3 ft, 60 s respectively, then find the unit of work.

    Solution
    Work = Force cross times distance
    So, dimensional formula of work is, straight W space equals open square brackets straight M to the power of 1 straight L squared straight T to the power of negative 2 end exponent close square brackets
    Therefore unit of work is given by, 
    u =1000lb x (3ft) x (60s)-2
      equals fraction numerator 1000 cross times 9 over denominator open parentheses 60 close parentheses squared end fraction fraction numerator lb. ft squared over denominator straight s squared end fraction equals space 2.5 lb. ft squared divided by straight s squared 
    Question 216
    CBSEENPH11016604
    Wired Faculty App

    If velocity of light in air (3x108m/s), acceleration due to gravity (9.8m/s2) and density of mercury (13600kg/m3) be chosen as fundamental units, then find the unit of mass, length and time.

    Solution

    Given that,
    V e l o c i t y space o f space l i g h t space equals space open square brackets L to the power of 1 T to the power of negative 1 end exponent close square brackets equals 3 cross times 10 to the power of 8 space m divided by s space space space space space space space space space space space space space space space space space space space space space space space space space space space space... left parenthesis 1 right parenthesis A c c e l e r a t i o n space d u e space t o space g r a v i t y equals space open square brackets L to the power of 1 T to the power of negative 2 end exponent close square brackets space equals space 9.8 m divided by s squared space space space space space space space space space space space space space... left parenthesis 2 right parenthesis D e n s i t y space o f space m e r c u r y comma space open square brackets M to the power of 1 L to the power of negative 3 end exponent close square brackets equals 13600 space k g divided by m cubed space space space space space space space space space space space space space space space space space space space space space space space... left parenthesis 3 right parenthesis
    Now, dividing equation (1) by (2),
             {T} = 3.061 x 107s                  ....(4)
    From(1) and (4) 
                 L = 3 x 108 x T
                  
                    = 9.183 x 1015 m              ....(5)
    Therefore, from equation (3) and (5), we have
                 M = 13600 x L3
                    = 13600 x (9.183 X 1015)3
                M = 1.053 x 1052 kg

    Question 217
    CBSEENPH11016607
    Wired Faculty App

    Derive an expression for terminal velocity of a body of radius ‘r’ and weight W falling in a viscous media of coefficient of viscosity η. 

    Solution

    Let       straight v space straight alpha space straight r to the power of straight a  
              space space straight alpha space straight W to the power of straight b
             <pre>uncaught exception: <b>mkdir(): Permission denied (errno: 2) in /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/util/sys/Store.class.php at line #56mkdir(): Permission denied</b><br /><br />in file: /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/util/sys/Store.class.php line 56<br />#0 [internal function]: _hx_error_handler(2, 'mkdir(): Permis...', '/home/config_ad...', 56, Array) #1 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/util/sys/Store.class.php(56): mkdir('/home/config_ad...', 493) #2 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/plugin/impl/FolderTreeStorageAndCache.class.php(110): com_wiris_util_sys_Store->mkdirs() #3 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/plugin/impl/RenderImpl.class.php(231): com_wiris_plugin_impl_FolderTreeStorageAndCache->codeDigest('mml=<math xmlns...') #4 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/plugin/impl/TextServiceImpl.class.php(59): com_wiris_plugin_impl_RenderImpl->computeDigest(NULL, Array) #5 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/service.php(19): com_wiris_plugin_impl_TextServiceImpl->service('mathml2accessib...', Array) #6 {main}</pre>
    rightwards double arrow      straight nu space straight alpha space straight r to the power of straight a straight W to the power of straight b straight eta to the power of straight c
    rightwards double arrow       straight v equals straight k space straight r to the power of straight a straight w to the power of straight b straight eta to the power of straight c                             ...(1)
    Where k is constant of proportionality.
    Substituting the dimension formula of each quantity in equation (1),
    We get, 
    open square brackets straight M to the power of 0 straight L to the power of 1 straight T to the power of 1 close square brackets equals open square brackets straight L close square brackets to the power of straight a open square brackets straight M to the power of 1 straight L to the power of 1 straight T to the power of negative 2 end exponent close square brackets to the power of straight b open square brackets straight M to the power of 1 straight L to the power of negative 1 end exponent straight T to the power of negative 1 end exponent close square brackets to the power of straight c 
    rightwards double arrow   open square brackets straight M to the power of 0 straight L to the power of 1 straight T to the power of 1 close square brackets equals open square brackets straight M to the power of straight b plus straight c end exponent straight L to the power of straight a plus straight b minus straight c end exponent straight T to the power of negative 2 straight b minus straight c end exponent close square brackets
    Equating the powers of M, L and T on both the sides, 
                       b + c = 0
    rightwards double arrow          a + b - c = 1
    rightwards double arrow             -2b - c = -1
    On solving the above equations, we get
      a = -1,     b = 1,  c = -1
    Subsituting for a, b and c  in  (1), we get
    straight nu equals straight k space straight r to the power of negative 1 end exponent straight w space space space straight eta to the power of negative 1 end exponent straight nu space equals space straight k straight omega over straight omega  
    This is the required expression for coefficient of viscosity. 

    Question 218
    CBSEENPH11016609
    Wired Faculty App

    Derive the expression for viscous force acting on spherical body of radius r moving with velocity v through viscous liquid of coefficient of viscosity η.

    Solution

    Viscous force acting on spherical body of radius r moving with velocity v through viscous liquid of coefficient of viscosity straight eta. 
    Let        F italic space alpha italic space r to the power of a 
               
                  straight alpha space straight nu to the power of straight b
              
               <pre>uncaught exception: <b>mkdir(): Permission denied (errno: 2) in /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/util/sys/Store.class.php at line #56mkdir(): Permission denied</b><br /><br />in file: /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/util/sys/Store.class.php line 56<br />#0 [internal function]: _hx_error_handler(2, 'mkdir(): Permis...', '/home/config_ad...', 56, Array) #1 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/util/sys/Store.class.php(56): mkdir('/home/config_ad...', 493) #2 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/plugin/impl/FolderTreeStorageAndCache.class.php(110): com_wiris_util_sys_Store->mkdirs() #3 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/plugin/impl/RenderImpl.class.php(231): com_wiris_plugin_impl_FolderTreeStorageAndCache->codeDigest('mml=<math xmlns...') #4 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/plugin/impl/TextServiceImpl.class.php(59): com_wiris_plugin_impl_RenderImpl->computeDigest(NULL, Array) #5 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/service.php(19): com_wiris_plugin_impl_TextServiceImpl->service('mathml2accessib...', Array) #6 {main}</pre>
    rightwards double arrow       straight F space straight alpha space straight r to the power of straight a straight v to the power of straight b straight eta to the power of straight c
    rightwards double arrow   space space space space space straight F space equals space straight k space straight r to the power of straight a straight v to the power of straight b straight eta to the power of straight c space space space space space space space space space space space space space space space space space space space.... left parenthesis 1 right parenthesis

    where k is constant of proportionality.

    Substituting the dimension formula of each quantity in equation (1),
    open square brackets straight M to the power of 1 straight L to the power of 1 straight T to the power of negative 2 end exponent close square brackets equals open square brackets straight M to the power of 0 straight L to the power of 1 straight T to the power of 0 close square brackets to the power of straight a open square brackets straight M to the power of 0 straight L to the power of 1 straight T to the power of negative 1 end exponent close square brackets to the power of straight b open square brackets straight M to the power of 1 straight L to the power of negative 1 end exponent straight T to the power of negative 1 end exponent close square brackets to the power of straight c
    rightwards double arrow open square brackets straight M to the power of 1 straight L to the power of 1 straight T to the power of negative 2 end exponent close square brackets equals open square brackets straight M to the power of 0 straight L to the power of straight a plus straight b minus straight c end exponent straight T to the power of negative straight b minus straight c end exponent close square brackets
    Equating the powers of M, L and T on both the sides, we get
                           c = 1
                       a + b - c = 1
                        - b - b = - 2
    On solving  we get, 
                        a = 1,    b = 1,    c = 1
    Subsituting for a , b and c in (1), we get
                              straight F space equals space straight k space straight r space straight nu space straight eta, is the viscous force.


    Question 219
    CBSEENPH11016616
    Wired Faculty App

    If velocity of light C, gravitational constant G and Planck's constant are chosen as fundamental units, then what are the dimensions of mass?

    Solution
    Let dimensional formula of mass in terms of C,G and h be
    space space straight m equals open square brackets straight C to the power of straight a straight G to the power of straight b straight h to the power of straight c close square brackets space space space space space space space space space space space space space space space space... left parenthesis 1 right parenthesis 
    Substituting the dimensions on both sides of equation (1), 
    space space space space open square brackets straight M to the power of 1 straight L to the power of 0 straight T to the power of 0 close square brackets equals open square brackets straight M to the power of 0 straight L to the power of 1 straight T to the power of negative 1 end exponent close square brackets to the power of straight a open square brackets straight M to the power of negative 1 end exponent straight L cubed straight T to the power of negative 2 end exponent close square brackets to the power of straight b open square brackets straight M to the power of 1 straight L squared straight T to the power of negative 1 end exponent close square brackets to the power of straight c equals space space open square brackets straight M to the power of 1 straight L to the power of 0 straight T to the power of 0 close square brackets equals open square brackets straight M to the power of negative straight b plus straight c end exponent straight L to the power of straight a plus 3 straight b plus 2 straight c end exponent straight T to the power of negative straight a minus 2 straight b minus straight c end exponent close square brackets  
    Comparing the dimensions of M. L. and T on both the sides, we get
                  - b + c = 1                            ...(2) 
                a + 3b + 2c = 0                      ...(3) 
                -a - 2b - c = 0                          ...(4)
    Adding (3) and (4), we get  
             b + c = 0                                    ...(5)
    On solving (2) and (5)
       b = -1/2,   and
        c = 1/2
    Substituting for b and c in (3), we get
                   a = -1/2
    Substitute for a, b and c in (1), we get
        <pre>uncaught exception: <b>mkdir(): Permission denied (errno: 2) in /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/util/sys/Store.class.php at line #56mkdir(): Permission denied</b><br /><br />in file: /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/util/sys/Store.class.php line 56<br />#0 [internal function]: _hx_error_handler(2, 'mkdir(): Permis...', '/home/config_ad...', 56, Array) #1 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/util/sys/Store.class.php(56): mkdir('/home/config_ad...', 493) #2 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/plugin/impl/FolderTreeStorageAndCache.class.php(110): com_wiris_util_sys_Store->mkdirs() #3 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/plugin/impl/RenderImpl.class.php(231): com_wiris_plugin_impl_FolderTreeStorageAndCache->codeDigest('mml=<math xmlns...') #4 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/plugin/impl/TextServiceImpl.class.php(59): com_wiris_plugin_impl_RenderImpl->computeDigest(NULL, Array) #5 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/service.php(19): com_wiris_plugin_impl_TextServiceImpl->service('mathml2accessib...', Array) #6 {main}</pre>
    Question 220
    CBSEENPH11016618
    Wired Faculty App

    A gas bubble from an explosion under water, oscillates with period T proportional to padbEc, where p is pressure, d is density and E is energy of an explosion. Find the value of a, b and c.  

    Solution

    Given, time period is proportional to straight T space alpha space straight p to the power of straight a straight d to the power of straight b space straight E to the power of straight c
    Equating both sides of above equation dimensionally, we get
    open square brackets straight M to the power of 0 straight L to the power of 0 straight T to the power of 1 close square brackets space equals open square brackets straight M to the power of 1 straight L to the power of negative 1 end exponent straight T to the power of negative 2 end exponent close square brackets to the power of straight a open square brackets straight M to the power of 1 straight L to the power of negative 3 end exponent straight T to the power of 0 close square brackets to the power of straight b open square brackets straight M to the power of 1 straight L squared straight T squared close square brackets to the power of straight c space space open square brackets straight M to the power of 0 straight L to the power of 0 straight T to the power of 1 close square brackets space equals space open square brackets straight M to the power of straight a plus straight b plus straight c end exponent straight L to the power of negative straight a minus 3 straight b plus 2 straight c end exponent straight T to the power of negative 2 straight a minus 2 straight c end exponent close square brackets


    Now, omparing the dimensions of M, L and T on both sides, we get
                       a + b + c = 0                      ....(1)
                  - a - 3b + 2c = 0                      ....(2)
                          - 2a - 2c = 0                     ....(3)
    Solving (1), (2) and (3), we get
                  straight a equals negative 5 over 6 comma space straight b equals 1 half space and space straight c space equals 1 third
    Question 221
    CBSEENPH11016622
    Wired Faculty App

    Find the unit of power in a system of unit, in which unit of mass is 20kg, unit of force is 10N and unit of energy is 5J.
    Solution
    The dimensional formula of power in terms of mass, force and energy.
    Let the dimensional formula of power in terms of mass, force and energy is [MaFbEc].
    i.e. Power, P =  [MaFbEc]
    Equating both sides of above equation dimensionally,
    space space space space space space space space space space space space space space space space space space space space space space space space space space space space space open square brackets straight M to the power of 1 straight L squared straight T to the power of negative 3 end exponent close square brackets space space space space space space space space space space space space space space space space space space space space space space space space equals open square brackets straight M to the power of 1 straight L to the power of 0 straight T to the power of 0 close square brackets to the power of straight a open square brackets straight M to the power of 1 straight L to the power of 1 straight T to the power of negative 2 end exponent close square brackets to the power of straight b open square brackets straight M to the power of 1 straight L squared straight T to the power of negative 2 end exponent close square brackets to the power of straight c rightwards double arrow open square brackets straight M to the power of 1 straight L squared straight T to the power of negative 3 end exponent close square brackets equals open square brackets straight M to the power of straight a plus straight b plus straight c end exponent straight L to the power of 0 plus straight b plus 2 straight c end exponent straight T to the power of negative 2 straight b minus 2 straight c end exponent close square brackets
    Comparing the dimensions of M, L and T on both sides, we get
        a + b+ c = 1                    ...(1)
           b + 2c = 2                    ...(2)
       -2b -2c = -3                      ...(3)
    Solving (1), (2) and (3), we get
    <pre>uncaught exception: <b>mkdir(): Permission denied (errno: 2) in /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/util/sys/Store.class.php at line #56mkdir(): Permission denied</b><br /><br />in file: /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/util/sys/Store.class.php line 56<br />#0 [internal function]: _hx_error_handler(2, 'mkdir(): Permis...', '/home/config_ad...', 56, Array) #1 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/util/sys/Store.class.php(56): mkdir('/home/config_ad...', 493) #2 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/plugin/impl/FolderTreeStorageAndCache.class.php(110): com_wiris_util_sys_Store->mkdirs() #3 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/plugin/impl/RenderImpl.class.php(231): com_wiris_plugin_impl_FolderTreeStorageAndCache->codeDigest('mml=<math xmlns...') #4 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/plugin/impl/TextServiceImpl.class.php(59): com_wiris_plugin_impl_RenderImpl->computeDigest(NULL, Array) #5 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/service.php(19): com_wiris_plugin_impl_TextServiceImpl->service('mathml2accessib...', Array) #6 {main}</pre>
    <pre>uncaught exception: <b>mkdir(): Permission denied (errno: 2) in /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/util/sys/Store.class.php at line #56mkdir(): Permission denied</b><br /><br />in file: /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/util/sys/Store.class.php line 56<br />#0 [internal function]: _hx_error_handler(2, 'mkdir(): Permis...', '/home/config_ad...', 56, Array) #1 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/util/sys/Store.class.php(56): mkdir('/home/config_ad...', 493) #2 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/plugin/impl/FolderTreeStorageAndCache.class.php(110): com_wiris_util_sys_Store->mkdirs() #3 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/plugin/impl/RenderImpl.class.php(231): com_wiris_plugin_impl_FolderTreeStorageAndCache->codeDigest('mml=<math xmlns...') #4 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/plugin/impl/TextServiceImpl.class.php(59): com_wiris_plugin_impl_RenderImpl->computeDigest(NULL, Array) #5 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/service.php(19): com_wiris_plugin_impl_TextServiceImpl->service('mathml2accessib...', Array) #6 {main}</pre>
    Substituting the value of units of m, F and E in equation (4), we get
    <pre>uncaught exception: <b>mkdir(): Permission denied (errno: 2) in /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/util/sys/Store.class.php at line #56mkdir(): Permission denied</b><br /><br />in file: /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/util/sys/Store.class.php line 56<br />#0 [internal function]: _hx_error_handler(2, 'mkdir(): Permis...', '/home/config_ad...', 56, Array) #1 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/util/sys/Store.class.php(56): mkdir('/home/config_ad...', 493) #2 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/plugin/impl/FolderTreeStorageAndCache.class.php(110): com_wiris_util_sys_Store->mkdirs() #3 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/plugin/impl/RenderImpl.class.php(231): com_wiris_plugin_impl_FolderTreeStorageAndCache->codeDigest('mml=<math xmlns...') #4 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/plugin/impl/TextServiceImpl.class.php(59): com_wiris_plugin_impl_RenderImpl->computeDigest(NULL, Array) #5 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/service.php(19): com_wiris_plugin_impl_TextServiceImpl->service('mathml2accessib...', Array) #6 {main}</pre>
    In new system, the unit of power is 5W.
    Question 222
    CBSEENPH11016624
    Wired Faculty App

    Name the digit that can be both significant and non-significant.

    Solution
    Zero.
    Question 223
    CBSEENPH11016625
    Wired Faculty App

    Is there any digit which is always nonsignificant?

    Solution
    No. There is no digit which is always non-significant. 
    Question 224
    CBSEENPH11016626
    Wired Faculty App

    Is there any digit, which is always significant?

    Solution
    Yes, all the non-zero digits are always significant.
    Question 225
    CBSEENPH11016627
    Wired Faculty App

    Under what conditions the non-zero digit is not significant?

    Solution
    Non-zero digits are always significant. 
    Question 226
    CBSEENPH11016628
    Wired Faculty App

    What information do we get from the significant figure of measurement? 

    Solution
    Significant figures gives us information about the accuracy of measurement. More are the significant figures in the measurement greater is the accuracy of measurement. 
    Question 227
    CBSEENPH11016629
    Wired Faculty App

    Find the significant zero in the number 0.0301

    Solution
    The zero lying between 3 and 1 is significant.
    Question 228
    CBSEENPH11016630
    Wired Faculty App

    How many significant digits are there in 9x109.

    Solution

    9 is the only significant digit present in the required number. 

    Question 229
    CBSEENPH11016631
    Wired Faculty App

    What are the significant digits in the measurement of 0 008070 kg?

    Solution
    There are 4 significant digits in the number. 
    Question 230
    CBSEENPH11016632
    Wired Faculty App

    Find the number of significant zeros in the number 16300.

    Solution

    There are no significant zero's in the given number. 

    Question 231
    CBSEENPH11016633
    Wired Faculty App

    What is the accuracy of time on cesium clock?

    Solution
    The time clock is 1012. 
     Accuracy of time on cesium clock is one. 
    Question 232
    CBSEENPH11016634
    Wired Faculty App

    Express 204 in five significant digits.

    Solution
    20400x10–2. 
    Question 233
    CBSEENPH11016635
    Wired Faculty App

    What is the order of time taken by light to cross the nucleus?

    Solution
    10–22 sec. 
    Question 234
    CBSEENPH11016636
    Wired Faculty App

    The radius of the circle is r = 1.72 cm. In how many significant figures will you express the area of circle? (Area of circle is ?r2.)

    Solution
    Given, radius of the circle, r = 1.72 cm
    The number of significant figures is 3 here. . 
    Question 235
    CBSEENPH11016637
    Wired Faculty App

    Find the number of significant digits in 100.001?

    Solution
    There are 6 significant digits.
    Question 236
    CBSEENPH11016638
    Wired Faculty App

    Two pieces of gold have masses 2.17g and 2.157g. What is the difference in the masses of two pieces?

    Solution
    Mass of 1 gold = 2.17 g
    Mass of second gold = 2.157 g
    Difference in their masses = 0.01g 
    Question 237
    CBSEENPH11016639
    Wired Faculty App

    Round of the number 6.7890 into three significant digits.

    Solution
    The number in three significant digits is 6.79. 
    Question 238
    CBSEENPH11016640
    Wired Faculty App

    Round of the number 702876.45 into three significant digits.

    Solution
    7.03x105.
    Question 239
    CBSEENPH11016641
    Wired Faculty App

    Does a change in the units of a measurement has any effect upon its significant figure?

    Solution
    No, change in the units of a measurement has no effect upon it's significant figure. 
    Question 240
    CBSEENPH11016642
    Wired Faculty App

    What do you mean by order of magnitude?

    Solution
    The approximation to the nearest power of ten of a number is called the order of magnitude. 
    Question 241
    CBSEENPH11016643
    Wired Faculty App

    What is the order of magnitude of radius of the earth in meters?

    Solution

    The order of magnitude of the radius of the earth in metres is 7. 

    Question 242
    CBSEENPH11016644
    Wired Faculty App

    What is the order of the magnitude of human being in meters?

    Solution
    The order of the magnitude of human beings in meters is zero.
    Question 243
    CBSEENPH11016646
    Wired Faculty App

    Define least count.

    Solution
    Least count is the minimum measurement that can be made with the instrument. 
    Question 244
    CBSEENPH11016647
    Wired Faculty App

    How does the accuracy of measurement depend on the least count of instrument?

    Solution
    With decrease in least count of the instrument, the accuracy of measurement increases. 
    Question 245
    CBSEENPH11016648
    Wired Faculty App

    Can we use one instrument to measure the physical quantity of all the magnitudes?

    Solution
    No. Different instruments are to be used to measure the physical quantity of different magnitudes. 
    Question 246
    CBSEENPH11016649
    Wired Faculty App

    What is the least count of simple scale?

    Solution
    0.001m or 0.1 cm is the least count of a simple scale. 
    Question 247
    CBSEENPH11016650
    Wired Faculty App

    What is hysteresis?

    Solution
    Hysteresis is the inability of the instrument not to show identical behaviour while taking the measurement in ascending and descending order.
    Question 248
    CBSEENPH11016651
    Wired Faculty App

    What is a screw gauge?

    Solution
    Screw gauge is an instrument used to measure very small length such as thickness of page, radius of very thin wire etc.
    Question 249
    CBSEENPH11016652
    Wired Faculty App

    On what principle the screw gauge works?

    Solution
    Screw gauge works on the principle of screw. 
    When we rotate the head ‘H’ by means of safety device ‘D’ through one complete rotation, the distance moved by the screw for every complete rotation is constant.
    Question 250
    CBSEENPH11016653
    Wired Faculty App

    What is the principle of screw?

    Solution
    When a screw (fitted in a nut having uniform threading) is rotated, the linear distance by which the screw advances is proportional to the number of rotations given to the screw. The constant linear distance moved moved by the screw over one complete rotation is called the Pitch of the screw. 
    Question 251
    CBSEENPH11016654
    Wired Faculty App

    What is the pitch of screw gauge?

    Solution
    Pitch of the screw is defined as the distance between two consecutive threads on a screw and is equal to the distance by which the screw advances ,when one complete rotation is given to it.
    Question 252
    CBSEENPH11016655
    Wired Faculty App

    How is pitch measured?

     
    Solution
    Pitch is the ratio of the distance moved on the linear scale after giving 'n' number of complete rotation to the screw to the number of rotation. 
    Question 253
    CBSEENPH11016656
    Wired Faculty App

    Why is screw gauge called a micrometer?

    Solution
    Screw gauge is also called a micrometre because it can measure the lengths of the order of 1 micro meter. 
    Question 254
    CBSEENPH11016657
    Wired Faculty App

    What is spherometer?

    Solution
    Spherometer is an instrument used to measure the radius of curvature of spherical surfaces.
    Question 255
    CBSEENPH11016658
    Wired Faculty App

    On what principle is spherometer based on?

    Solution
    Spherometer works on the principle of a screw. It precisely measures small lengths. 
    Question 256
    CBSEENPH11016659
    Wired Faculty App

    For what purposes do we generally use spherometer?

    Solution
    Spherometer is generally used to measure the small thickness of plates and radii of curvature of spherical surfaces. 
    Question 257
    CBSEENPH11016660
    Wired Faculty App

    Define absolute error.

    Solution
    Absolute error is defined as the difference between actual value and measured value.
    Question 258
    CBSEENPH11016661
    Wired Faculty App

    What is random error?

    Solution

    The random errors are those errors, which occur irregularly and hence are random with respect to sign and size. Random errors can arise due to random and unpredictable fluctuations in experimental conditions. 


    Question 259
    CBSEENPH11016662
    Wired Faculty App

     Is the random error of constant sign?

    Solution
    No, random error does not have a constant sign. 
    Question 260
    CBSEENPH11016663
    Wired Faculty App

    Is constant error of the type random error?

    Solution
    No, constant error and random error are two different types of error. 
    Question 261
    CBSEENPH11016664
    Wired Faculty App

    Give two examples of systematic error.

    Solution
    Error due to loss of heat by radiation and error due to buoyancy are the examples of systematic error.
    Question 262
    CBSEENPH11016665
    Wired Faculty App

    Can you minimize the random error by applying some suitable correction?

    Solution
    Random errors arise due to unpredictable fluctuations in experimental conditions. Hence, suitable measures cannot be taken to reduce the errors. 
    Question 263
    CBSEENPH11016666
    Wired Faculty App

    What is gross error?

    Solution
    Gross error is due to the carelessness of the observer. 
    Question 264
    CBSEENPH11016668
    Wired Faculty App

    What do you mean by relative error?

    Solution
    Relative error is the ratio of mean absolute error to the mean value of the quantity measure.
    Relative error measures the uncertainty in one part.
    Question 265
    CBSEENPH11016669
    Wired Faculty App

    The mean diameter of a thin brass rod is to be measured by vernier callipers.

    Why is a set of 100 measurements of the diameter expected to yield a more reliable estimate than a set of 5 measurements only?
    Solution
    A set of 100 measurements will yield a more reliable estimate because as the number of observations increases, the probable error decreases.
    Question 266
    CBSEENPH11016670
    Wired Faculty App

    What is the relation between arial magnification and linear magnification?

    Solution
    Arial magnification is the square of linear magnification.
    Question 267
    CBSEENPH11016671
    Wired Faculty App

    What is the dimensional formula of relative error?

    Solution
    The relative error is a dimensionless quantity because it is a ratio of the error. 
    Question 268
    CBSEENPH11016672
    Wired Faculty App

    If A = a ± ∆a and B = b ± ∆b, then what is the maximum possible absolute error in A+B ?

    Solution
    The maximum possible absolute error in A+B is ∆a + ∆b
    Question 269
    CBSEENPH11016673
    Wired Faculty App

    If A = a ± ∆a and B = b ± ∆b, then what is the maximum possible absolute error in A–B ?

    Solution
     The maximum possible absolute error is ∆a + ∆b. 
    Question 270
    CBSEENPH11016674
    Wired Faculty App

    If A = a ± α% and B = b ± β%, then what is the maximum possible percentage error in AB ?

    Solution
    The maximum possible percentage error in AB = (a+P)%
    Question 271
    CBSEENPH11016675
    Wired Faculty App

    If A = a ± α% and B = b± β%, then what is the maximum possible percentage error in A/B ?

    Solution
    Maximum possible percentage error in A/B is (α+β)%
    Question 272
    CBSEENPH11016676
    Wired Faculty App

    If A= 12.00 ± 0.23 and B = 5.00 ±0.12, then what is the possible absolute error in A+B?

    Solution
    The absolute error in A+B = ± 0.35. 
    Question 273
    CBSEENPH11016677
    Wired Faculty App

    If the percentage error in A is 1.2%, then what is the percentage error in A2?

    Solution
    Given,
    Percentage error in A = 1.2 %
    Therefore, percentage error in A2 =  2.4%.
    Question 274
    CBSEENPH11016678
    Wired Faculty App

    Does the percentage error in a physical quantity depend upon the system of unit used?

    Solution
    No, percentage error in a physical quantity depend upon the system of unit used.
    Question 275
    CBSEENPH11016679
    Wired Faculty App

     State various rules for counting significant digits.

    Solution

    Various rules for counting significant figures are: 
    (i) All non-zero digits are significant.
    e.g. In a number 112.36, there are five significant figures.

    (ii) All the zeros occurring between two nonzero digits are significant.
    e.g. In a number 101.006, there are six significant figures.

    (iii) The extreme right zeros in a number with a decimal point are significant.
    e.g. In 2.400 there are four significant digits.

    (iv) If the number is less than 1, then the zeros, right to decimal point and left to non-zero digit are significant.
    e.g. In 0.0098, there are two significant digits.

    (v) The extreme right zeros in whole number are not significant if it is merely a number.
    e.g. 40300 have three significant digits. The last two zeros are not significant.

    (vi) The extreme right zeros in whole number are significant if it comes from some measurement.
    e.g. 40300 cm has five significant digits. The last two zeros are significant.

    Question 276
    CBSEENPH11016680
    Wired Faculty App

    Under what conditions the zeros are not significant?

    Solution

    The conditions under which zeroes are non-significant are:
    (i) When the first non-zero digit occurs after decimal, then all the zeros between decimal and first non-zero digit are non-significant.

    (ii) The zeros occurring on the extreme right in a whole number are not significant.

     

    Question 277
    CBSEENPH11016681
    Wired Faculty App

    Under what conditions the zeros are significant?

    Solution

    The conditions under which zeroes are significant are: 
    (i) When the zero lies between two non-zero digits, then it is significant.

    (ii) The extreme right zeros after decimal point in a fractional number are significant.

    Question 278
    CBSEENPH11016682
    Wired Faculty App

    What is the difference between vernier constant and least count?

    Solution
    The least count is the minimum measurement that an instrument can make. Every measuring instrument has least count. The size of the smallest divison on a scale is called as the least count. 
    Vernier constant is the special name given for the least count of vernier calipers. It is the ratio of smallest division of main scale to the number of divisions of vernier scale.
    Question 279
    CBSEENPH11016683
    Wired Faculty App

    What is vernier calliper? What are the factors on which its accuracy depends?

    Solution
    Vernier calliper is an instrument used to measure the small lengths/diameters/depths. Vernier calliper consists of two scales; one scale is ordinary centimeter scale, called main scale. The second scale called vernier scale can be slided on main scale.
    The accuracy of measurement of vernier scale depends on the value of one division on vernier scale. The vernier scale is constructed so that it is spaced at a constant fraction of the fixed main scale. So for a decimal measuring device each mark on the vernier is spaced nine tenths of those on the main scale. 
    Question 280
    CBSEENPH11016684
    Wired Faculty App

    What do you mean by vernier constant? Find its value for the vernier calliper whose n divisions on vernier scale coincide with (n–1) divisions of main scale.

    Solution
    Vernier constant of a vernier calliper is equal to difference in the value of one main scale division and one vernier scale division. It is also equal to least count of the instrument.
    Here n vernier scale divisions coincides with (n–1) main scale divisions. So, 
             n VSD = (n - 1) MSD
             1 VSD = fraction numerator straight n minus 1 over denominator straight n end fraction MSD
    space space therefore Vernier constant (VC) = 1 MSD - 1 VSD
                                      space space space space space equals space 1 space MSD space minus space fraction numerator straight n minus 1 over denominator straight n end fraction MSD 
                                           = 1 over straight n MSD
    Question 281
    CBSEENPH11016685
    Wired Faculty App

    A screw gauge has a pitch of 1.0 mm and 200 divisions on the circular scale. Do you think it is possible to increase the accuracy of the screw gauge arbitrarily by increasing the number of divisions on the circular scale?

    Solution
    Pitch of the screw gauge = 1.0 mm
    No. of divisions on the circular scale = 200 
    On increasing the number of divisions on the circular scale the least count of the screw gauge will decrease. Hence, accuracy of the screw gauge increases. 
    Question 282
    CBSEENPH11016686
    Wired Faculty App

    Define constant error and imperfectness error.

    Solution

    An error that repeats itself in all measurements is called a constant error. 

    Imperfectness error is the error that cannot be removed even if its reasons are known. Imperfectness errors are caused by unprdictable fluctuations during experiment. 

    Question 283
    CBSEENPH11016687
    Wired Faculty App

    What is systematic error? What are different types of systematic errors?

    Solution
    The systematic errors are those errors that tend to be in one direction, either positive or negative.
     
    Systematic error can be categorised into: 
    i) Instrumental errors that arise from the errors due to imperfect design or calibration of the measuring instrument, zero error in the instrument, etc.
    ii) Imperfection in experimental technique or procedure - External conditions (such as changes in temperature, humidity, wind velocity, etc.) during the experiment may systematically affect the measurement. 
    iii) Personal errors that arise due to an individuals bias, lack of proper setting of the apparatus or individuals carelessness in taking observations without observing proper precautions, etc. 
    Question 284
    CBSEENPH11016689
    Wired Faculty App

     What is random error? How it can be minimized?

    Solution
    The random errors are those errors, which occur irregularly and hence are random with respect. to sign and size. These can arise due to random and unpredictable fluctuations in experimental conditions (e.g. unpredictable fluctuations in temperature, voltage supply, mechanical vibrations of experimental set-ups, etc), personal (unbiased) errors by the observer taking readings, etc.
    The error can be minimized y taking the average of many observed values. Despite taking certain mesaures, it is hard to reduce exactly random errors. 
    Question 285
    CBSEENPH11016690
    Wired Faculty App

    How does the error combine together in case of addition or subtraction?

    Solution
    When we add or subtract two or more quantities then maximum absolute error in result is the sum of the maximum absolute errors in each quantity. 
    Question 286
    CBSEENPH11016695
    Wired Faculty App

    Name different types of errors which come into play while using the measuring instrument to make measurement?
    Solution
    Different types of error that come into play while using the measuring instrument are:
    (a) Imperfectness error, 
    (b) constant error,
    (c) random error,
    (d) systematic error, and 
    (e) gross error.
    Question 287
    CBSEENPH11016716
    Wired Faculty App

    It is claimed that two cesium clocks, if allowed to run for 100 years, free from any disturbance, may differ by only about 0.02 s. What does this imply for the accuracy of the standard cesium clock in measuring a time-interval of 1 s ?
    Solution
    When the time interval of 1 sec is measured, the accuracy in measurement is given by, 
    fraction numerator increment straight T over denominator straight T end fraction space equals space fraction numerator 0.02 over denominator 100 end fraction space space space space space space space space space space space equals space fraction numerator 0.02 over denominator 100 cross times 365 cross times 24 cross times 60 cross times 60 end fraction space space space space space space space space space space space equals space fraction numerator 0.02 over denominator 3.15 cross times 10 to the power of 7 end fraction space space space space space space space space space space equals 6.35 cross times 10 to the power of negative 10 end exponent space s space space space space space space space space space space space equals 1 over 10 to the power of 11 
    Therefore, accuracy is 1 part in 1011. 
    Question 288
    CBSEENPH11016722
    Wired Faculty App

    State the number of significant figures in the following : (a) 0.007 m (b) 2.64 cross times space 10 to the power of 24 kg (c) 0.2370 g cm-3 d) 6.320 J e)   6.032 N m-2 
    f)  0.0006032 m2
    Solution
    The number of significant figures are: 
    (a) One (b) Three (c) Four d) Three e) Four f) seven
    Question 289
    CBSEENPH11016723
    Wired Faculty App

    The dimensions of a block are 4.156x1.23x1.123. Express its volume to appropriate significant figures.

    Solution

    Volume of the block is given by Error converting from MathML to accessible text. 
    Volume of block is, V = 4.156 x 1.23 x 1.123 
                                    = 5.7406412 units 
    Out of three multiplied numbers, 1.23 has least number of significant figures i.e. three.
    Therefore, on rounding the volume to three significant figures we get,
                            V = 5.74 units.

    Question 290
    CBSEENPH11016725
    Wired Faculty App

     The length of plank is measured by simple scale and found to be 15.0 cm. What is the accuracy in one part?

    Solution
    The length of plank = 15.0 cm
    Here, accuracy in the measurement of 15.0 cm = 0.1 cm
    space space therefore Accuracy in one part = fraction numerator 0.1 over denominator 15.0 end fraction equals 1 over 150 
    Question 291
    CBSEENPH11016752
    Wired Faculty App

    Add 123.51 m and 21.6m.

    Solution
    Accuracy of sum is limited to accuracy of least accurate term.
    In 123.51 m and 21.6 m, the least accurate term is 21.6 m because it is known to one-tenth of meter only.
    Therefore, the sum of 123.51 m and 21.6 m should be accurate to one-tenth of meter only.
    space space space 123.51 space straight m plus space space 21.6 space straight m ___________ space space 145.11 space straight m ___________
    By reducing the result to one-tenth of meter, we get 145.1 m. 
    Question 292
    CBSEENPH11016756
    Wired Faculty App

    The mass of a box measured by a grocerís balance is 2.300 kg. Two gold pieces of masses 20.15 g and 20.17 g are added to the box. What is
    (a) the total mass of the box,
    (b) the difference in the masses of the pieces to correct significant figures ? 
    Solution
    a) The total mass of the box is equal to the sum of the masses of box and constituents in the box.
    i.e.       M = 2.3 + 0.02015 + 0.02017
                   = 2.34032 kg 
    Mass of the box is least accurate to one tenth of kg only. Therefore, the result must not be more accurate than one tenth of kg.
    space space therefore Total mass of the box, M = 2.3 kg
    b) Difference in masses of the pieces = 0.02 g 
    Question 293
    CBSEENPH11016757
    Wired Faculty App

    Two gold pieces weigh 20.15 g and 20.17 g. What is the difference in the masses of the pieces to correct significant figures?

    Solution
    Mass of  first piece = 20.15 g
    MAss of second piece = 20.17 g 
    Therefore,
    Difference in the masses of two pieces is,
    incrementm = (20.17 - 20.15) g
            = 0.02 g 
    Since the masses of pieces are known upto second decimal place of gm and difference is also upto second decimal place, hence the difference in the masses is 0.02 g.
    Question 294
    CBSEENPH11016759
    Wired Faculty App

    What is the order of the magnitude of the ratio of radius of the earth to the height of a building 5m high?

    Solution

    We have,
    Radius of earth = 6.4 space straight x space 10 to the power of 6 space end exponent straight m 
    Height of building = 5 m
    Now, 
    fraction numerator Radius space of space the space earth over denominator Height space of space building space end fraction equals fraction numerator 6.4 space straight x space 10 to the power of 6 straight m over denominator 5 straight m end fraction
                              space space space space equals space 1.28 space straight x space 10 to the power of 6 almost equal to 10 to the power of 6
    So, ratio is of the order of magnitude = 6

    Question 295
    CBSEENPH11016769
    Wired Faculty App

    Let a physical quantity X be measured in an experiment and the experiment be repeated for n times. If measured values are X1 X2......Xn then find the range in which all the measured values will lie.

    Solution

    When there are 'n' measured values, we cannot take any value as the true value.
    Thus, true value is the mean of all the observed values.
    straight X space equals space fraction numerator straight X subscript 1 plus straight X subscript 2 plus......... space plus space straight X subscript straight n over denominator straight n end fraction 
    The absolute error in measured values is,
    <pre>uncaught exception: <b>mkdir(): Permission denied (errno: 2) in /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/util/sys/Store.class.php at line #56mkdir(): Permission denied</b><br /><br />in file: /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/util/sys/Store.class.php line 56<br />#0 [internal function]: _hx_error_handler(2, 'mkdir(): Permis...', '/home/config_ad...', 56, Array) #1 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/util/sys/Store.class.php(56): mkdir('/home/config_ad...', 493) #2 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/plugin/impl/FolderTreeStorageAndCache.class.php(110): com_wiris_util_sys_Store->mkdirs() #3 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/plugin/impl/RenderImpl.class.php(231): com_wiris_plugin_impl_FolderTreeStorageAndCache->codeDigest('mml=<math xmlns...') #4 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/plugin/impl/TextServiceImpl.class.php(59): com_wiris_plugin_impl_RenderImpl->computeDigest(NULL, Array) #5 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/service.php(19): com_wiris_plugin_impl_TextServiceImpl->service('mathml2accessib...', Array) #6 {main}</pre> 
    Now the mean absolute error is ,
    straight capital delta space straight X equals space fraction numerator ΔX subscript 1 plus ΔX subscript 2 plus.......... plus ΔX subscript straight n space over denominator straight n end fraction 
    This means absolute error is the maximum possible error in the measurement.
    Thus, all measured values of measurement lie in the range, 
    straight X plus-or-minus ΔX
    i.e., straight X minus ΔX space less than space straight X subscript 1 space less than space straight X space plus ΔX

    Question 296
    CBSEENPH11016775
    Wired Faculty App

    Define the least count of screw gauge.

    Solution
    Least count of screw gauge is the distance by which the screw advances when it is rotated by one division on the circular scale.
    Mathematically, it is given by
    Least count =  fraction numerator Pitch over denominator No. space of space div. space on space circular space scale end fraction
    Question 297
    CBSEENPH11016779
    Wired Faculty App

    What is the least count of a screw gauge, whose circular scale is divided into 100 equal parts and the pitch of the screw is 1 mm?

    Solution
    Number of divisions on circular scale, N = 100
    Pitch of screw gauge is, d = 1 mm
    space space therefore Least count = straight d over straight N 
                            equals fraction numerator 1 mm over denominator 100 end fraction equals 0.01 space mm 
                            equals space fraction numerator 1 space mm over denominator 100 end fraction equals 0.01 mm
    Question 298
    CBSEENPH11016781
    Wired Faculty App

    What is the formula to find out the radius of curvature of a spherical surface?

    Solution

    Radius of curvature of a spherical surface is given by, 
                                straight R space equals space fraction numerator straight l squared over denominator 6 straight h end fraction plus straight h over 2 
    where,
    l is average distance between the three fixed legs of spherometer, and 
    h is the height of the central screw above the plane surface. 

    Question 299
    CBSEENPH11016784
    Wired Faculty App

    The least count of a stopwatch is 0.2 sec. The time period of 20 oscillations of a pendulum is measured to be 25.0 sec. What is the percentage error in this measurement?

    Solution
    The least count of stopwatch is = 0.2 s
    Time period of 20 oscillations = <pre>uncaught exception: <b>mkdir(): Permission denied (errno: 2) in /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/util/sys/Store.class.php at line #56mkdir(): Permission denied</b><br /><br />in file: /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/util/sys/Store.class.php line 56<br />#0 [internal function]: _hx_error_handler(2, 'mkdir(): Permis...', '/home/config_ad...', 56, Array) #1 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/util/sys/Store.class.php(56): mkdir('/home/config_ad...', 493) #2 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/plugin/impl/FolderTreeStorageAndCache.class.php(110): com_wiris_util_sys_Store->mkdirs() #3 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/plugin/impl/RenderImpl.class.php(231): com_wiris_plugin_impl_FolderTreeStorageAndCache->codeDigest('mml=<math xmlns...') #4 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/plugin/impl/TextServiceImpl.class.php(59): com_wiris_plugin_impl_RenderImpl->computeDigest(NULL, Array) #5 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/service.php(19): com_wiris_plugin_impl_TextServiceImpl->service('mathml2accessib...', Array) #6 {main}</pre>
    Percentage error in measurement = <pre>uncaught exception: <b>mkdir(): Permission denied (errno: 2) in /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/util/sys/Store.class.php at line #56mkdir(): Permission denied</b><br /><br />in file: /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/util/sys/Store.class.php line 56<br />#0 [internal function]: _hx_error_handler(2, 'mkdir(): Permis...', '/home/config_ad...', 56, Array) #1 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/util/sys/Store.class.php(56): mkdir('/home/config_ad...', 493) #2 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/plugin/impl/FolderTreeStorageAndCache.class.php(110): com_wiris_util_sys_Store->mkdirs() #3 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/plugin/impl/RenderImpl.class.php(231): com_wiris_plugin_impl_FolderTreeStorageAndCache->codeDigest('mml=<math xmlns...') #4 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/plugin/impl/TextServiceImpl.class.php(59): com_wiris_plugin_impl_RenderImpl->computeDigest(NULL, Array) #5 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/service.php(19): com_wiris_plugin_impl_TextServiceImpl->service('mathml2accessib...', Array) #6 {main}</pre>
                                                       = 0.8%
    Question 300
    CBSEENPH11016787
    Wired Faculty App

    Two gold pieces weigh 20.15 g and 20.17 g. What is the difference in the masses of the pieces to correct significant figures?

    Solution
    Mass of first gold piece = 20.17 g
    Mass of second gold piece = 20.15 g
    Difference in the masses of two pieces is,
                     Δm space equals space open parentheses 20.17 minus 20.15 close parentheses straight g equals 0.02 straight g 
    Here, masses of pieces are known upto second decimal place.
    Difference is also upto second decimal place.
    Hence the difference in the masses is 0.02g.      
     
    Question 301
    CBSEENPH11016789
    Wired Faculty App

    Discuss the process of combination of errors when two physical quantities are multiplied.

    Solution

    Combination error when two quantities are multiplied, is given by
    Let X = AB 
    Let ∆A, ∆B and ∆X be the absolute errors in A, B and X respectively.
    Therefore the above equation with error can be written as, 
    space space space space space space space space space space space space space space space space straight X plus-or-minus straight capital delta equals left parenthesis straight A plus-or-minus ΔA right parenthesis left parenthesis straight B plus-or-minus ΔB right parenthesis space space space space space rightwards double arrow space space space space straight X open parentheses 1 plus-or-minus ΔX over straight X close parentheses equals AB open parentheses 1 plus-or-minus ΔA over straight A close parentheses open parentheses 1 plus-or-minus ΔB over straight B close parentheses space space space space space 

    space space space space space space space space space space space space rightwards double arrow space space space space space space straight X open parentheses 1 plus-or-minus ΔX over straight X close parentheses equals AB open parentheses 1 plus-or-minus ΔA over straight A plus-or-minus ΔB over straight B plus-or-minus ΔA over straight A ΔB over straight B close parentheses rightwards double arrow space space space space space open parentheses 1 plus-or-minus ΔX over straight X close parentheses equals open parentheses 1 plus-or-minus ΔX over straight X plus-or-minus ΔB over straight B plus-or-minus ΔA over straight A ΔB over straight B close parentheses
     <pre>uncaught exception: <b>mkdir(): Permission denied (errno: 2) in /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/util/sys/Store.class.php at line #56mkdir(): Permission denied</b><br /><br />in file: /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/util/sys/Store.class.php line 56<br />#0 [internal function]: _hx_error_handler(2, 'mkdir(): Permis...', '/home/config_ad...', 56, Array) #1 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/util/sys/Store.class.php(56): mkdir('/home/config_ad...', 493) #2 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/plugin/impl/FolderTreeStorageAndCache.class.php(110): com_wiris_util_sys_Store->mkdirs() #3 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/plugin/impl/RenderImpl.class.php(231): com_wiris_plugin_impl_FolderTreeStorageAndCache->codeDigest('mml=<math xmlns...') #4 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/plugin/impl/TextServiceImpl.class.php(59): com_wiris_plugin_impl_RenderImpl->computeDigest(NULL, Array) #5 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/service.php(19): com_wiris_plugin_impl_TextServiceImpl->service('mathml2accessib...', Array) #6 {main}</pre> are small quantities.

    Therefore the product ΔA over straight A. ΔB over straight B will be very small and hence can be neglected.
    Thus the above equation reduce to,
    space space space space space space space plus-or-minus ΔX over straight X equals plus-or-minus ΔA over straight A plus-or-minus ΔB over straight B 
    Maximum value of fractional error in X is 
    ΔX over straight X equals ΔA over straight A plus ΔB over straight B
    Multiplying both sides by 100
    <pre>uncaught exception: <b>mkdir(): Permission denied (errno: 2) in /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/util/sys/Store.class.php at line #56mkdir(): Permission denied</b><br /><br />in file: /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/util/sys/Store.class.php line 56<br />#0 [internal function]: _hx_error_handler(2, 'mkdir(): Permis...', '/home/config_ad...', 56, Array) #1 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/util/sys/Store.class.php(56): mkdir('/home/config_ad...', 493) #2 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/plugin/impl/FolderTreeStorageAndCache.class.php(110): com_wiris_util_sys_Store->mkdirs() #3 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/plugin/impl/RenderImpl.class.php(231): com_wiris_plugin_impl_FolderTreeStorageAndCache->codeDigest('mml=<math xmlns...') #4 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/plugin/impl/TextServiceImpl.class.php(59): com_wiris_plugin_impl_RenderImpl->computeDigest(NULL, Array) #5 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/service.php(19): com_wiris_plugin_impl_TextServiceImpl->service('mathml2accessib...', Array) #6 {main}</pre> 
    i.e. Maximum possible %age error in X = Maximum possibe %age error in A + Maximum possible %age error in B.

    Question 302
    CBSEENPH11016790
    Wired Faculty App

    Compute the error in An, if percentage error in the measurement of A is α%.

    Solution

    Here, X =An 
    Let ∆A and ∆X be the absolute errors in A and X respectively.
    Therefore the above equation with error can be written as

    space space space space space space space space space space space space space space space space space space open parentheses straight X plus-or-minus ΔX close parentheses equals open parentheses straight A plus-or-minus ΔA close parentheses to the power of straight N rightwards double arrow space space space space space space straight X open parentheses 1 plus-or-minus ΔX over straight X close parentheses equals straight A to the power of straight n open parentheses 1 plus-or-minus ΔA over straight A close parentheses to the power of straight n rightwards double arrow space space space space space open parentheses 1 plus-or-minus ΔX over straight X close parentheses equals space space open parentheses 1 plus-or-minus straight n space ΔX over straight X close parentheses rightwards double arrow space space space space space space space ΔX over straight X equals straight n ΔA over straight A 
    Multiplying both sides by 100, 
    <pre>uncaught exception: <b>mkdir(): Permission denied (errno: 2) in /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/util/sys/Store.class.php at line #56mkdir(): Permission denied</b><br /><br />in file: /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/util/sys/Store.class.php line 56<br />#0 [internal function]: _hx_error_handler(2, 'mkdir(): Permis...', '/home/config_ad...', 56, Array) #1 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/util/sys/Store.class.php(56): mkdir('/home/config_ad...', 493) #2 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/plugin/impl/FolderTreeStorageAndCache.class.php(110): com_wiris_util_sys_Store->mkdirs() #3 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/plugin/impl/RenderImpl.class.php(231): com_wiris_plugin_impl_FolderTreeStorageAndCache->codeDigest('mml=<math xmlns...') #4 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/plugin/impl/TextServiceImpl.class.php(59): com_wiris_plugin_impl_RenderImpl->computeDigest(NULL, Array) #5 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/service.php(19): com_wiris_plugin_impl_TextServiceImpl->service('mathml2accessib...', Array) #6 {main}</pre> 
    i.e. Maximum possible %age error in X = n (Maximum possible %age error in A)

    Question 303
    CBSEENPH11016792
    Wired Faculty App

    Two forces F2,= (50±0.35)N and F2= (35±0.16)N are acting simultaneously on a body. Find the net resultant force on the body if they are acting in: (i) same direction (ii) opposite direction

    Solution
    (i) When the two forces are acting in same direction, then
    F = F1 + F2 
       equals space open parentheses 50 plus-or-minus 0.35 close parentheses straight N plus open parentheses 35 plus-or-minus 0.16 close parentheses straight N equals space open square brackets open parentheses 50 plus 35 close parentheses plus-or-minus open parentheses 0.35 plus 0.16 close parentheses close square brackets straight N equals space left parenthesis 85 plus-or-minus 0.51 right parenthesis straight N
    (ii) When the two forces are acting in opposite direction, then
    F = F1 - F2
       equals space open parentheses 50 plus-or-minus 0.35 close parentheses straight N minus open parentheses 35 plus-or-minus 0.16 close parentheses straight N equals space open square brackets open parentheses 50 minus 35 close parentheses plus-or-minus open parentheses 0.35 plus 0.16 close parentheses close square brackets straight N equals space left parenthesis 15 plus-or-minus 0.51 right parenthesis straight N
     
    Question 305
    CBSEENPH11016796
    Wired Faculty App

    What are the full forms of RADAR and SONAR?

    Solution

    RADAR stands for Radio Detection and Ranging.

    SONAR stands for Sound Navigation and Ranging.

    Question 306
    CBSEENPH11016799
    Wired Faculty App

    The dimensions of a slab are 5.43cm x 2.93cmx 1.87cm. Express its volume in appropriate significant.

    Solution

    Volume of a slab = 5.43 cross times 2.93 cross times space1.87 
                             = 29.75 cm3
    The volume in appropriate significant figure = 29.8 cm3 

    Question 307
    CBSEENPH11016801
    Wired Faculty App

    A cylinder has a length of 6.92 cm and a radius of 2.54 cm. Express its surface area significant figures.

    Solution

    Given, 
    Length of the cylinder, l = 6.92 m 
    Radius of the cylinder = 2.54 cm 
    Surface area of the cylinder is given by, A = 2 πrh space plus space πr to the power of 2 space end exponent+ πr squared

    Area = 150.89 cm2 

    Question 308
    CBSEENPH11016802
    Wired Faculty App

    What is the number of significant figures in a measurement 0.00985?

    Solution

    Three significant figures are there in the number 0.00985. 

    Question 309
    CBSEENPH11016803
    Wired Faculty App

    Which of the following measurements are not done by screw gauge:

    0.3200cm, 5.4cm, 1.22cm, 0.00005mm
    Solution

    The measurements which do not require us to use screw gauge are 0.3200 cm, 5.4 cm and 1.22 cm. 

    Since, 0.00005 mm is a very small quantity, using screw gauge will help us to measure it precisely. 
    Question 310
    CBSEENPH11016820
    Wired Faculty App

    The centripetal force F acting on a particle moving uniformly in a circle may depend upon mass (m), velocity (u) and radius (r) of the circle. Derive the formula for F using the method of dimensions.

    Solution

    Let F = K ma vb rc               ... (i)
    where, 
    K is a dimensionless constant of proportionality and
    a,b c are the powers of m,v and r respectively to represent F. 
    Now, writing the dimensions of various quantities in (i), we get
    [M1 L 1 T-2] = Ma [LT-1]b Lc
                      = Ma Lb T--b Lc
                      = Ma Lb+c T-b
    On applying the principle of homogeneity of dimensions, we get
    a = 1 ,
    b+c = 1               ... (ii)
    -b = -2 or b = 2 
    From equation (ii),
    c = 1-b = 1-2 = -1
    Putting these values in (i), we get
    F = K m1 v2 r-1 
    rightwards double arrow space space straight F space equals space straight K thin space mv squared over straight r 
    This is the required relation.

    Question 311
    CBSEENPH11016857
    Wired Faculty App

    Pick up the dimensional constant quantities from the following.

    Surface energy, gas constant, Rydberg constant, ?, e, velocity of light in water.
    Solution

    The dimensional constant quantities are:
    Gas constant, Rydberg constant, velocity of light in water.

    Question 312
    CBSEENPH11016865
    Wired Faculty App

    Name two physical quantities that have same dimensional formula as that of angular velocity.

    Solution

    Frequency and velocity gradient have the same dimensional formula as that of angular velocity. 
    The formula is [MoLoT-1]

    Question 313
    CBSEENPH11017326
    Wired Faculty App

    Define centre of mass.

    Solution
    Centre of mass of a body or a system of bodies is a point at which the entire mass of the body or system is supposed to be concentrated. 
    Question 314
    CBSEENPH11017327
    Wired Faculty App

    What is the need of centre of mass?

    Solution
    Newton’s second law of motion is strictly applicable to point masses only. To apply the Newton's law of motion to rigid bodies, the concept of centre of mass is introduced.
    The concept of centre of mass of a system enables us to discuss overall motion of the system by replacing the system by an equivalent single point object. 
    Question 315
    CBSEENPH11017328
    Wired Faculty App

    Is it necessary for centre of mass to lie within the body?

    Solution
    No, centre of mass needs not to lie within the body. It is not necessary that the total mass of the system be actually present at the centre.
    The position of the centre of mass is calculated using the usual Newtonian type of equations of motion. 
    Question 316
    CBSEENPH11017329
    Wired Faculty App

    Is it necessary that there should be matter at the centre of mass of system?

    Solution
    No, it is not necessary that there be matter at the centre of mass of the system.
    For e.g., if two equal point masses are separated by certain distance, the centre of mass lies at the mid point of two point masses and there is no mass at that point.
    Question 317
    CBSEENPH11017330
    Wired Faculty App

    What is the significance of defining the center of mass of a system?

    Solution
    The motion of n particle system can be reduced to one particle motion.
    An equivalent single point object would enable us to discuss the overall motion of the system. 
    Question 318
    CBSEENPH11017332
    Wired Faculty App

    A body at rest explodes in three fragments. What is the velocity of the centre of mass of exploded part?

    Solution
    Zero.
    In the absence of external force, the centre of mass moves with constant velocity. Since before explosion the body was at rest, therefore, the velocity of the centre of mass is zero before an explosion. Hence it will also be zero after the explosion.
    Question 319
    CBSEENPH11017338
    Wired Faculty App

    A system consists of two particles of masses m1 and m2 having position vectors 
    rightwards arrow for straight F subscript 1 of space and space rightwards arrow for straight r subscript 2. of. What is the position vector of the centre of mass of system?
    Solution

    Given, the system consists of two masses m1 and m2. Position vectors of the masses m1 and m2 are r1 and r2 respectively.
    Therefore, the position vector of the centre of mass of the system is given by, 
                      space space space space space fraction numerator straight m subscript 1 rightwards arrow for straight r subscript 1 of plus straight m subscript 2 rightwards arrow for straight r subscript 2 of over denominator straight m subscript 1 plus straight m subscript 2 end fraction

    Question 320
    CBSEENPH11017339
    Wired Faculty App

    Under what conditions the centre of mass of two bodies lies midway between the two bodies?

    Solution
    When the masses of two bodies are equal then centre of mass lies midway between the two bodies.
    Question 321
    CBSEENPH11017340
    Wired Faculty App

    The position vectors of two bodies of equal masses are rightwards arrow for straight r subscript 1 of and space rightwards arrow for straight r subscript 2 of. What is the position vector of their centre of mass?
    Solution

    Given position vectors of two bodies are r1 and r2. 
    Therefore, the position vector of their centre of mass is given by,
                          r =  fraction numerator rightwards arrow for straight r subscript 1 of plus rightwards arrow for straight r subscript 2 of over denominator 2 end fraction

    Question 322
    CBSEENPH11017341
    Wired Faculty App

    Can centre of mass and centre of gravity be at the same point?

    Solution
    Yes, when a body is placed in a uniform gravitational field, both the centre of mass and the centre of gravity coincide.
    Question 323
    CBSEENPH11017342
    Wired Faculty App

     Where does the centre of mass of uniform triangular lamina lie?

    Solution
    The centre of mass of uniform triangular lamina lies at the point of intersection of medians.
    Question 324
    CBSEENPH11017343
    Wired Faculty App

    Where does the centre of mass of uniform cone or pyramid lie?

    Solution
    The centre of mass of uniform cone or pyramid lies on the axis at a distance 3h/4 from the vertex of the cone, where h is the height of the cone.
    Question 325
    CBSEENPH11017344
    Wired Faculty App

    Electron and proton at rest are released. What is the velocity of centre of mass when the two approach towards each other?

    Solution
    When an electron and proton at rest are released, the velocity of the centre of mass is zero, when the two approach towards each other. This is because, in the absence of external force, the centre of mass moves with constant velocity.
    Question 326
    CBSEENPH11017345
    Wired Faculty App

    Does the position of centre of mass of rigid body change w.r.t. body when in motion?

    Solution
    No, with respect to the motion of the body, the centre of mass of the rigid body does not change. 
    Question 327
    CBSEENPH11017346
    Wired Faculty App

    Which will be at greater height from the surface of the earth, for a 5 km tall body-centre of gravity or centre of mass?

    Solution
    Centre of mass will be at a greater height than the centre of gravity. For taller objects, the centre of gravity acts in such a way as to topple the object. So, the centre of mass is concentrated at a point higher so that the body does not topple. 
    Question 328
    CBSEENPH11017347
    Wired Faculty App

    When the projectile explodes in flight, the momentum vector of each fragment changes. What is the total change in momentum of all the fragments due to internal forces due to which projectile explodes?

    Solution
    The change in momentum of the fragments will be zero due to the internal forces. Internal forces exist in pairs inside a system and cancels out. 
    Question 329
    CBSEENPH11017348
    Wired Faculty App

    A body at rest explodes into two fragments of unequal masses. Which fragment will have greater speed?

    Solution
    Given, that a body at rest will explode into two fragments of unequal masses.
    Lighter fragment will have a greater velocity than the heavier one because, in the absence of external force, the total momentum of the system remains conserved. Hence, the velocity of the system will remain constant and at rest.  
    Question 330
    CBSEENPH11017349
    Wired Faculty App

    Two bodies of masses m1 and m2 are moving such that their momcntums are equal and opposite. What is the velocity of centre of mass of system of two bodies?

    Solution
    The centre of mass of the two systems will move with constant velocities. Here in this case, the velocity is zero because the momentum of the masses are equal and opposite. 
    Question 331
    CBSEENPH11017350
    Wired Faculty App

    A bomb of mass M at rest explodes into many fragments so that kinetic energy released is K. What is the velocity of centre of mass of fragments?

    Solution
    Before the explosion happened, the bomb is at rest. Therefore, the velocity of the centre of mass of fragment after the explosion is zero. Velocity remains constant. 
    Question 332
    CBSEENPH11017351
    Wired Faculty App

    A projectile is fired from O so that it follows the parabola path OPQ. If the projectile explodes in mid air and fragments fly in different directions, then what path the centre of mass will follow?

    Solution
    The centre of mass will continue to follow parabola path OPQ. The position of the centre of mass is independent of  the path taken by the mass. Thus, the position remains the same.
    Question 333
    CBSEENPH11017353
    Wired Faculty App

    The nucleus ZXA at rest undergoes a decay. If the α-particle emits with velocity 2.4x107 m/s, then find the velocity of centre of mass of a particle and daughter nuclei Z-2 Y A-4. 

    Solution
    Since the parent nucleus is at rest before decay, therefore the velocity of the centre of mass after decay is zero.
    Question 334
    CBSEENPH11017354
    Wired Faculty App

    Do the internal forces affect the motion of centre of mass?

    Solution
    No, the internal forces between the particles forming the system cancel out in pairs. Thus, there is no effect of internal forces on the motion of the centre of mass.
    Question 335
    CBSEENPH11017355
    Wired Faculty App

    What is the acceleration of centre of mass of system consisting of N particles when no force is acting on the system?

    Solution
    When no force is acting on the system, acceleration of the centre of mass of the system consisting of N particles is zero.
    Question 336
    CBSEENPH11017356
    Wired Faculty App

    Is it necessary that there be mass at centre of mass?

    Solution
    No, the centre of mass may or may not lie within the body. i.e., it may be possible that at the position of the centre of mass there is no mass.
    For e.g. in the case of a ring, the centre of mass is at the centre and there is no mass of ring at the centre of a ring.
    Question 337
    CBSEENPH11017357
    Wired Faculty App

     Define centre of mass and centre of gravity.

    Solution
    Centre of mass of a body is the point where the whole mass of the body is supposed to be concentrated.
    Centre of gravity of a body is the point where whole weight of the body is supposed to act or concentrated.
    Question 338
    CBSEENPH11017362
    Wired Faculty App

    Derive an expression for the position vector of centre of mass of a body consisting of two particles in terms of the their position vectors. Deduce the result for n particles.

    Solution

    Consider that  the body consists of two particles of massesspace space space space space space space straight m subscript 1 space and space straight m subscript 2 having position vector rightwards arrow for straight r subscript 1 of and space rightwards arrow for straight r subscript 2 of respectively.
    The moment of masses of particles constituting the body about O is, 
    space space space space space space space space space space space space space space space space space space space space space space space space space straight m subscript 1 rightwards arrow for straight r subscript 1 of plus straight m subscript 2 rightwards arrow for straight r subscript 2 of space space space space space space space space space space space space space space space space space space space space space space space space space space space space... left parenthesis 1 right parenthesis 
                       
    Let rightwards arrow for straight R of be the position vector of the center of mass.
    The moment of mass of center of mass about origin O is, 
    <pre>uncaught exception: <b>mkdir(): Permission denied (errno: 2) in /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/util/sys/Store.class.php at line #56mkdir(): Permission denied</b><br /><br />in file: /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/util/sys/Store.class.php line 56<br />#0 [internal function]: _hx_error_handler(2, 'mkdir(): Permis...', '/home/config_ad...', 56, Array) #1 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/util/sys/Store.class.php(56): mkdir('/home/config_ad...', 493) #2 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/plugin/impl/FolderTreeStorageAndCache.class.php(110): com_wiris_util_sys_Store->mkdirs() #3 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/plugin/impl/RenderImpl.class.php(231): com_wiris_plugin_impl_FolderTreeStorageAndCache->codeDigest('mml=<math xmlns...') #4 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/plugin/impl/TextServiceImpl.class.php(59): com_wiris_plugin_impl_RenderImpl->computeDigest(NULL, Array) #5 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/service.php(19): com_wiris_plugin_impl_TextServiceImpl->service('mathml2accessib...', Array) #6 {main}</pre>                                   ...(2)
    Since moment of mass of the body about any point is equal to the moment of mass of CM about the same point, therefore from equations (1) and (2), we get
    open parentheses straight m subscript 1 plus straight m subscript 2 close parentheses rightwards arrow for straight R of equals straight m subscript 1 rightwards arrow for straight r subscript 1 of plus straight m subscript 2 rightwards arrow for straight t subscript 2 of  

    rightwards double arrow         space space rightwards arrow for straight R of equals fraction numerator straight m subscript 1 rightwards arrow for straight r subscript 1 of plus straight m subscript 2 rightwards arrow for straight r subscript 2 of over denominator straight m subscript 1 plus straight m subscript 2 end fraction 

    For a system of n particles, the position of center of mass is given by, 

    rightwards arrow for straight R of equals fraction numerator straight m subscript 1 rightwards arrow for straight r subscript 1 of plus straight m subscript 2 rightwards arrow for straight r subscript 2 of plus...... plus straight m subscript straight n plus rightwards arrow for straight r subscript straight n of over denominator straight m subscript 1 plus straight m subscript 2 plus....... plus straight m subscript straight n end fraction  

    Question 339
    CBSEENPH11017378
    Wired Faculty App

    A binary system consists of two bodies of masses m1 and m2 separated by distance r. What is the distance of centre of mass from mass m1 and m2?

    Solution

    Let stack r subscript 1 with rightwards harpoon with barb upwards on top space space a n d space stack r subscript 2 with rightwards harpoon with barb upwards on top be the position vectors of massesstraight m subscript 1 space and space straight m subscript 2 w.r.t. the centre of mass.

    The moment of mass of the system about centre of mass  is zero.
    Therefore, 
                          m1r1 + m2r2 = 0

    That is, 
    m1 r1 = -  m2 r2 , or 
    m1 r1 = m2 r2                       ... (1) 
    Also, position vector is given by
    R = r1 + r2                             ...(2)
    On solving (1) and (2), we have
    straight r subscript 1 space equals space fraction numerator straight m subscript 2 straight R over denominator straight m subscript 1 space plus space straight m subscript 2 end fraction space space space space space space space space and space space space space space straight r subscript 2 space equals fraction numerator straight m subscript 1 space straight R over denominator straight m subscript 1 space plus space straight m subscript 2 end fraction space space space 

    Question 340
    CBSEENPH11017386
    Wired Faculty App

    In an arbitrary molecule XY, the distance between the nuclei of X and Y is 2.4 Å. If the mass of X is 23 times the mass of Y, then what will be the distance of centre of mass from X atom?

    Solution
    Let m be the mass of atom Y.
    Therefore, the mass of atom X will be 23m.
    Distance of the centre of mass from atom X is given by,  
            

    straight r subscript straight x equals fraction numerator rm subscript straight y over denominator straight m subscript straight x plus straight m subscript straight y end fraction equals fraction numerator rm over denominator 23 straight m plus straight m end fraction equals 1 over 24 straight r space space space space equals fraction numerator 2.4 over denominator 24 end fraction straight A with dot on top space equals space 0.1 straight A with dot on top space therefore space space space space space space space space space space space space straight r subscript straight x equals 0.1 straight A with dot on top
    Question 341
    CBSEENPH11017389
    Wired Faculty App

    The distance between the centers of carbon and oxygen atoms in CO molecule is 1.124 Å. Calculate the distance of centre of mass of molecule from oxygen atom.

    Solution
    The distance of centre of mass from oxygen atom is, 

    The distance of the centre of mass of molecule from oxygen atom is, 
    r2 = fraction numerator straight m subscript straight c over denominator straight m subscript straight c space plus space straight m subscript straight o end fraction space cross times space 1.124 space space straight A with straight o on top 
        = 12 over 28 space cross times space 1.124 space straight A with straight o on top space equals space 0.483 space space straight A with straight o on top 

    Question 343
    CBSEENPH11017404
    Wired Faculty App

    Two particles of masses 100gm and 300 gm have position coordinates (2,5,13) and (-6,4,-2) respectively. Find the position coordinates of centre of mass.

    Solution
    The position coordinates of centre of mass are given by, 
    X = fraction numerator straight m subscript 1 straight x subscript 1 space plus space straight m subscript 2 straight x subscript 2 over denominator straight m subscript 1 space plus space straight m subscript 2 end fraction 
    Y = fraction numerator straight m subscript 1 straight y subscript 1 space plus space straight m subscript 2 straight y subscript 2 over denominator straight m subscript 1 space plus space straight m subscript 2 end fraction 
    Z = fraction numerator straight m subscript 1 straight z subscript 1 space plus space straight m subscript 2 straight z subscript 2 over denominator straight m subscript 1 space plus space straight m subscript 2 end fraction
    Substituting the respective values, we get 
    straight X space equals space fraction numerator 100 space cross times space 2 space plus space 300 space cross times space left parenthesis negative 6 right parenthesis over denominator 100 plus 30 end fraction space equals space minus 4 space straight Y space equals space fraction numerator 100 space cross times space 5 space plus space 300 space cross times space 4 over denominator 100 space plus space 300 end fraction space equals space 17 over 4 straight Z space equals space fraction numerator 100 space cross times space 13 space plus space 300 space cross times space left parenthesis negative 2 right parenthesis over denominator 100 space plus space 300 end fraction space equals space 7 over 4 
    Therefore, the position coordinates of centre of mass are (negative 4 comma space 17 over 4 comma space 7 over 4) 
    Question 344
    CBSEENPH11017432
    Wired Faculty App

    Center of mass of a system of three particles of masses 2kg, 4kg and 6kg is at P ↔ (-2,0,4). If 6kg mass is removed from system, the center of mass of system shifts to Q ↔ (1,2,4). Find the position co-ordinates of 6kg mass.

    Solution

    Consider, stack straight r subscript 1 with rightwards harpoon with barb upwards on top space comma space stack r subscript 2 with rightwards harpoon with barb upwards on top space a n d stack space r subscript 3 with rightwards arrow on top space be the position vectors of masses 2 kg, 4 kg and 6 kg respectively. The position coordinates of the centre of mass of 2 kg, 4 kg and 6 kg is (-2, 0, 4). 
    Therefore position vector of the centre of mass of 2  kg, 4 kg and 6 kg is, 
    R1 = -2straight i with hat on top + 0 straight j with hat on top + 4 straight k with hat on top 
    Also, 
    straight R with rightwards harpoon with barb upwards on top subscript 1 space equals space fraction numerator 2 space stack r subscript 1 with rightwards harpoon with barb upwards on top space plus space 4 space stack r subscript 2 with rightwards harpoon with barb upwards on top space plus space 6 stack space r subscript 3 with rightwards arrow on top over denominator 2 space plus space 4 space plus space 6 end fraction 
    Therefore, 
    fraction numerator 2 space stack straight r subscript 1 with rightwards harpoon with barb upwards on top space plus space 4 space stack straight r subscript 2 with rightwards harpoon with barb upwards on top space plus 6 stack space straight r subscript 3 with rightwards harpoon with barb upwards on top over denominator 2 space plus space 4 space plus space 6 end fraction space equals space minus space 2 space straight i with rightwards arrow on top space plus space 0 space straight j with hat on top space plus space 4 space straight k with hat on top space 
    Therefore,
    2 space stack straight r subscript 1 with rightwards harpoon with barb upwards on top space plus space 4 space stack straight r subscript 2 with rightwards harpoon with barb upwards on top space plus space 6 space stack straight r subscript 3 with rightwards harpoon with barb upwards on top space equals space minus space 24 space straight i with hat on top space plus space 0 space straight j with hat on top space plus space 48 space straight k with hat on top   ... (1) 
    The position coordinates of centre of mass of 2kg and 4kg is (1, 2, 4).
    Therefore position vector of centre of mass of 2kg and 4 kg  is, 
    stack straight R subscript 2 with rightwards harpoon with barb upwards on top space equals space straight i with hat on top space plus space 2 space straight j with hat on top space plus space 4 space straight k with hat on top space space space space space space space space space space space space space space space space space space... space left parenthesis 2 right parenthesis thin space 
    Also,
    stack R subscript 2 space with rightwards harpoon with barb upwards on top space equals space fraction numerator 2 space stack r subscript 1 with rightwards harpoon with barb upwards on top space plus space 4 space stack r subscript 2 with rightwards harpoon with barb upwards on top over denominator 2 space plus space 4 end fraction space 
    therefore space fraction numerator 2 space stack straight r subscript 1 with rightwards harpoon with barb upwards on top space plus space 4 space stack straight r subscript 2 with rightwards harpoon with barb upwards on top over denominator 2 space plus space 4 end fraction space equals space straight i with hat on top space plus space 2 straight j with hat on top space plus space 4 space straight k with hat on top space 2 space stack straight r subscript 1 with rightwards harpoon with barb upwards on top space plus space 4 space stack straight r subscript 2 with rightwards harpoon with barb upwards on top space equals space 6 space space straight i with hat on top space plus space 12 space straight j with hat on top space plus space 24 space straight k with hat on top space space space space space space space space... space left parenthesis 2 right parenthesis 
    From (2) from (1), we get
    6 space stack straight r subscript 3 with rightwards harpoon with barb upwards on top space equals space minus space 30 space straight i with hat on top space minus space 12 space straight j with hat on top space plus space 24 space straight k with hat on top stack straight r subscript 3 with rightwards harpoon with barb upwards on top space equals space minus 5 space straight i with hat on top space minus space 2 space straight j with hat on top space plus space 4 space straight k with hat on top  
    The position vector of 6kg mass is,
    stack straight r subscript 3 with rightwards harpoon with barb upwards on top space equals space minus space 5 space i with hat on top space minus space 2 space j with hat on top space plus space 24 space k with hat on top
    Therefore, position coordinate of 6 kg is ( -5, -2, 4).

    Question 345
    CBSEENPH11017433
    Wired Faculty App

    A thin uniform flat plate of mass 4-5 kg is in the form of L shape lamina as shown in figure.

    Find the position of center of mass.

    Solution
    As shown in the figure below, the lamina consists of three squares, each of side 2 unit. 
    Therefore, the coordinates of the centre of mass of three squares are C1 ↔ (1,1), C2 ↔ (3,1) and C3 ↔ (1,3).
    Let m be the mass of each square.
    Therefore, coordinates of centre of mass of lamina are, 

    straight X space equals space fraction numerator straight m subscript 1 straight x subscript 1 plus straight m subscript 2 straight x subscript 2 plus straight m subscript 3 straight x subscript 3 over denominator straight m subscript 1 plus straight m subscript 2 plus straight n 1 subscript 3 end fraction space space space space equals space fraction numerator straight m cross times 1 plus straight m cross times 3 plus straight m cross times 1 over denominator straight m plus straight m plus straight m end fraction space space space equals 5 over 3 space unit space space straight Y space equals space fraction numerator straight m subscript 1 straight y subscript 1 plus straight m subscript 2 straight y subscript 2 plus straight m subscript 3 straight y subscript 3 over denominator straight m subscript 1 plus straight m subscript 2 plus straight n 1 subscript 3 end fraction space space space space space space equals space fraction numerator straight m cross times 1 plus straight m cross times 1 plus straight m cross times 3 over denominator straight m plus straight m plus straight m end fraction space space space space equals 5 over 3 space unit space 
    Question 346
    CBSEENPH11017434
    Wired Faculty App

    Discuss the motion of center of mass of a system consisting of n particles and show that center of mass moves with constant velocity in the absence of external force.

    Solution

    Consider a system of n particles of masses m1, m2, ..... mn having position vectors stack straight r subscript 1 with rightwards harpoon with barb upwards on top comma space stack r subscript 2 with rightwards harpoon with barb upwards on top comma space.... stack space r subscript n with rightwards harpoon with barb upwards on top. The position of the centre of mass of the system is, 
    straight R with rightwards harpoon with barb upwards on top space equals space fraction numerator m subscript 1 stack r subscript 1 with rightwards harpoon with barb upwards on top space plus space m subscript 2 stack r subscript 2 with rightwards harpoon with barb upwards on top space plus space.... plus space m subscript n stack r subscript n with rightwards harpoon with barb upwards on top over denominator m subscript 1 space plus space m subscript 2 plus space..... space plus space m subscript 2 end fraction space
    If M is the total mass of the system, then 
    straight R with rightwards harpoon with barb upwards on top space equals space fraction numerator straight m subscript 1 space stack straight r subscript 1 with rightwards harpoon with barb upwards on top space plus space straight m subscript 2 stack straight r subscript 2 with rightwards harpoon with barb upwards on top space..... space straight m subscript straight n stack straight r subscript straight n with rightwards harpoon with barb upwards on top over denominator straight M end fraction straight M straight R with rightwards harpoon with barb upwards on top space equals space straight m subscript 1 stack straight r subscript 1 with rightwards harpoon with barb upwards on top space plus space straight m subscript 2 stack straight r subscript 2 with rightwards harpoon with barb upwards on top space.... space straight m subscript straight n space stack straight r subscript straight n with rightwards harpoon with barb upwards on top space space space space space space space space space space space space space space space space space space space... space left parenthesis 1 right parenthesis thin space Differentiating space left parenthesis 1 right parenthesis thin space both space sides space straight w. straight r. straight t space to space straight t comma space straight M space fraction numerator straight d straight R with rightwards harpoon with barb upwards on top over denominator dt end fraction space equals space straight m subscript 1 space dr subscript 1 over dt space plus space straight m subscript 2 space dr subscript 2 over dt space plus space.... plus space straight m subscript straight n fraction numerator straight d stack straight r subscript straight n with rightwards harpoon with barb upwards on top over denominator dt end fraction space rightwards double arrow space straight M space straight V with rightwards harpoon with barb upwards on top space equals space straight m subscript 1 space stack straight v subscript 1 with rightwards harpoon with barb upwards on top space plus space straight m subscript 2 space stack straight v subscript 2 with rightwards harpoon with barb upwards on top space....... space straight m subscript straight n stack straight v subscript straight n with rightwards harpoon with barb upwards on top space space space space space space... space left parenthesis 2 right parenthesis thin space Differentiating space left parenthesis 2 right parenthesis space both space sides space straight w. straight r. straight t space to space straight t comma space straight M space fraction numerator straight d straight V with rightwards harpoon with barb upwards on top over denominator dt end fraction space equals space straight m subscript 1 space fraction numerator straight d stack straight v subscript 1 with rightwards harpoon with barb upwards on top over denominator dt end fraction space plus space straight m subscript 2 space fraction numerator straight d stack straight v subscript 2 with rightwards harpoon with barb upwards on top over denominator dt end fraction space.... space straight m subscript straight n space fraction numerator straight d stack straight v subscript straight n with rightwards harpoon with barb upwards on top over denominator dt end fraction space straight M space straight A with rightwards harpoon with barb upwards on top space equals space straight m subscript 1 space stack straight a subscript 1 with rightwards harpoon with barb upwards on top space space plus space straight m subscript 2 stack space straight a subscript 2 with rightwards harpoon with barb upwards on top space plus space... straight m subscript straight n space stack straight a subscript straight n with rightwards harpoon with barb upwards on top space space space space space space space space space space equals stack space straight F subscript 1 with rightwards harpoon with barb upwards on top space plus space stack straight F subscript 2 with rightwards harpoon with barb upwards on top plus space.... space plus space stack straight F subscript straight n with rightwards arrow on top space space space space space space space space space space space space space space space space space space space space space space... space left parenthesis space 3 right parenthesis thin space where comma space straight F with rightwards harpoon with barb upwards on top subscript straight a space equals space space straight m subscript 1 space stack straight a subscript 1 with rightwards harpoon with barb upwards on top space comma space stack straight F subscript 2 with rightwards harpoon with barb upwards on top space equals space straight m subscript 2 space stack space straight a subscript 2 with rightwards harpoon with barb upwards on top space...... space stack straight F subscript straight n with rightwards harpoon with barb upwards on top space equals space straight m subscript straight n space stack straight a subscript straight n with rightwards harpoon with barb upwards on top space
    are the net forces (internal + external) acting on masses m1, m2 ..... and mn respectively. 
    The net force on the system of particles is, 
    straight F with rightwards harpoon with barb upwards on top space equals space straight F with rightwards harpoon with barb upwards on top subscript 1 space plus space straight F with rightwards harpoon with barb upwards on top subscript 2 space.... space plus stack space straight F with rightwards harpoon with barb upwards on top subscript straight n space space space space space equals space left parenthesis space stack straight F subscript 1 with rightwards harpoon with barb upwards on top to the power of straight e space plus stack space straight F subscript 1 with rightwards harpoon with barb upwards on top space to the power of straight i space right parenthesis space plus space space left parenthesis space stack straight F subscript 2 with rightwards harpoon with barb upwards on top to the power of straight e space plus stack space straight F subscript 2 with rightwards harpoon with barb upwards on top space to the power of straight i space right parenthesis..... space left parenthesis space left parenthesis space stack straight F subscript straight n with rightwards harpoon with barb upwards on top to the power of straight e space plus stack space straight F subscript straight n with rightwards harpoon with barb upwards on top space to the power of straight i space right parenthesis space space space space space space equals space left parenthesis stack space straight F with rightwards harpoon with barb upwards on top subscript 1 to the power of straight e space plus stack space straight F with rightwards harpoon with barb upwards on top subscript 2 to the power of straight e space.... space straight F with rightwards harpoon with barb upwards on top subscript straight n to the power of straight e right parenthesis space plus space left parenthesis thin space straight F with rightwards harpoon with barb upwards on top subscript 1 to the power of straight i space plus space straight F with rightwards harpoon with barb upwards on top subscript 2 to the power of straight i.... space straight F with rightwards harpoon with barb upwards on top subscript straight n to the power of straight i right parenthesis space space space space space space equals space left parenthesis stack space straight F subscript 1 with rightwards harpoon with barb upwards on top to the power of straight e space plus stack space straight F with rightwards harpoon with barb upwards on top subscript 2 to the power of straight e space..... stack space straight F with rightwards harpoon with barb upwards on top subscript straight n to the power of straight e right parenthesis thin space space space space equals stack space straight F with rightwards harpoon with barb upwards on top subscript ext space space space space space space space space space space space space space space space space space space space space space space space space space... space left parenthesis 4 right parenthesis

    Question 347
    CBSEENPH11017437
    Wired Faculty App

    In which of following cases, there is matter at centre of mass:

    (i) Uniform solid sphere

    (ii) Hollow sphere

    (iii) Ring

    (iv) Uniform annular disc

    (v) Disc

    (vi) Solid cube?
    Solution
    In uniform solid sphere, disc and solid cube there is matter at the centre of mass, while there is no matter at centre of mass in hollow sphere, ring and annular disc.
    Question 348
    CBSEENPH11017438
    Wired Faculty App

    Does a homogenous disc has any moment of mass about the centre of the disc?

    Solution
    The moment of the mass of a body about the centre of mass is zero. So, at the centre of the homogeneous disc, the moment of mass of the disc is zero. And, the centre of mass of the homogeneous disc lies at the centre of the disc. 
    Question 349
    CBSEENPH11017439
    Wired Faculty App

    Explain the motion of the earth-moon system around the sun using centre of mass concept.

    Solution
    The Earth-moon system forms a binary system. Both the Earth and the moon revolve in circles about the common centre of mass. The centre of mass of the earth-moon system revolves around the sun in an elliptical path around the sun.
    Question 350
    CBSEENPH11017440
    Wired Faculty App

    A child sits stationary at one end of a long trolley moving uniformly with a speed v on a smooth horizontal floor. If the child gets up and runs about on the trolley in any manner, what is the speed of the centre of mass of the (trolley + child) system?

    Solution
    In the absence of external forces, the centre of mass moves with constant velocity. That is, the velocity of the centre of mass of trolley and child remains unchanged, i.e., v. 
    When the child is moving on the trolley, the force exerted by the child on the trolley is the internal force. 
    Question 351
    CBSEENPH11017441
    Wired Faculty App

    What is the nature of motion of centre of mass of an isolated system?

    Solution
    An isolated system is a system which does not experience any external force. In the absence of external force, the centre of mass moves with a constant velocity. Thus, the motion of the centre of mass of an isolated system is in uniform motion.
    Question 353
    CBSEENPH11017453
    Wired Faculty App

    How would you reduce the two-body problem to a single body problem?

    or

    How would you reduce the problem of two body system to a problem of single body system?
    Solution

    Consider two particles of masses m1 and m2 having position vectors r with rightwards harpoon with barb upwards on top subscript 1 space and space r with rightwards harpoon with barb upwards on top subscript 2 respectively.
    Let F with rightwards harpoon with barb upwards on top subscript 12 be the force experienced by m2 due to m1.
    Therefore, 
                         F with rightwards harpoon with barb upwards on top subscript 12 equals negative space F with rightwards harpoon with barb upwards on top subscript 21
     
    Now the equation of motion of m1 is,  
    space space space space space space space space straight m subscript 1 space fraction numerator straight d squared straight r with rightwards harpoon with barb upwards on top subscript 1 over denominator dt squared end fraction equals space straight F with rightwards harpoon with barb upwards on top subscript 21 space equals space minus space straight F with rightwards harpoon with barb upwards on top subscript 12 space rightwards double arrow space space space space space space space fraction numerator straight d squared straight r with rightwards harpoon with barb upwards on top subscript 1 over denominator dt squared end fraction space equals fraction numerator negative space straight F with rightwards harpoon with barb upwards on top subscript 12 space over denominator straight m subscript 1 end fraction space space space space space space space space space space space space space... left parenthesis 1 right parenthesis 

    Similarly, the equation of motion of m2 is given by
              fraction numerator straight d squared straight r with rightwards harpoon with barb upwards on top subscript 1 over denominator dt squared end fraction space equals fraction numerator negative space straight F with rightwards harpoon with barb upwards on top subscript 12 space over denominator straight m subscript 2 end fraction space space
    Subtracting (1) from (2), we get 
    space space space space space space space space space fraction numerator straight d squared left parenthesis straight r with rightwards harpoon with barb upwards on top subscript 2 space minus space straight r with rightwards harpoon with barb upwards on top subscript 1 right parenthesis over denominator dt squared end fraction space equals space open square brackets 1 over straight m subscript 1 space plus space 1 over straight m subscript 2 close square brackets space straight F with rightwards harpoon with barb upwards on top subscript 12 space rightwards double arrow space fraction numerator straight d squared straight r with rightwards harpoon with barb upwards on top over denominator dt squared end fraction space space equals space 1 over straight mu space straight F with rightwards harpoon with barb upwards on top subscript 12 rightwards double arrow space straight F with rightwards harpoon with barb upwards on top subscript 12 space equals space straight mu space fraction numerator straight d squared straight r with rightwards harpoon with barb upwards on top over denominator dt squared end fraction space space space space space space space space space space space space space space space space space space space space... space left parenthesis 3 right parenthesis space
    where, <pre>uncaught exception: <b>mkdir(): Permission denied (errno: 2) in /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/util/sys/Store.class.php at line #56mkdir(): Permission denied</b><br /><br />in file: /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/util/sys/Store.class.php line 56<br />#0 [internal function]: _hx_error_handler(2, 'mkdir(): Permis...', '/home/config_ad...', 56, Array) #1 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/util/sys/Store.class.php(56): mkdir('/home/config_ad...', 493) #2 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/plugin/impl/FolderTreeStorageAndCache.class.php(110): com_wiris_util_sys_Store->mkdirs() #3 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/plugin/impl/RenderImpl.class.php(231): com_wiris_plugin_impl_FolderTreeStorageAndCache->codeDigest('mml=<math xmlns...') #4 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/plugin/impl/TextServiceImpl.class.php(59): com_wiris_plugin_impl_RenderImpl->computeDigest(NULL, Array) #5 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/service.php(19): com_wiris_plugin_impl_TextServiceImpl->service('mathml2accessib...', Array) #6 {main}</pre> is called reduced mass of
    the system.
    Since equation (3) signifies the equation of motion of single particles, thus the two body problem is reduced to single-body problem.

    Question 354
    CBSEENPH11017456
    Wired Faculty App

    Discuss the motion of moon around the earth and derive its equation of motion.

    Solution

    Earth and the moon form a two-body system.
    Let M and m be the masses of the earth and moon respectively.
    Let straight r with rightwards harpoon with barb upwards on top be the position vector of moon with respect to earth.
    Here, the force on the moon is gravitational force of earth and given by, 
                    rightwards arrow for F of equals negative fraction numerator G M m over denominator r squared end fraction r with hat on top 
    The equation of motion of moon is,
    space F with rightwards harpoon with barb upwards on top space equals space straight mu space fraction numerator straight d squared straight r over denominator dt squared end fraction
    where, μ is the reduced mass of the moon.
    Reduced mass of the earth is given by,
                  straight mu equals fraction numerator Mm over denominator straight M plus straight m end fraction
    Substituting the values of straight mu space and space F with rightwards harpoon with barb upwards on top in the equation of motion, we get 
    negative fraction numerator G M m over denominator straight r squared end fraction space stack straight r space with hat on top space equals space fraction numerator M m over denominator straight M plus straight m end fraction fraction numerator d squared r over denominator d t squared end fraction rightwards double arrow space space space space fraction numerator d squared r over denominator d t squared end fraction space space equals space minus space fraction numerator straight G left parenthesis straight M plus straight m right parenthesis over denominator straight r squared end fraction space straight r with hat on top
    This is the required equation of motion. 

    Question 355
    CBSEENPH11017457
    Wired Faculty App

    Four uniform and identical meter sticks are stacked on a table with 10cm, 60cm, 30cm, and 50cm segments over the edge as shown in figure. Would the meter sticks remains on the table or topple down?

    Solution
    The metre sticks which are given are identical.
    The centre of each stick is located at its mid-point.
    Consider, edge O of the table as the reference point.
    The distances of centre of gravity of sticks S1, S2, S3, and S4 are,
    x1, =40cm, x2 = -10cm, x3 =20cm and x4 =0 cm respectively.
    The condition for the meter sticks for not to topple from the table is that, the centre of gravity (CG) of the system of sticks should lie on the right from the edge O.
    Let m be the mass of each stick.
    The CG of system from the O is,
    straight X space equals fraction numerator mx subscript 1 plus mx subscript 2 plus mx subscript 3 plus mx subscript 4 over denominator straight m plus straight m plus straight m plus straight m end fraction space space space space space equals space fraction numerator straight x subscript 1 plus straight x subscript 2 plus straight x subscript 3 plus straight x subscript 4 over denominator 4 end fraction space space space space space equals space fraction numerator 40 minus 10 plus 20 plus 0 over denominator 4 end fraction space space space space equals 12.5 cm 

    Since, the CG of the system lies on the table, the system of sticks will not topple over the edge.

    Question 356
    CBSEENPH11017459
    Wired Faculty App

    Uniform and identical meter sticks are stacked, so that 20cm of each stick extends beyond the meter stick beneath as shown in figure. How many maximum numbers of sticks can be placed one over the other without making them to fall over?

    Solution

    The given metre sticks are identical in nature.
    The centre of gravity of each stick is located at its mid-point as shown in the fig.

    Let, n be the maximum numbers of sticks that can be placed one over the other without making them to topple. 
    Let the edge O of the lowest stick S1, which is on the floor be the reference point.
    The distances of centre of gravity of sticks S2, S3, S4......Sn are x2 = -30cm, x3 =-10cm, x4 = 10cm .........x = -30 + 20(n-2) = (20n -70)cm respectively.

    Let m be the mass of each stick.
    The position of centre of gravity of the sticks S2, S3, S4......Sn from the O is,
    straight X space equals fraction numerator mx subscript 2 plus mx subscript 3 plus mx subscript 4..... mx subscript straight n over denominator straight m plus straight m plus straight m plus....... plus straight m end fraction space space space space equals fraction numerator straight x subscript 2 plus straight x subscript 3 plus straight x subscript 4....... straight x subscript straight n over denominator straight n minus 1 end fraction space space space space equals fraction numerator negative 30 minus 10 plus 10..... plus left parenthesis 20 straight n minus 70 right parenthesis over denominator straight n minus 1 end fraction space space space space equals space left parenthesis 10 straight n minus 50 right parenthesis space cm

    The condition for sticks not topple is that the position of CG of sticks S2, S3, S4......Sn should lie on left from edge of the S1.

    i.e.,    X ≤ 0

    rightwards double arrow     10n - 50 ≤ 0

    rightwards double arrow      n ≤ 5

    rightwards double arrow      n ≤ 5

    Hence, at the maximum, five sticks can be stacked one over other. 

    Question 357
    CBSEENPH11017462
    Wired Faculty App

    Uniform and identical sticks each of length 25cm are stacked, so that 8cm of each stick extends beyond the stick beneath as shown in Figure.


    How many maximum numbers of sticks can be placed one over the other without making them to fall over?

    Solution
    The given meter sticks are identical to each other.
    The centre of gravity of each stick is located at its mid point.
    The system is as shown in the figure below. 

     

    Let 'n' be the maximum numbers of sticks that can be placed one over the other without making them topple. 
    Let edge O of the lowest stick S1, on the floor, is the reference point.
    The distances of centre of gravity of sticks S2, S3, S4......Sn are x2 = -4.5cm, x3 =3.5cm, x4= 11.5cm, ..., xn= -4.5+(8n.2) = (8n -20.5)cm respectively.
    Let m be the mass of each stick.
    The position of CG of sticks S2 S3, S4......Sn from the O is,

    straight X space equals fraction numerator m x subscript 2 plus m x subscript 3 plus m x subscript 4....... m x subscript straight n end subscript over denominator straight m plus straight m plus straight m plus...... plus straight m end fraction space space space space equals space fraction numerator straight x subscript 2 plus straight x subscript 3 plus straight x subscript 4..... straight x subscript straight n over denominator straight n minus 1 end fraction space space space space equals fraction numerator negative 4.5 plus 3.5 plus 11.5..... plus left parenthesis 8 straight n minus 20.5 right parenthesis over denominator straight n minus 1 end fraction space space space space equals left parenthesis 4 straight n minus 12.5 right parenthesis space c m

    The position of CG of sticks S2, S3, S4......Sn should lie on left from edge of the S1 is the condition for the sticks to not topple. 

    i.e.,         X ≤ 0

    rightwards double arrow       4n -12 . 5 ≤ 0

    rightwards double arrow                    n ≤ 3.125

    rightwards double arrow                     n = 3

    Hence on a maximum, three sticks can be stacked one over other. 

    Question 358
    CBSEENPH11017465
    Wired Faculty App

    A man of mass m stands on a rope ladder, which is tied to the free balloon of mass M as shown in figure. The balloon is initially at rest. The man starts to climb up the ladder with speed v relative to ladder. Find the velocity of balloon.

    Solution

    Let v with rightwards harpoon with barb upwards on top space and space u with rightwards harpoon with barb upwards on top be the velocities of the ballon and man w.r.t. ground.
    Therefore velocity of man w.r.t. ballon is, 
    space space space space space space v with rightwards harpoon with barb upwards on top space equals space u with rightwards harpoon with barb upwards on top space minus space V with rightwards arrow on top space rightwards double arrow space v with rightwards harpoon with barb upwards on top space equals space u with rightwards harpoon with barb upwards on top space plus space v with rightwards harpoon with barb upwards on top
    Initially, the whole system is at rest.
    Therefore the center of mass C of system is at rest.
    When the man starts climbing up the ladder, he exerts force on the ladder in downward direction and he gets an equal and opposite reaction in upward direction.
    Since, there is no external force on the system, therefore when the man starts climbing up, the center of mass of system remains at rest at C.
    straight v with rightwards harpoon with barb upwards on top subscript cm space equals space 0 space space space space space space... space left parenthesis 1 right parenthesis thin space Also comma space straight v with rightwards harpoon with barb upwards on top subscript cm space equals space fraction numerator straight m space straight u with rightwards harpoon with barb upwards on top space plus space straight M straight v with rightwards harpoon with barb upwards on top over denominator straight m space plus space straight M end fraction space space space space space space space space... space left parenthesis 2 right parenthesis
    From equation (1) and (2), we have 
    space space space space space space space space space space space space space straight m space straight u with rightwards harpoon with barb upwards on top plus straight M space straight V with rightwards harpoon with barb upwards on top equals 0 rightwards double arrow space space space straight m left parenthesis u with rightwards harpoon with barb upwards on top space plus space v with rightwards harpoon with barb upwards on top right parenthesis plus straight M space straight v with rightwards arrow on top equals 0 rightwards double arrow space space space space space space space space space space space space v with rightwards harpoon with barb upwards on top space equals negative fraction numerator straight m space straight v with rightwards harpoon with barb upwards on top over denominator straight m plus straight M end fraction
    Negative velocity indicates that the ladder will move down.

    Question 359
    CBSEENPH11017468
    Wired Faculty App

    A circular disc of radius R is removed from a bigger disc of radius 2R from one edge of the disc. Find the position of center of mass of residue disc.
    Solution
    The shape of the residue disc, after removing a circular disc from a bigger disc is as shown in the figure. 
                          
    Let <pre>uncaught exception: <b>mkdir(): Permission denied (errno: 2) in /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/util/sys/Store.class.php at line #56mkdir(): Permission denied</b><br /><br />in file: /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/util/sys/Store.class.php line 56<br />#0 [internal function]: _hx_error_handler(2, 'mkdir(): Permis...', '/home/config_ad...', 56, Array) #1 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/util/sys/Store.class.php(56): mkdir('/home/config_ad...', 493) #2 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/plugin/impl/FolderTreeStorageAndCache.class.php(110): com_wiris_util_sys_Store->mkdirs() #3 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/plugin/impl/RenderImpl.class.php(231): com_wiris_plugin_impl_FolderTreeStorageAndCache->codeDigest('mml=<math xmlns...') #4 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/plugin/impl/TextServiceImpl.class.php(59): com_wiris_plugin_impl_RenderImpl->computeDigest(NULL, Array) #5 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/service.php(19): com_wiris_plugin_impl_TextServiceImpl->service('mathml2accessib...', Array) #6 {main}</pre> be the surface mass density of the disc. 
    The mass of full disc is, 
                 <pre>uncaught exception: <b>mkdir(): Permission denied (errno: 2) in /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/util/sys/Store.class.php at line #56mkdir(): Permission denied</b><br /><br />in file: /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/util/sys/Store.class.php line 56<br />#0 [internal function]: _hx_error_handler(2, 'mkdir(): Permis...', '/home/config_ad...', 56, Array) #1 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/util/sys/Store.class.php(56): mkdir('/home/config_ad...', 493) #2 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/plugin/impl/FolderTreeStorageAndCache.class.php(110): com_wiris_util_sys_Store->mkdirs() #3 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/plugin/impl/RenderImpl.class.php(231): com_wiris_plugin_impl_FolderTreeStorageAndCache->codeDigest('mml=<math xmlns...') #4 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/plugin/impl/TextServiceImpl.class.php(59): com_wiris_plugin_impl_RenderImpl->computeDigest(NULL, Array) #5 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/service.php(19): com_wiris_plugin_impl_TextServiceImpl->service('mathml2accessib...', Array) #6 {main}</pre> 
    The mass of cut out section is,  
                straight m subscript 1 equals πR squared straight sigma 
    The mass of the residue section is,   
                straight m subscript 2 equals straight M minus straight m subscript 1 equals 3 πR squared straight sigma
    The bigger disc consists of the cut out disc of radius R and residue portion.
    The centre of mass of bigger disc lies at its center C.
    The centre of mass of removed disc lies at C1, at a distance R from C.
    Let the centre of mass of residue disc lies at C2 As the moment of masses about CM is zero, therefore moment of cutout disc and residue disc about C is zero.
    i.e.,     straight m subscript 1 space stack CC subscript 1 with rightwards harpoon with barb upwards on top plus straight m subscript 2 space stack CC subscript 2 with rightwards harpoon with barb upwards on top space equals space 0
    rightwards double arrow        stack C C subscript 2 with rightwards harpoon with barb upwards on top space equals space straight m subscript 1 over straight m subscript 2 stack C C subscript 1 with rightwards harpoon with barb upwards on top equals negative 1 third stack C C subscript 1 with rightwards harpoon with barb upwards on top 
    The centre of mass of residue part lies at a distance 1 third straight R on the line joining CC1 on the other side of C as that of C1. 
    Question 360
    CBSEENPH11017469
    Wired Faculty App

    Is moment of inertia a scalar or a vector?

    Solution
    Moment of inertia is neither scalar nor vector, but it is a tensor.
    Question 361
    CBSEENPH11017470
    Wired Faculty App

    What is the moment of inertia of a point mass about an axis at a distance r?

    Solution
     Moment of Inertia of a point mass about an axis at a distance r = mr2.
    Question 362
    CBSEENPH11017471
    Wired Faculty App

    Is moment of inertia constant for a given body?

    Solution
    The value of I depends on the follows factors: 
    i) position of the axis of rotation,
    ii) orientation of the axis of rotation,
    iii) shape of the body,
    iv) size of the body,
    v) distribution of mass of the body about the axis of rotation.  
    Therefore, the moment of inertia of a given body is different about a different axis.
    Question 363
    CBSEENPH11017472
    Wired Faculty App

    Which quantity, in rotational motion, plays the same role as mass in linear motion?

    Solution
    Moment of inertia (I) plays the same role in rotational motion as mass (m) plays in linear motion.
    Question 364
    CBSEENPH11017473
    Wired Faculty App

    A disc of metal is melted and recast in the form of a solid sphere. What will happen to the moment of inertia about vertical axis passing through center?

    Solution
    The moment of inertia will decrease about it's vertical axis passing through the centre. 
    Moment of Inertia of a solid sphere of mass M and radius R = 2 over 5 MR squared 
    Question 365
    CBSEENPH11017474
    Wired Faculty App

    There are two cylinders of same mass and same radius. One cylinder is solid and other is hollow. Which of them has more moment of inertia about its axis?

    Solution
    Hollow cylinder has more moment of inertia than the solid cylinder. This is because, for a hollow cylinder, the entire mass is at its periphery. That is, at a maximum distance from the centre. Hence, moment of inertia is less for a solid cylinder. 
    Question 366
    CBSEENPH11017475
    Wired Faculty App

    What is the moment of inertia of ring of mass m and radius r about its axis?

    Solution
    Moment of inertia of ring about its axis = mr2.
    Question 367
    CBSEENPH11017476
    Wired Faculty App

    What is the moment of inertia of ring of mass m and radius r about its diameter?
    Solution
    Consider two mutually perpendicular diameters AB and CD.
    Now, using the theorem of perpendicular axes, we have 
              IAB + ICD = MR2 
    That is, 
                Id + Id = MR2
    rightwards double arrow               Id = 1 half space M R squared
    The moment of inertia of ring about its diameter is 1/2 MR2. 
    Question 368
    CBSEENPH11017477
    Wired Faculty App

    What is the moment of inertia of ring of mass m and radius r about its tangent in the plane of ring?

    Solution
     
    In the figure given, the tangent to the circular ring in the plane of the ring is shown. 
    Let, the tangent be represented by XY and diameter be represented by AB. 
    Using the theorem of parallel axes, we have 
    IXY = IAB + M (R)2 
          = 1 half space MR squared space plus space MR squared space equals space 3 over 2 space MR squared 
    i.e., IXY = 3 over 2 space MR squared
    Therefore, the moment of inertia of ring about tangent in the plane of the ring is 3/2 MR2.
    Question 369
    CBSEENPH11017478
    Wired Faculty App

    What is the moment of inertia of disc of mass m and radius r about its axis?

    Solution
                   
    Consider, M is the mass of a uniform circular disc of radius R with centre O. 
    Let, O be the centre of the ring. And YY' is the length of the axis. 
    So the axis is represented by YOY'. 
    Now, 
    Surface area of the disc = straight pi space straight R squared
    Mass space per space unit space are space of space the space disc space equals space fraction numerator straight M over denominator straight pi space straight R squared end fraction space  
    Consider a small element of the disc, which would be a circular strip of radius x and width dx, say. 
    Length of this element = 2straight pix 
    Surface are of this element = 2 space straight pi space straight x space dx
    Therefore, 
    Mass space of space this space element space equals space fraction numerator straight M over denominator straight pi space straight R squared end fraction space left parenthesis 2 πx space dx right parenthesis space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space equals space fraction numerator 2 straight M space straight x space dx over denominator straight R squared end fraction space Moment space of space inertia space of space this space element space of space the space disc space about space YOY apostrophe space equals space mass space straight x space left parenthesis distance right parenthesis squared space space space space space space space space equals space fraction numerator 2 straight M space straight x space dx over denominator straight R squared end fraction space straight x space left parenthesis straight x right parenthesis squared space equals space fraction numerator 2 straight M space straight x cubed space dx over denominator straight R squared end fraction The space small space element space may space lie space anywhere space from space the space centre space of space disc space left parenthesis straight x space equals space 0 right parenthesis thin space to space left parenthesis straight x space equals space straight R right parenthesis. Therefore comma space straight I space equals space integral subscript straight x space equals space 0 end subscript superscript straight x space equals space straight R end superscript space fraction numerator 2 straight M space straight x cubed space dx over denominator straight R squared end fraction space equals space fraction numerator 2 straight M over denominator straight R squared end fraction space open square brackets straight x to the power of 4 over 4 close square brackets subscript straight x space equals space 0 end subscript superscript straight x space equals space straight R end superscript space straight I space equals space 2 over 4 straight M over straight R squared space left parenthesis space straight R to the power of 4 space minus space 0 right parenthesis space straight I thin space equals space 1 half space MR squared
    Therefore, the moment of inertia of disc about its axis is 1/2 mr2 .
    Question 370
    CBSEENPH11017479
    Wired Faculty App

    What is the moment of inertia of disc of mass m and radius r about its diameter?

    Solution
    Moment of inertia of a uniform circular disc about any diameter of the disc is, 
    Id = 1 half space straight I space equals space 1 half open square brackets 1 half MR squared close square brackets space equals space 1 fourth space MR squared space 
    Question 371
    CBSEENPH11017480
    Wired Faculty App

    What is the moment of inertia of disc of mass m and radius r about its tangent in the plane of disc?

    Solution
    The moment of inertia of disc about tangent in the plane of disc = 5 over 4 MR squared .
    Question 372
    CBSEENPH11017481
    Wired Faculty App

    What is the moment of inertia of solid sphere of mass m and radius r about its diameter?

    Solution
    The moment of inertia of solid sphere about its diameter = 2/5 mr2 .
    Question 373
    CBSEENPH11017482
    Wired Faculty App

    What is the moment of inertia of hollow sphere of mass M and radius R about its diameter?

    Solution
    The moment of inertia of hollow sphere about its diameter = 2/3 MR2.
    Question 374
    CBSEENPH11017483
    Wired Faculty App

    What is the moment of inertia of solid sphere of mass m and radius r about its tangent?

    Solution
     The moment of inertia of solid sphere about its diameter = 7/5 MR2 
    Question 375
    CBSEENPH11017484
    Wired Faculty App

    What is the moment of inertia of hollow sphere of mass m and radius r about its tangent?

    Solution
    Moment of inertia of hollow sphere about its diameter = 5/3 mr2
    Question 376
    CBSEENPH11017485
    Wired Faculty App

    What is the moment of inertia of rod of mass m and length / about an axis passing through mid point and perpendicular to its length?

    Solution
    Moment of Inertia of a rod of mass 'm' and length 'l' about an axis passing through the mid-point and perpendicular it's length = Error converting from MathML to accessible text..
    Question 377
    CBSEENPH11017486
    Wired Faculty App

    About which axis the moment of inertia of ring is minimum?

    Solution
    The moment of inertia of ring about diameter is minimum. This is because the entire mass of the ring is concentrated at it's periphery. 
    Question 378
    CBSEENPH11017487
    Wired Faculty App

    About which axis the moment of inertia of cylinder is minimum?

    Solution
    The moment of inertia of cylinder is minimum about its axis of symmetry. Because the mass is minimum at the symmetry. 
    Question 379
    CBSEENPH11017488
    Wired Faculty App

    Does the moment of inertia depend on the state of motion of rotating body?

    Solution
    No, the moment of inertia does not depend on the state of body. It is same about a given axis whether the body is at rest or rotating. 
    Question 380
    CBSEENPH11017489
    Wired Faculty App

    A cannon ball and a marble ball roll from rest down an incline. Which goes to the bottom first? 

    Solution

    Both the balls will reach at the same time. Because, the acceleration due to gravity is independent of the mass. 

    Question 381
    CBSEENPH11017490
    Wired Faculty App

    Define radius of gyration.

    Solution
    The radius of gyration of a body about a given axis is the perpendicular distance of a point P from the axis, where if whole mass of the body were concentrated, as the body shall have the same moment of inertia as it has with the actual distribution of mass. 
    Radius of gyration is represented by K. 
    Question 382
    CBSEENPH11017491
    Wired Faculty App

    Is the radius of gyration a constant quantity?

    Solution
    No, radius of gyration depends on the distance of axis of rotation from the body.
    Question 383
    CBSEENPH11017492
    Wired Faculty App

    A body consists of 'n' identical particles. What is the radius of gyration of the body about a given axis?

    Solution
    The radius of gyration about a given axis is the root mean square distance of constituent particles from the axis. 
    Question 384
    CBSEENPH11017494
    Wired Faculty App

    A disc and a ring have same mass and same radius. Which of them has greater value of radius of gyration about their axis?

    Solution
    The ring has greater value of radius of gyration than disc. The mass of the body with respect to the axis of rotation for a ring is greater. 
    Question 385
    CBSEENPH11017495
    Wired Faculty App

    What is the ratio of radius of gyration of solid sphere about diameter to its radius?

    Solution

    Relation between moment of inertia and radius of gyration, I = MK2
    And, for a solid sphere, moment of inertia about the diameter = 2 over 5 space MR squared 
    Therefore, 
    Radius of gyration of a solid sphere about the diameter to it's radius =square root of 2 divided by 5 end root R

    Question 386
    CBSEENPH11017496
    Wired Faculty App

    Give an example in which the radius of gyration is equal to half of the radius of body.

    Solution
    The radius of gyration of disc about diameter is half the radius of disc.
    Question 387
    CBSEENPH11017497
    Wired Faculty App

    Why does a flywheel has most of its mass concentrated at the rim?

    Solution
    The Concentration of the mass at the rim increases moment of Inertia (I) of the flywheel. Is such a wheel gains or loses some K.E of rotation, 1 half straight I space straight omega squared, it brings about a relatively smaller change in its angular speed straight omega. Hence, such a flywheel helps in maintaining uniform rotation.  
    Question 388
    CBSEENPH11017499
    Wired Faculty App

    Why is it more difficult to revolve the stone by tying it to a longer string than that tying it to a shorter string?

    Solution
    It is difficult to revolve the stone by tying it to a longer string than tying it to a shorter string because the moment of inertia of stone tied with longer string is more than that tied with smaller string.
    Question 389
    CBSEENPH11017500
    Wired Faculty App

    How does moment of inertia differ from inertia?

    Solution

    Inertia is the ability due to which a body itself does not change the state of rest or uniform motion in a straight line in the absence of external force.
    Moment of inertia is the ability due to which the body itself does not change the state of rest or uniform circular motion in the absence of external torque. 

    Mass is the measure of inertia whereas moment of inertia depends upon mass and radius of gyration.

    Question 390
    CBSEENPH11017502
    Wired Faculty App

    Why moment of inertia of a body is always referred to about an axis?

    Solution
    Moment of Inertia of a body is always referred to about an axis because by changing the axis of rotation, the distance of constituent particles from the axis of rotation also changes. Therefore moment of inertia of body about different axis is different. That is, reference axis is required in order to find the moment of inertia. 
    Question 391
    CBSEENPH11017505
    Wired Faculty App

    Define moment of inertia and radius of gyration. If a body consists of n particles of equal masses then show that radius of gyration is equal to rms of the distance of constituent particles from axis of rotation.

    Solution

    Moment of inertia is the ability of body by virtue of which the body remains in the state of rest or uniform circular motion unless external torque is applied on the body.

    Consider a rigid body consisting of n particles of masses m1 m2, m3 .......... mn situated at distances r1, r2, r3 ............. rn respectively from the axis of rotation AB as shown in figure.
                            
    The moment of inertia of rigid body about AB axis is,
    straight I equals space straight m subscript 1 straight r subscript 1 squared plus straight m subscript 2 straight r subscript 2 squared plus...... plus straight m subscript straight n straight r subscript straight n squared 
     equals sum from straight i equals 1 to straight n of straight m subscript 1 straight r subscript 1 squared space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space... left parenthesis 1 right parenthesis

    Radius of gyration of a body about an axis is the distance, at which whole of mass of the body is supposed to be concentrated, so that it would have the same moment of inertia as that of body.
    It is denoted by K.
    If M is the mass of body, then

    I =MK2                               ...(2) 

    From (1) and (2), we get 
    space space space space space space MK squared equals straight m subscript 1 straight r subscript 1 squared plus straight m subscript 2 straight r subscript 2 squared plus......... plus straight m subscript straight n straight r subscript straight n squared space rightwards double arrow space space space space straight K space equals space square root of fraction numerator straight m subscript 1 straight r subscript 1 squared plus straight m subscript 2 straight r subscript 2 squared plus......... plus straight m subscript straight n straight r subscript straight n squared over denominator straight M end fraction end root 
    If all the particles are identical having each mass m, then
             M = nm
    rightwards double arrow space space space space mnK squared equals straight m left parenthesis straight r subscript 1 squared plus straight r subscript 2 squared plus......... plus straight r subscript straight n squared right parenthesis space rightwards double arrow space space space space space space space space space straight K space equals space square root of fraction numerator straight r subscript 1 squared plus straight r subscript 2 squared plus...... plus straight r subscript straight n squared over denominator straight n end fraction end root  
    Hence the result


    Question 392
    CBSEENPH11017507
    Wired Faculty App

    State and prove the theorem of perpendicular axes.

    Solution
    According to the theorem of perpendicular axis, the moment of inertia of a plane lamina, about any axis perpendicular to the plane of the lamina is equal to the sum of the moment of inertia of lamina about two mutually perpendicular axis lying in the plane of lamina and intersecting each other at a point where the perpendicular axis passes through the body. 

                     
    Proof:
    Let X and Y be two perpendicular axes lying in plane lamina intersecting at O and Z be perpendicular to the plane of lamina passing through O.
    Let a point mass m1 be located at (xi, yj) and ri be the distance of mi from Z axis.
    T h e r e f o r e comma space space space space space space space divided by subscript straight z equals sum straight m subscript 1 straight r subscript 1 squared space space space space space space space space space space space equals sum straight m subscript 1 left parenthesis straight x subscript 1 squared plus straight y subscript 1 squared right parenthesis space space space space space space space space space space space equals space sum straight m subscript 1 straight x subscript 1 squared plus sum straight m subscript 1 straight y subscript 1 squared space space space space space space space space space space space equals straight I subscript straight y plus straight I subscript straight x space S o comma space space straight I subscript straight z space equals space straight I subscript straight y space plus space straight I subscript straight x
    Question 393
    CBSEENPH11017509
    Wired Faculty App

    State and prove the theorem of parallel axis.

        

    Solution
    Statement:
    The theorem of parallel axis states that the moment of inertia of a body about an axis parallel to axis passing through centre of mass is equal to the sum of the moment of inertia of body about an axis passing through centre of mass and product of mass and square of distance between the two axes. 
     
    Proof:
    Let, IC be the moment of inertia of about an axis passing through the centre of mass i.e. about AB and I be the moment of inertia about axis A' B' at a distance h. 

    Then,
    I =Ic + Mh2
    Consider, a particle of mass m at a distance r from the centre of gravity of the body.  


    Therefore comma space Distance space from space straight A apostrophe space straight B apostrophe space equals space straight r plus straight h space therefore space straight I space space equals space sum straight m left parenthesis straight r plus straight h right parenthesis squared space space space space space equals space sum straight m left parenthesis straight r squared plus straight h squared plus 2 rh right parenthesis space space space space space equals sum mr squared plus sum mh squared plus sum 2 mrh space space space space space equals space straight I subscript straight c plus straight h squared sum straight m plus 2 straight h sum mr space space space space space equals space straight I subscript straight c plus Mh squared plus 0 space space space space space space space space space space space space space space space space space space space space space space open square brackets because space sum mr equals 0 close square brackets space space space straight I space equals space straight I subscript straight v plus Mh squared 
    Hene the result.

    Question 394
    CBSEENPH11017510
    Wired Faculty App

    Derive the expression for moment of inertia of the thin rod about an axis through its center and perpendicular to its length.
    Solution
    Consider a rod of length 'L' and mass 'M'.
    Let 'λ' be the linear mass density. 

    <pre>uncaught exception: <b>mkdir(): Permission denied (errno: 2) in /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/util/sys/Store.class.php at line #56mkdir(): Permission denied</b><br /><br />in file: /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/util/sys/Store.class.php line 56<br />#0 [internal function]: _hx_error_handler(2, 'mkdir(): Permis...', '/home/config_ad...', 56, Array) #1 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/util/sys/Store.class.php(56): mkdir('/home/config_ad...', 493) #2 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/plugin/impl/FolderTreeStorageAndCache.class.php(110): com_wiris_util_sys_Store->mkdirs() #3 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/plugin/impl/RenderImpl.class.php(231): com_wiris_plugin_impl_FolderTreeStorageAndCache->codeDigest('mml=<math xmlns...') #4 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/plugin/impl/TextServiceImpl.class.php(59): com_wiris_plugin_impl_RenderImpl->computeDigest(NULL, Array) #5 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/service.php(19): com_wiris_plugin_impl_TextServiceImpl->service('mathml2accessib...', Array) #6 {main}</pre>
    Let a small line element dx be at a distance x from O. 
    Moment of inertia of this small line element of mass λdx about O is, 
     
    To find the total moment of inertia of rod, integrate from x = -L/2 to x = L/2.
    I = Error converting from MathML to accessible text. = integral subscript negative straight L divided by 2 end subscript superscript straight L divided by 2 end superscript lambda space x squared space d x space  
      equals space straight lambda space open vertical bar straight x cubed over 3 close vertical bar subscript negative straight L divided by 2 end subscript superscript straight L divided by 2 end superscript space space equals space straight lambda over 3 open square brackets open parentheses straight L over 2 close parentheses cubed minus space open parentheses fraction numerator negative straight L over denominator 2 end fraction close parentheses cubed close square brackets space equals space fraction numerator straight M over denominator 3 straight L end fraction open square brackets straight L cubed over 8 space plus space straight L cubed over 8 close square brackets equals space 1 over 12 space ML squared  
    Question 395
    CBSEENPH11017511
    Wired Faculty App

    Derive the expression for moment of inertia of thin rod about an axis through its end and perpendicular to its length.

    Solution
    Consider a rod of length L and mass M.
    Let λ be the linear mass density.
    therefore space space space space space straight lambda space equals space straight M over straight L

    Let a small line element dx be at a distance x from O.
    Moment of inertia of this small line element of mass λdx about O is,

    dl = λdx x2 

    To find the total moment of inertia of rod, integrate from x = 0 to x = L
    straight I space equals integral dI equals integral from 0 to straight I of λx squared dx space space equals straight lambda open vertical bar straight x cubed over 3 close vertical bar subscript 0 superscript straight I space space space equals fraction numerator straight M over denominator 3 straight L end fraction open square brackets straight L cubed minus 0 cubed close square brackets space space space space equals 1 third ML squared 
    Question 396
    CBSEENPH11017512
    Wired Faculty App

    Find the moment of inertia of ring of mass 'M' and radius R about an axis passing through the centre and perpendicular to its plane.

    Solution
    Let AB be the axis passing through the centre and perpendicular to the plane of the ring.
    Consider an arbitrary line element of length dx at point P on the ring.
    The moment of inertia of the line element dx of mass λdx about AB axis is, 
    dI = straight lambda space dx space straight R squared 
     
    To find the moment of inertia of the ring about AB axis, integrate over the entire ring 
    straight I space equals space integral space dI space equals contour integral space straight lambda space dx space straight R squared space space space space space space space space space space space space space space space space space equals space straight lambda space straight R squared space left parenthesis 2 πR right parenthesis space space space space space space space space space space space space space space space space space equals space straight lambda space left parenthesis 2 πR right parenthesis thin space straight R squared space space space space space space space space space space space space space space space space space equals space MR squared space straight i. straight e. comma space space space space space space straight I space equals space MR squared 
    Question 397
    CBSEENPH11017515
    Wired Faculty App

    If moment of inertia of ring about an axis passing through centre and perpendicular to the plane of ring is I then find the moment of inertia about:

    (i) diameter

    (ii) tangent parallel to diameter using theorem of parallel or perpendicular axis.
    Solution

    (i) Moment of inertia of ring about diameter:
    The moment of Inertia of ring about all the diameters is the same because the ring is symmetric w.r.t. 
    Let Id be the moment of inertia of ring about its diameter.
    According to the theorem of perpendicular axis,
    Id + Id = Ia = I 
    rightwards double arrow space space space space space space space space space space straight I subscript straight d space equals space straight I over 2 
     

    (ii) Moment of infertia of ring about tangent parallel to diameter:

    Let M be the mass and R be the radius of ring.
    Then, moment of inertia about axis passing through the centre and perpendicular to plane is I = MR2.
    Let It be the moment of inertia of ring about tangent parallel to diameter.
    Using the theorem of parallel axis,
    straight I subscript straight t equals straight I subscript straight d plus M R squared space space space equals space 1 half straight I plus straight I space space space equals 3 over 2 straight I

    Question 398
    CBSEENPH11017516
    Wired Faculty App

    Find the moment of inertia of disc of mass M and radius R about an axis passing through centre and perpendicular to its plane.

    Solution
    Let σ is the surface mass density of the disc.

    Therefore, 
                  straight M space equals space straight sigma space straight pi space straight R squared 

    Let AB be the axis passing through the center and perpendicular to the plane of disc. 
    Consider an arbitrary ring of radius x and thickness dx concentric with disc.
    The mass of the ring of radius x and thickness dx is,

                   dm = σ2πxdx

    The moment of inertia of this elementary ring about AB axis is,

                     dl = dmx2 = σ2πx3 dx

    To find the moment of inertia of the disc about AB axis, integrate the above equation from x - 0 to x = R. 
    Error converting from MathML to accessible text. 

    Question 399
    CBSEENPH11017517
    Wired Faculty App

    What is rigid body?

    Solution
    A body is said to be rigid if the relative position of constituent particles do not change on applying any external force on it. Rigid body has a definite size and shape.  The size and shape do not change during the motion of the body. 
    Question 400
    CBSEENPH11017518
    Wired Faculty App

    What is general motion of a rigid body?

    Solution
    When the body has both translational and rotational motions, a rigid body is said to be in general motion. 
    Question 401
    CBSEENPH11017519
    Wired Faculty App

    Define rotational motion.

    Solution
    If all the particles of a body move in a circle with their centers on the axis of rotation, the motion is sais to be a rotational motion. 
    Question 402
    CBSEENPH11017520
    Wired Faculty App

    A rigid body is rotating about an axis. Different particles are at different distances from the axis. Do different particles possess different angular velocities?

    Solution
    No, different particles possess same angular velocity and equal to angular velocity of the rigid body. 
    Question 403
    CBSEENPH11017521
    Wired Faculty App

    A rigid body is rotating about an axis. Different particles are at different distances from the axis. Do different particle possess same linear velocities like angular velocity?

    Solution
    No, different particles move with different velocities if they are at different distance from the axis. 
    Question 404
    CBSEENPH11017522
    Wired Faculty App

    What points on a cylinder rolling over a surface move rectilinearly?

    Solution
    A rectilinear motion is exhibited by all the points lying on the axis of the cylinder. 
    Question 405
    CBSEENPH11017523
    Wired Faculty App

    What is moment of momentum called?

    Solution
    Angular momentum.
    Question 406
    CBSEENPH11017524
    Wired Faculty App

    What does angular momentum measure?

    Solution
    The angular motion or rotational motion exhibited by the body is measured by angular momentum. 
    Question 407
    CBSEENPH11017525
    Wired Faculty App

     Is the summation of all the forces acting on a body zero be the sufficient condition for the body to be in rotational equilibrium?

    Solution
    No, the summation of all the forces acting on the body zero does not mean that the torque produced by these forces is also zero. 
    Rotational and angular motion can be produced even when the external forces acting on the body is zero. 
    For eg., When a fan is switched on, the centre of the fan remains unmoved, while the fan rotates with an angular acceleration. Since, the centre of mass of the fan remains at rest, the vector sum of all the external forces acting on the fan is zero. 
    Question 408
    CBSEENPH11017528
    Wired Faculty App

    Under what condition angular momentum of a moving body is zero?

    Solution

    Angular momentum is given by, 
               stack L thin space with rightwards harpoon with barb upwards on top space equals r with rightwards harpoon with barb upwards on top space x space p with rightwards harpoon with barb upwards on top space
    Angular momentum of moving body is zero if either r = 0 or straight r with rightwards harpoon with barb upwards on top space vertical line vertical line thin space straight p with rightwards harpoon with barb upwards on top. 

    Question 409
    CBSEENPH11017529
    Wired Faculty App

    A body is rotating in XY plane in anticlockwise direction. What is the direction of angular momentum?

    Solution
    The direction of angular momentum is along the positive direction of Z axis. 
    Question 410
    CBSEENPH11017530
    Wired Faculty App

    What is moment of force or torque?

    Solution
    Moment of force is the turning effect of a force about the axis of rotation. 
    Question 411
    CBSEENPH11017531
    Wired Faculty App

    Write the expression for torque in vector form.

    Solution

    Expression of torque is given by, 
    straight tau with rightwards harpoon with barb upwards on top space equals space r with rightwards harpoon with barb upwards space on top cross times space F with rightwards harpoon with barb upwards on top space

    Question 412
    CBSEENPH11017532
    Wired Faculty App

    How is torque related with angular momentum?

    Solution
    Torque is the rate of change of angular momentum.

    i.e.,   straight tau with rightwards harpoon with barb upwards on top space equals space fraction numerator d L with rightwards harpoon with barb upwards on top over denominator d t end fraction 
    Question 413
    CBSEENPH11017533
    Wired Faculty App

    What is the torque produced by radial component of force acting on the body?

    Solution
    The radial component of force produces zero torque.
    When force is applied at the free end of the road, straight theta space equals space 0 to the power of straight o.
    Therefore, 
    Torque = 0 
    Question 414
    CBSEENPH11017534
    Wired Faculty App

    A body is rotating. Is it necessarily that some torque is acting on the body?

    Solution
    No. If a body is rotating with constant angular velocity, then no torque is required. 
    Question 415
    CBSEENPH11017535
    Wired Faculty App

    Does the angular momentum of electron remain conserved when electron rotates around nucleus?

    Solution
    Yes, the angular momentum remains conserved because the torque exerted by electrostatic force on an electron is zero.
    Question 416
    CBSEENPH11017536
    Wired Faculty App

    Does every force produce torque?

    Solution
    No, the force that passes through the axis of rotation does not produce any torque.
    Along the axis, straight theta space equals space 0 to the power of straight o. 
    Therefore, torque is equal to zero. 
    Question 417
    CBSEENPH11017537
    Wired Faculty App

    Does the angular momentum of a planet revolving around a star remain constant over the entire orbit even if the orbit is highly elliptical?

    Solution
    Yes, the angular moment of a planet revolving around the star remains conserved. The angular momentum lost os gained by the sun and vice versa. Hence, angular momentum always remains conserved. 
    Question 418
    CBSEENPH11017538
    Wired Faculty App

    A particle of mass m is moving with constant velocity parallel to X-axis. Does it possess angular momentum w.r.t. origin?

    Solution
    Yes. The particle possesses angular momentum about the origin. This is because for any particle moving with a constant velocity, angular momentum will remain conserved. 
    Question 419
    CBSEENPH11017539
    Wired Faculty App

    When angular momentum of a system of particles is conserved?

    Solution
    The angular momentum of a system of particles is conserved when no torque is acting on it. 
    Question 420
    CBSEENPH11017540
    Wired Faculty App

    If torque on a body changes the momentum of body from 3 units to 9 units in 5 seconds, then what is the magnitude of torque on the body?

    Solution

    Initial torque = 3 units 
    Final torque = 5 units
    Time taken to change the momentum of the body = 5 seconds
    Therefore, the magnitude of torque on the body is, straight tau space equals fraction numerator 9 minus 3 over denominator 5 end fraction equals 1.2 space unit

    Question 421
    CBSEENPH11017541
    Wired Faculty App

    A man standing on the turn table suddenly raises his hands. What will happen?

    Solution

    When a person on a turntable will raise his hands, there is an increase in the moment of inertia. Hence, the angular velocity of the turntable will decrease.

    Question 422
    CBSEENPH11017542
    Wired Faculty App

    A body tied with string revolves uniformly in a circle. Does the angular momentum of body change from its initial value if the string is cut suddenly?

    Solution
    No, the length of the string will not affect the angular momentum of the body. Therefore, the angular momentum of body will remain the same.
    Question 423
    CBSEENPH11017543
    Wired Faculty App

    A body is in rotational motion. Is it necessary that it experience any torque?

    Solution
    No, a body rotating with uniform angular velocity does not experience any torque. 
    Question 424
    CBSEENPH11017544
    Wired Faculty App

    What do you mean by areal velocity?

    Solution
    The area swept by position vector of a particle in one second is known as the areal velocity. 
    Question 425
    CBSEENPH11017545
    Wired Faculty App

    A force F is applied on a body at a point whose position vector is rightwards arrow for straight r of. What is the turning effect due to radial component of force?
    Solution
    straight theta space equals space 0 to the power of straight o for force due to the radial component. Therefore, the turning effect of the radial component of force is zero.
    Question 426
    CBSEENPH11017546
    Wired Faculty App

    Name the law from which Kepler's second law can be deduced?

    Solution
    Kepler's second law can be deduced from the law of conservation of angular momentum.
    Question 427
    CBSEENPH11017547
    Wired Faculty App

    A person is standing on ground near the edge of a rotating table. If he suddenly sits near the edge of table, how angular velocity of table will be changed?

    Solution
    If a person who is standing near the edge of the rotating table suddenly sits at the edge of the table, the angular velocity of the table will decrease. Because moment of Inertia would increase now for the table with increase in mass. 
    Question 428
    CBSEENPH11017548
    Wired Faculty App

    A dancer spinning on ice folds his arm. What will happen?

    Solution
    When a dancer who is spinning folds his arms, the moment of inertia decreases. Therefore, the spinning rate of the dancer becomes fast.
    Question 429
    CBSEENPH11017549
    Wired Faculty App

    A physical quantity is represented as a product of moment of inertia and angular acceleration. What quantity is this?

    Solution
    The physical quantity is torque. Torque is the product of moment of inertia and angular acceleration. 
    Question 431
    CBSEENPH11017551
    Wired Faculty App

    A revolving body has angular momentum numerically equal to its moment of inertia. What is angular velocity of the body?

    Solution
    Relation between angular momentum, moment of inertia and angular velocity is given by, 
                          L = I straight omega
    Angular momentum is numerically equal to moment of inertia. 
    Therefore, 
    Angular velocity, straight omega = 1 rad/s.
    Question 432
    CBSEENPH11017552
    Wired Faculty App

    When moment of inertia of a body is numerically equal to twice the kinetic energy of the body?

    Solution
    Kinetic space Energy space of space the space body comma space straight K. straight E space equals space 1 half space straight I space straight omega squared space That space is comma space 2 space straight K. straight E space equals space straight I space straight omega squared 
    Therefore, when the body revolves with angular velocity 1 rad/s, moment of inertia of a body is numerically equal to twice the kinetic energy of the body.
    Question 433
    CBSEENPH11017553
    Wired Faculty App

    A body rotating about given axis with unit angular velocity has kinetic energy K. What is the moment of inertia of the rotating body about given axis?

    Solution

    Kinetic energy is given by, 
     straight K space equals space 1 half I thin space straight omega squared  
    Given, angular velocity, straight omega = 1 rad/sec
    So, moment of inertia of the rotating body is,
                               I = 2K

    Question 434
    CBSEENPH11017555
    Wired Faculty App

    A body of radius R and radius of gyration K is rolling without slipping. What fraction of the total kinetic energy is rotational kinetic energy?

    Solution

    Radius of the body = R
    Radius of gyration = K 
    Fraction of the total kinetic energy to rotational energy is, fraction numerator straight K squared over denominator straight K squared plus straight R squared end fraction

    Question 435
    CBSEENPH11017556
    Wired Faculty App

    A body of radius R and radius of gyration K is rolling without slipping. What fraction of the total kinetic energy is translational kinetic energy?

    Solution

    The fraction of the total kinetic energy to translational kinetic energy is, fraction numerator straight R squared over denominator straight K squared plus straight R squared end fraction

    Question 438
    CBSEENPH11017559
    Wired Faculty App

    A body is rolling down the inclined plane without slipping. How does the acceleration of rolling body depend on its radius?

    Solution
    The acceleration of the rolling body is independent of the radius and mass of the body.  
    Question 439
    CBSEENPH11017560
    Wired Faculty App

    Does the acceleration of a body rolling down the inclined plane depend on the shape of the rolling body?

    Solution
    Yes, the acceleration of body rolling down the inclined plane depends on the shape of the rolling body.
    Question 440
    CBSEENPH11017561
    Wired Faculty App

    A solid sphere and a hollow sphere both roll down from the same point from the same inclined plane without slipping. Which of the two spheres reach the ground earlier and with greater speed?

    Solution
    Rotational inertia for solid sphere = 2 over 5 space M R squared
    Rotational Inertia for hollow sphere = 2 over 3 space M R squared
    Kinetic energy is due to both rotational as well as translational motion. But, here the rotational motion of solid sphere is less than the hollow sphere. This implies the translational component of solid will be higher. Hence, the solid sphere will move down with a greater velocity. 
    Question 441
    CBSEENPH11017562
    Wired Faculty App

    Does the acceleration of cylinder rolling down the inclined plane depend on the mass of the body?

    Solution
    No, acceleration of the rolling body is independent of the mass of the body. 
    Question 442
    CBSEENPH11017563
    Wired Faculty App

    At what angular velocity, the angular momentum of body is numerically equal to its kinetic energy?

    Solution
    The angular momentum of a body is numerically equal to its kinetic energy when its angular velocity is 2 rad/s. 
    K.E of a body = 1 half space straight I space straight omega squared 
    Question 444
    CBSEENPH11017565
    Wired Faculty App

    What is the acceleration of hollow cylinder of mass m and radius r rolling down the inclined plane without slipping?

    Solution
    The acceleration of the hollow cylinder of mass m and radius r rolling down the inclined plane = 1/2 g sin θ.
    Question 445
    CBSEENPH11017567
    Wired Faculty App

    What should be the minimum coefficient of friction of rough inclined plane so that the hollow sphere rolls down without slipping?

    Solution
    For rolling without slipping, the minimum value of coefficient of friction between the cylinder and the inclined plane is given by, 

    straight mu space equals space straight F over straight R space equals space fraction numerator 1 space mg space sin space straight theta over denominator 3 space mg space cos space straight theta end fraction space equals space 1 third space tan space straight theta 
    Question 446
    CBSEENPH11017568
    Wired Faculty App

    What is rotational kinematics and rotational dynamics?

    Solution

    Rotational kinematics is the study of rotational motion without considering the cause of rotational motion.
    That is, without taking into consideration the torque acting on a body.

    Rotational dynamics is the study of rotational motion by considering the cause of rotational motion.

    Question 447
    CBSEENPH11017570
    Wired Faculty App

    Torque and work both are defined as force timTorque and work both are defined as force times the distance, then how do they differ?es the distance, then how do they differ?

    Solution

    Torque:
    Change in rotational motion is apparently produced by torque. And torque is eventually produced by two equal and opposite forces. 
    Torque is equal to the product of one force and the perpendicular distance between two forces.
    Work:
    When the force acting on a body displaces the body, work is said to be done. The body can be displaced by only single force.   
    Work is equal to the product of displacement and component of force in the direction of displacement.

    Torque is a vector quantity while work is a scalar.

    Question 448
    CBSEENPH11017571
    Wired Faculty App

    Comment on statement: in pure rotational motion, the instantaneous velocity of all the particles is same. Explain your answer.

    Solution
    The statement is wrong.
    Magnitude of the instantaneous velocity of an arbitrary particle = rω.
    Direction is along the tangent to the circular path followed by the particle.
    The particles are at different distances from the axis of rotation and have different magnitude of velocity and the particles which are at the same distance have different direction of velocity.
    Therefore, the instantaneous velocity of different particles is different either in magnitude or direction or both.
    Question 449
    CBSEENPH11017572
    Wired Faculty App

    The cap of pen or a tap can be easily opened with the help of two fingers than with one finger. Why?

    Solution
    A single force is applied with the help of one finger. And, two forces in the opposite direction is applied with the help of two fingers.
    Rotational motion produced by a single force is less than two equal and opposite forces. Therefore, it is easy to open the cap with the help of two fingers rather than one finger.
    Question 450
    CBSEENPH11017573
    Wired Faculty App

    Why it is more difficult to revolve a stone by tying it to a longer string than by tying it to shorter string?

    Solution
    The torque required to revolve the stone with a given angular acceleration is directly proportional to the  moment of inertia of the stone.
    Moment of inertia of stone about an axis is directly proportional to the square of the length of the string.
    Thus, more torque is required to revolve a stone tied with a longer string than tying it with a shorter string.  
    Question 451
    CBSEENPH11017577
    Wired Faculty App

    The moment of inertia of disc about an axis passing through centre and perpendicular to the plane of disc is1/2 MR2 , where M is mass and R is radius of disc. Find the moment of inertia about:

    (i) diameter

    (ii) tangent parallel to diameter.
    Solution
    (i) Moment of inertia of ring about diameter: 
        
     

    Disc is symmetric with respect to all of its diameters, therefore the moment of inertia of disc about all the diameters is same.
    Let Id be the moment of inertia of ring about its diameter.
    According to the theorem of perpendicular axis,
    straight I subscript straight d plus straight I subscript straight d equals straight I subscript straight a equals 1 half MR squared therefore space space straight I subscript straight d equals 1 fourth MR squared

    (ii) Moment of inertia of disc about tangent parallel to diameter:

    Let A' B' be the tangent parallel to diameter AB. By using the theorem of parallel axis,
    straight I subscript straight d equals straight I subscript straight d plus MR squared space space space space equals 1 fourth MR squared plus MR squared space space space space space equals 5 over 4 MR squared 

    Question 452
    CBSEENPH11017578
    Wired Faculty App

    The moment of inertia of disc about an axis passing through centre and perpendicular to the plane of disc is 1/2 MR2, where M is mass and R is radius of disc. Find the moment of inertia about an axis perpendicular to tangent. 
    Solution
    Let I be the moment of inertia about an axis perpendicular to tangent. 

                        
    The line perpendicular to tangent is the axis parallel to the axis passing through centre and perpendicular to the plane of the disc.
    Using theorem of parallel axis, the moment of inertia about an axis perpendicular to tangent is,
    straight I equals straight I subscript straight a plus MR squared space equals 1 half MR squared plus MR squared space equals 3 over 2 MR squared
    Question 455
    CBSEENPH11017588
    Wired Faculty App

    Calculate the radius of gyration of hollow sphere about its diameter and tangent.

    Solution

    Let M be the mass and R be the radius of hollow sphere.

    (i) Radius of gyration about diameter:

    Moment of inertia of hollow sphere about diameter is, 
                     straight I equals 2 over 3 MR squared
    If Kd is radius of gyration about diameter then,
    space space space space space space space space M K subscript straight d squared equals 2 over 3 M R squared rightwards double arrow space space space space space space space straight K subscript straight d equals square root of 2 over 3 end root straight R

    (ii) Radius of gyration about tangent:

    Moment of inertia of hollow sphere about tangent can be calculated by using the theorem of parallel axis.
    That is, 
               straight I subscript 1 equals straight I subscript straight d plus MR squared equals 2 over 3 MR squared plus MR squared equals 5 over 3 MR squared
    If Kt is radius of gyration about tangent then,
    MK subscript straight t squared equals 5 over 3 MR squared space space space space space straight K subscript straight t equals square root of 5 over 3 end root straight R subscript 0

     

    Question 456
    CBSEENPH11017590
    Wired Faculty App

    A solid sphere and hollow shell of same mass have same moment of inertia about their diameters. What is the ratio of their diameters? 
    Solution
    Let Rs and Rb be the radii of solid sphere and hollow sphere respectively. Since solid sphere and hollow sphere have same moment of inertia,
    therefore space space space 2 over 3 MR subscript straight h squared equals 2 over 5 MR subscript straight s squared space rightwards double arrow space space space straight R subscript straight s squared over Rh squared equals 5 over 3 space rightwards double arrow space space space space straight R subscript straight s over straight R subscript straight h equals square root of 5 over 3 end root 
    This is the required ratio of their diameters. 
    Question 457
    CBSEENPH11017593
    Wired Faculty App

    Three mass points m1, m2 and m3 are located at the vertices of an equilateral triangle of length a. What is the moment of inertia of the system about an axis along the altitude of the triangle passing through m1?

    Solution
    Moment of inertia of the system about an axis along altitude AL passing through mass m1 is equal to the sum of the moment of inertia of constituting particle about AL.
                   

    Since m1 lies on axis AL, therefore moment of inertia of mass m1, is,

    I1 = 0
    Masses m2 and m3 are at a distance 1/2 a from AL axis,

    therefore space space straight I subscript 2 equals straight m subscript 2 open parentheses straight a over 2 close parentheses squared space space space and space space space space straight I subscript 3 equals straight m subscript 3 open parentheses straight a over 2 close parentheses squared
    Now the moment of interia of system is,
    straight I space equals straight I subscript 1 plus straight I subscript 2 plus straight I subscript 3 space space space equals space 0 plus straight m subscript 2 straight a squared over 4 plus straight m subscript 3 straight a squared over 4 equals left parenthesis straight m subscript 2 plus straight m subscript 3 right parenthesis straight a squared over 4 space space space equals left parenthesis straight m subscript 2 plus straight m subscript 3 right parenthesis straight a squared over 4 

    Question 458
    CBSEENPH11017597
    Wired Faculty App

     Find the moment of inertia of disc about a transverse axis through the centre of disc of radius 0.25m and thickness 10 cm. (Density of disc is 8000kg/m3.)
    Solution

    The mass of the disc, M= πr2t 
    The moment of inertia of disc about transverse axis through the centre of disc is, 
    straight I equals M r squared equals pi r to the power of 4 t space rho H e r e comma R a d i u s space o f space t h e space d i s c comma space space straight r equals 0.25 straight m Thickness comma space straight t equals 10 c m space equals space 0.1 straight m space D e n s i t y comma space straight rho equals 8000 k g divided by straight m cubed space Therefore comma space M o m e n t space o f space I n e r t i a comma space straight I space equals space pi r to the power of 4 t space rho space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space equals space 22 over 7 cross times left parenthesis 0.25 right parenthesis to the power of 4 cross times 0.1 cross times 7700 space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space equals 9.453 space k g m squared space space space space space space space space

    Question 459
    CBSEENPH11017599
    Wired Faculty App

    Four solid spheres each of mass M and radius R are placed at the vertex of square of side 4R. Find the moment of inertia of configuration about one of the side of square.

    Solution
    The moment of inertia of sphere at A about axis CD is, 
    straight I subscript 1 equals 2 over 5 MR squared plus straight M left parenthesis 4 straight R right parenthesis squared space space space equals 16.4 space MR squared

                      
    The moment of inertia of sphere at B about axis CD is, 
    straight I subscript 1 space equals 2 over 5 MR squared plus straight M left parenthesis 4 straight R right parenthesis squared space space space equals 16.4 space MR squared 
    The moment of inertia of sphere at C about CD axis is, 
    straight I subscript 1 equals 2 over 5 MR squared equals 0.4 space MR squared 
    The moment of inertia of sphere at D about CD axis is, 
    straight I subscript 1 equals 2 over 5 MR squared equals 0.4 space MR squared 

    Therefore moment of inertia of system about CD axis is,

    I = I1 + I2 + I3 + I4 = 33.6 MR2 


    Question 460
    CBSEENPH11017600
    Wired Faculty App

    What is the ratio of the masses of solid sphere and hollow sphere of equal radii so that radius of gyration is same about diameters?
    Solution
    The radius of gyration is independent of the mass but depends on upon the distribution of mass.
    Therefore, solid sphere and hollow sphere of equal radii can never have an equal radius of gyration. This is because the distribution of mass in the solid and hollow sphere is different. The mass in a hollow sphere is concentrated at its periphery than at its centre. 
    Question 461
    CBSEENPH11017601
    Wired Faculty App

    Two satellites of equal masses are orbiting the earth in different orbits. Will their moment of inertia be same or different?

    Solution
    The satellite orbiting around the earth can be considered as particle.
    The moment of inertia about earth = MR2.
    Therefore, the satellite in outer orbit has more moment of inertia than that of satellite in inner orbit. 
    Question 462
    CBSEENPH11017602
    Wired Faculty App

    Two spheres, one solid and other hollow have same radii and mass. Which sphere has more moment of inertia and why?

    Solution
    The hollow sphere has more moment of inertia than solid. It is because, in hollow sphere more mass is distributed at a larger distance than solid sphere. The mass is concentrated at the periphery of the hollow disc. 
    Question 463
    CBSEENPH11017603
    Wired Faculty App

    If we shift the axis parallel to it, about which axis the moment of inertia of body is least?

    Solution

    According to theorem of parallel axis, the moment of inertia of a body about an axis parallel to axis passing through centre of mass of body is,

                         I = Ic + Mh2
    For an axis passing through the centre of mass of the body, minimum value of h is zero. Hence, the moment of inertia is minimum about this axis. 

    Question 464
    CBSEENPH11017604
    Wired Faculty App

    Show that radius of gyration of ring about an axis passing through centre and perpendicular to the plane of ring is equal to radius of ring.

    Solution

    The moment of inertia of ring about an axis passing through centre and perpendicular to the plane of ring is,

          I=MR2                                ...(1)

    If K is radius of gyration then

          I =MK2                                ...(2)

    From (1) and (2),
            K=R

    Hence, the radius of gyration of ring about an axis passing through the centre and perpendicular to the plane of ring is equal to the radius of ring. 

    Question 465
    CBSEENPH11017611
    Wired Faculty App

    From a circular disc of radius R and mass 9M, a small disc of radius R/3 is removed from the disc as shown in figure. Find the moment of inertia of remaining disc about an axis perpendicular to the plane of the disc and passing through center.

    Solution
    The moment of inertia of complete disc about a perpendicular axis passing through center O is,
     straight I subscript 1 equals 1 half open parentheses 9 straight M close parentheses straight R squared space space equals 9 over 2 MR squared  
    The mass of cut out disc if radius R/3 is,
    straight m space equals fraction numerator 9 straight M over denominator pi R squared end fraction straight pi open parentheses straight R over 3 close parentheses squared space space space space space equals straight M 
    Now, using the theorem of parallel axis, the moment of inertia of cut out disc about the perpendicular axis passing through center O is,
    straight I subscript 2 space equals 1 half straight M open parentheses straight R over 3 close parentheses squared plus straight M open parentheses fraction numerator 2 straight R over denominator 3 end fraction close parentheses squared space space space space equals 1 half MR squared  

    The moment of inertia of residue disc is, 

    straight I equals straight I subscript 1 minus straight I subscript 2 space space equals 9 over 2 MR squared minus 1 half MR squared space space space equals 4 MR squared 
    Question 466
    CBSEENPH11017629
    Wired Faculty App

    Two masses m1 and m2 are separated by distance r. Find the moment of inertia of system about an axis perpendicular to the line joining the two masses and passing through their centre of mass.
    Solution
    Let centre of mass lies at a distance r1 from mass m1 and r2 from mass m2
                   
    As the moment of masses about the center of mass is zero, therefore,

    space space space space space space space space space space space space space space straight m subscript 1 straight r subscript 1 equals straight m subscript 2 straight r subscript 2 space space space space space space space space space space space space space space space space space space space space space space space space space... left parenthesis 1 right parenthesis space Also comma space space space space space space space space space straight r equals straight r subscript 1 plus straight r subscript 2 space space space space space space space space space space space space space space space space space space space space space space space space space space... left parenthesis 2 right parenthesis 
    From (1) and (2), we get
    <pre>uncaught exception: <b>mkdir(): Permission denied (errno: 2) in /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/util/sys/Store.class.php at line #56mkdir(): Permission denied</b><br /><br />in file: /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/util/sys/Store.class.php line 56<br />#0 [internal function]: _hx_error_handler(2, 'mkdir(): Permis...', '/home/config_ad...', 56, Array) #1 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/util/sys/Store.class.php(56): mkdir('/home/config_ad...', 493) #2 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/plugin/impl/FolderTreeStorageAndCache.class.php(110): com_wiris_util_sys_Store->mkdirs() #3 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/plugin/impl/RenderImpl.class.php(231): com_wiris_plugin_impl_FolderTreeStorageAndCache->codeDigest('mml=<math xmlns...') #4 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/plugin/impl/TextServiceImpl.class.php(59): com_wiris_plugin_impl_RenderImpl->computeDigest(NULL, Array) #5 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/service.php(19): com_wiris_plugin_impl_TextServiceImpl->service('mathml2accessib...', Array) #6 {main}</pre>
     
    Now, moment of inertia of the system about the centre of mass is,
    straight I equals straight m subscript 1 straight r subscript 1 squared plus straight m subscript 2 straight r subscript 2 squared space space equals straight m subscript 1 open parentheses fraction numerator straight m subscript 2 over denominator straight m subscript 1 plus straight m subscript 2 end fraction straight r close parentheses squared plus straight m subscript 2 open parentheses fraction numerator straight m subscript 1 over denominator straight m subscript 1 plus straight m subscript 2 end fraction straight r close parentheses space space equals fraction numerator straight m subscript 1 straight m subscript 2 over denominator straight m subscript 1 plus straight m subscript 2 end fraction straight r squared  
    Question 467
    CBSEENPH11017638
    Wired Faculty App

    What is torque? Derive an expression for it in vector form.

    Solution
    Torque is a force that produces the turning or rotational motion in a body.
    Torque is produced by a force perpendicular to the line joining the point about which the body rotates and the point of application of force.
    Torque is a vector quantity and its direction is along the axis of rotation.
    Consider a particle moving in KY plane along  the curve PQ under the influence of a force rightwards arrow for straight F of. 
    Let any instant t, the particle is at straight A left right arrow left parenthesis straight x comma straight y right parenthesis and at instant t+dt, the particle is at straight B left right arrow left parenthesis straight x plus dx comma straight y plus dy right parenthesis. 
    Displacement undergone by the particle, stack A B with rightwards harpoon with barb upwards on top space equals dx space straight i with hat on top space plus dy space straight j with hat on top.
    Let Fx and Fy be the components of force rightwards arrow for F of, when the particle is at A. 
                          
    The work done by force in displacing the body from A to B is,
    dW space equals straight F with rightwards harpoon with barb upwards on top space. space AB with rightwards harpoon with barb upwards on top space space space space space space equals left parenthesis straight F subscript straight x space straight i with hat on top plus straight F subscript straight y space straight j with hat on top right parenthesis. left parenthesis dx space straight i with hat on top plus dy space straight j with hat on top right parenthesis space space space space space space space equals straight F subscript straight x dx plus straight F subscript straight y dy space space space space space space space space space space space space space space space space space space space space space space space space space space... left parenthesis 1 right parenthesis  
    From the above figure, we have
    space straight x equals straight r space cos space straight theta space comma space straight y equals straight r space sin space straight theta space rightwards double arrow space space space dx equals negative rsinθdθ equals negative ydθ space comma space and space space space space space space space space dy equals space rcosθ space equals space xdθ
    Substituting for dx and dy in equation(1), we get
    dW space equals space straight F subscript straight x left parenthesis negative ydθ right parenthesis space plus space straight F subscript straight y left parenthesis xdθ right parenthesis space space space space space space space space equals space left parenthesis xF subscript straight y minus yF subscript straight x right parenthesis dθ space space space space space space space space space space space space space space space space space space... left parenthesis 2 right parenthesis space Also comma space dW equals straight tau subscript straight z dθ space space space space space space space space space space space space space space space space space space space space space space space space space space... left parenthesis 3 right parenthesis
    From (1) and (2)
    straight tau subscript straight z equals space xF subscript straight y minus space straight y space straight F subscript straight x space space space space space space space space space space space space space space space space space space space... space left parenthesis 4 right parenthesis 
    where, τz is the torque produced by force.
    As the body revolves in XY plane, therefore torque by force will be along Z-axis.

    Similarly, if the body moves in YZ plane, then torque along X-axis will be given by

    τ = yFz ~ zFy                  ...(5)

    If the body moves in XZ plane, then torque along Y-axis will be given by,

    τy=zFx-xFz                      ...(6) 

    If the body moves in space, then the torque produced by force is,
    tau space equals straight tau subscript straight x straight i with hat on top plus straight tau subscript straight y stack space straight j with hat on top plus straight tau subscript straight z space straight k with hat on top with space on top 
       equals left parenthesis yF subscript straight z minus zF subscript straight y right parenthesis space straight i with hat on top space plus space left parenthesis zF subscript straight x minus xF subscript straight z right parenthesis space straight j with hat on top plus left parenthesis xF subscript straight y minus yF subscript straight x right parenthesis end subscript space straight k with hat on top
        space space space space equals open vertical bar table row cell straight i with hat on top end cell cell straight j with hat on top end cell cell straight k with hat on top end cell row straight x straight y straight z row cell straight F subscript straight x end cell cell straight F subscript straight y end cell cell straight F subscript straight z end cell end table close vertical bar rightwards arrow for straight tau of equals rightwards arrow for straight r of cross times rightwards arrow for straight F of 

    This is the required expression for torque. 
    Question 468
    CBSEENPH11017639
    Wired Faculty App

    Show that torque produced by force on the body is equal to product of the force and perpendicular distance of line of action of force from the axis of rotation or it is equal to the product of radial distance of point of action of force from the axis of rotation and transverse component of force.

    Solution

    Consider a particle moving along the curve PQ under the influence of a force straight F with rightwards harpoon with barb upwards on top.
    Let at any instant r, the particle be at A and its position vector is straight r with rightwards harpoon with barb upwards on top. 
     
    Torque is given by, 
    straight tau with rightwards harpoon with barb upwards on top space equals space r with rightwards harpoon with barb upwards on top space x space F with rightwards harpoon with barb upwards on top space The space magnitude space of space torque space is comma space straight tau with rightwards harpoon with barb upwards on top space equals space straight r space straight F space sin space straight ϕ space where comma space straight ϕ space be space the space angle space between space between space straight r with rightwards harpoon with barb upwards on top space and space straight F with rightwards harpoon with barb upwards on top. space Now comma space straight tau space equals space rF space sin space straight ϕ space space space space equals space straight r space left parenthesis straight F space sin space straight ϕ right parenthesis space space space space space equals space straight r space straight F subscript straight o  
    where,
    Fϕ is transverse component of force.
    Therefore, torque is equal to the product of radial distance and transverse component of force.

    Also,
    x   = τ Fsin ϕ
         = F(rsin ϕ)

     straight tau with rightwards harpoon with barb upwards on top  = Fd
    Therefore, torque is equal to the product of the magnitude of force and perpendicular distance of line of action of force from the axis of rotation.

    Question 469
    CBSEENPH11017642
    Wired Faculty App

    Define couple. Derive an expression for the moment of couple.

    Solution

    Couple is the set of two equal and opposite force having a different line of action. A couple produces the torque and tends to cause the rotational motion.
    Consider two equal and opposite forces acting at points A and B having position vectors r with rightwards harpoon with barb upwards on top subscript 1 space and stack space r subscript 2 with rightwards harpoon with barb upwards on top as shown in figure. 
      
    Moment of force negative space straight F with rightwards harpoon with barb upwards on top about O is, 
    space space space space space space space space space space space space space space space space space stack straight tau subscript 1 with rightwards harpoon with barb upwards on top space equals stack space straight r with rightwards harpoon with barb upwards on top subscript 1 space cross times space left parenthesis negative straight F with rightwards harpoon with barb upwards on top space right parenthesis space equals space minus straight r with rightwards harpoon with barb upwards on top subscript 1 space cross times space straight F with rightwards harpoon with barb upwards on top Moment space of space force space straight F with rightwards harpoon with barb upwards on top space about space straight O space is comma space stack straight tau subscript 2 with rightwards harpoon with barb upwards on top space equals straight r with rightwards harpoon with barb upwards on top subscript 2 space cross times space straight F with rightwards harpoon with barb upwards on top space Moment space of space couple colon space straight tau with rightwards harpoon with barb upwards on top space equals space straight tau with rightwards harpoon with barb upwards on top subscript 1 space plus space straight tau with rightwards harpoon with barb upwards on top subscript 2 space space space space equals space minus straight r with rightwards harpoon with barb upwards on top subscript 1 space cross times space straight F with rightwards harpoon with barb upwards on top space plus stack space straight r subscript 2 space with rightwards harpoon with barb upwards on top space cross times space straight F with rightwards harpoon with barb upwards on top space space space equals space left parenthesis straight r with rightwards harpoon with barb upwards on top subscript 2 space minus space straight r with rightwards harpoon with barb upwards on top subscript 1 right parenthesis cross times space straight F with rightwards harpoon with barb upwards on top space space equals space straight r with rightwards harpoon with barb upwards on top space cross times space straight F with rightwards harpoon with barb upwards on top space Magnitude space of space the space moment space of space couple colon space straight tau space equals space rF space sin space straight theta space equals space straight F space straight r space sin space straight theta space equals space straight F space straight d space
    where straight theta is between F and r and d is the perpendicular distance between the lines of action of two forces of couples. 
    Thus, the magnitude of the moment of couple is equal to magnitude of one of the force of couple and perpendicular distance between the lines of action of two forces of couple. 

    Question 470
    CBSEENPH11017646
    Wired Faculty App

    Under what condition torque acting on body by a given force is zero?

    Solution
    W e space k n o w space t h a t comma space space space space space space space space tau space equals space r with rightwards arrow on top x space F with rightwards harpoon with barb upwards on top space rightwards double arrow space space space straight tau space equals space r space Fsinθ space

    If sin θ = 0 or r = 0

    then, θ= 0°, 180°
    That is, Torque = 0.

    That is, when line of action of forces passes through the axis of rotation, torque is zero. 
    Question 471
    CBSEENPH11017648
    Wired Faculty App

    Define angular momentum. Derive an expression for it in vector form.

    Solution

    The angular momentum of a rotating body is the measure of the quantity of angular motion contained in the body. It is also called the moment of linear momentum of the body.

    Due to rotational inertia possessed by the body about the axis of rotation, the rotating body cannot change its angular momentum itself.
    The torque produced by applied force on the rotating body changes its angular momentum and rate of change of angular momentum is equal to torque on the body.
    i.e., straight tau space equals space fraction numerator straight d straight L with rightwards harpoon with barb upwards on top over denominator dt end fraction straight tau subscript straight x space straight i with hat on top space plus space straight tau subscript straight y space straight j with hat on top space plus space straight tau subscript straight z space straight k with hat on top space equals space fraction numerator straight d straight L with rightwards harpoon with barb upwards on top subscript straight x over denominator dt end fraction space straight i with hat on top space plus space fraction numerator straight d straight L with rightwards harpoon with barb upwards on top subscript straight y over denominator dt end fraction stack space straight j with hat on top space plus space fraction numerator straight d straight L with rightwards harpoon with barb upwards on top subscript straight z over denominator dt end fraction straight k with hat on top Therefore comma space straight tau subscript straight x space equals space fraction numerator straight d straight L with rightwards harpoon with barb upwards on top subscript straight x over denominator dt end fraction space straight tau subscript straight y space equals space fraction numerator straight d straight L with rightwards harpoon with barb upwards on top subscript straight y over denominator dt end fraction straight tau subscript straight z space equals space fraction numerator straight d straight L with rightwards harpoon with barb upwards on top subscript straight z over denominator dt end fraction
    Consider, a body moving in XY plane.
    The torque on the body revolving in XY plane is,

           τz =xFy-yFx                             ...(1)

    Let px and vx be the X-component,
    py and vy be the Y component of linear momentum & velocity of body in XY plane respectively.
    Now, 
    straight tau subscript straight z space equals space straight x space dp subscript straight y over dt space minus space straight y space dp subscript straight x over dt space space space space space equals space straight x space dmv subscript straight y over dt space minus space straight y space dmv subscript straight x over dt space space space equals space straight m space open parentheses straight x dv subscript straight y over dt space minus space straight y fraction numerator space dv subscript straight x over denominator dt end fraction close parentheses space space space equals space straight m open parentheses dxv subscript straight y over dt minus dyv subscript straight x over dt close parentheses open square brackets dxv subscript straight y over dt minus dyv subscript straight x over dt space equals straight x dv subscript straight y over dt space minus space straight y fraction numerator space dv subscript straight x over denominator dt end fraction space close square brackets space straight tau subscript straight z space equals open parentheses dxmv subscript straight y over dt space minus space dymv subscript straight x over dt close parentheses space space space space space equals space open parentheses dxp subscript straight y over dt minus dyp subscript straight x over dt close parentheses space space space space space equals space straight d over dt space left parenthesis space straight x space straight p subscript straight y space minus space yp subscript straight x right parenthesis Also comma space straight tau subscript straight z space equals space dL subscript straight z over dt space equals space straight d over dt open parentheses straight x space straight p subscript straight y space minus space yp subscript straight x close parentheses rightwards double arrow space straight L subscript straight z space equals space xp subscript straight y space minus space yp subscript straight x space Similarly comma space if space the space body space moves space in space YZ space plane comma space then space straight L subscript straight x space equals space yp subscript straight z space minus space zp subscript straight y space and space if space the space body space moves space in space XZ space plane comma space then comma space straight L subscript straight z space equals space zp subscript straight x space minus space xp subscript straight x space If space the space body space is space moving space in space space comma space then space the space total space angular space momentum space of space the space body space is comma space straight L with rightwards harpoon with barb upwards on top space equals space straight L subscript straight x space straight i with hat on top space plus space straight L subscript straight y space straight j with hat on top space plus space straight L subscript straight z space straight k with hat on top space space space space equals space space open parentheses space yp subscript straight z space minus space zp subscript straight y space close parentheses space straight i with hat on top space plus space left parenthesis zp subscript straight x space minus space xp subscript straight z right parenthesis space straight j with hat on top space plus space left parenthesis xp subscript straight y space minus space yp subscript straight x right parenthesis space straight k with hat on top space space space space equals open vertical bar table row straight i straight j straight k row straight x straight y straight y row cell straight p subscript straight x end cell cell straight p subscript straight y end cell cell straight p subscript straight z end cell end table close vertical bar space straight L with rightwards harpoon with barb upwards on top space equals space straight r with rightwards harpoon with barb upwards on top space straight x space straight p with rightwards harpoon with barb upwards on top

    Question 472
    CBSEENPH11017650
    Wired Faculty App

    In which condition do the centre of mass and centre of gravity of body coincide?

    Solution

    When the size of a body is small, the centre of mass and centre of gravity coincide. 

    Question 473
    CBSEENPH11017651
    Wired Faculty App

    Two particles A and B of mass 5m and 2m are separated by certain distance. Which of the two masses is nearer to centre of mass?

    Solution

    The mass which is heavier is closer to the centre of mass. That is, mass of 5m lies closer to the centre of mass. 

    Question 474
    CBSEENPH11017652
    Wired Faculty App

    A motor car is travelling at 30 m/s on a circular road of radius 500 m. It is increasing in speed at the rate of 2 m/s2. What is it's acceleration? 
    Solution

    Here, 
    Speed of motor car, v = 30 m/s 
    Radius of circular road, r = 500 m 
    Rate of increase in speed = 2 m/s 
    To find: Acceleration of the motor car? 
    We know that, 

    straight a subscript straight r space equals space straight v squared over straight r space equals space fraction numerator 30 space cross times space 300 over denominator 500 end fraction space equals space 1.8 space straight m divided by straight s squared space That space is comma space straight a space equals space square root of straight a subscript straight r squared plus space straight a subscript straight T squared end root space space space space equals space square root of left parenthesis 1.8 right parenthesis squared plus space left parenthesis 2 right parenthesis squared end root space equals space 2.7 space straight m divided by straight s squared

    Question 475
    CBSEENPH11017653
    Wired Faculty App

    Two persons jump from a spring board into a swimming pool. One person jumps straight and second person curls himself. Which of the two persons can make more revolutions before they hit the pool?

    Solution

    Second person will make more revolutions before they hit the pool because as the person curls up the moment of Inertia decreases. Therefore, the angular velocity of the person increases. 

    Question 476
    CBSEENPH11017654
    Wired Faculty App

    How does the moment of inertia of a given body about a particular axis change when the body increases its speed?

    Solution

    When the body increases it's speed, the moment of inertia of the body remains unchanged. 

    Question 477
    CBSEENPH11017655
    Wired Faculty App

    A swimmer intends to jump into water from a height. How should he adjust his body so that he is able to form a loop in air easily?
    Solution

    In order to form a loop in air easily and jump from a height, the swimmer should curl his body. This way the moment of inertia would decrease and hence the angular velocity of the swimmer increases. 

    Question 478
    CBSEENPH11017656
    Wired Faculty App

    What happens to the potential energy when a body rolls down an inclined plane? 

    Solution

    While rolling, the body has both translational as well as rotational motion. Therefore, the potential energy of the body is converted into translational and rotational kinetic energy. 

    Question 479
    CBSEENPH11017657
    Wired Faculty App

    Two particles P and Q are revolving in circular path of radii r and 2r with same angular velocity. Which of the two travels faster?

    Solution

    F = mstraight v squared over straight r 
    Velocity of the body is directly proportional to the radius. 
    Therefore, the particle Q with radius 2r will travel faster. 

    Question 480
    CBSEENPH11017659
    Wired Faculty App

    Show that angular momentum of a rotating body is equal to product of radial distance of point of particle from the axis of rotation and transverse component of linear momentum or it is equal to the product of the linear momentum and perpendicular distance of direction of motion from the axis of rotation.

    Solution

    Consider a particle moving in the space.
    Let at any instant t, the particle be at A.
    Let straight v with rightwards harpoon with barb upwards on top be the velocity of particle at A and straight r with rightwards harpoon with barb upwards on top be the position vector. 
    The angular momentum of particle is, 
    L with rightwards harpoon with barb upwards on top space equals space r with rightwards harpoon with barb upwards on top cross times space p with rightwards harpoon with barb upwards on top
    Magnitude of angular momentum, straight L space equals space straight r space straight p space sinϕ 
    where,
    straight ϕ be the angle which straight p with rightwards harpoon with barb upwards on top makes with straight r with rightwards harpoon with barb upwards on top. 
    Now,
        straight L equals rpsinϕ space space space equals straight r left parenthesis psinϕ right parenthesis space space space equals rp subscript straight ϕ space 
     

    where,
    pϕ is transverse component of angular momentum. 
    Therefore, angular momentum is equal to product of radial distance and transverse component of linear momentum.

    Also L = r p sin ϕ
             = sin ϕ)

             = pd 

    Therefore, angular momentum is equal to product of the magnitude of angular momentum and perpendicular distance of direction of motion from the axis of rotation.

    Question 481
    CBSEENPH11017663
    Wired Faculty App

    Show that angular momentum of body revolving about an axis is twice the mass of arial velocity.

    Solution
    Let a particle of mass m be revolving in XY plane about Z axis.
               
    Let, at any instant t particle be at A and at t + ∆t particle be at B.
    Let the radius vector sweep the area ∆A in time ∆t. 
    therefore space space space space capital delta A equals space A r e a space o f space straight capital delta space A O B space space space space space space stack increment A with rightwards harpoon with barb upwards on top equals 1 half r with rightwards harpoon with barb upwards on top space cross times space stack increment r with rightwards harpoon with barb upwards on top therefore space space A r i a l space v e l o c i t y comma space v with rightwards harpoon with barb upwards on top subscript a r end subscript space equals fraction numerator space space rightwards arrow for capital delta capital alpha of over denominator capital delta t end fraction space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space equals space 1 half fraction numerator rightwards arrow for straight r of cross times rightwards arrow for capital delta r of over denominator capital delta t end fraction space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space equals space 1 half rightwards arrow for straight r of cross times fraction numerator rightwards arrow for capital delta r of over denominator rightwards arrow for capital delta t of end fraction space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space equals 1 half r with rightwards harpoon with barb upwards on top cross times v with rightwards harpoon with barb upwards on top space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space equals space 1 half fraction numerator rightwards arrow for straight r of cross times straight m rightwards arrow for straight v of over denominator straight m end fraction space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space equals fraction numerator rightwards arrow for straight L of over denominator 2 straight m end fraction therefore space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space equals space 2 straight m fraction numerator space space rightwards arrow for capital delta capital alpha of over denominator capital delta t end fraction
    Question 482
    CBSEENPH11017664
    Wired Faculty App

    State Kepler’s laws of planetary motion.

    Solution

    Kepler’s three laws of planetary motion are:

    (i) All the planets revolve around the sun in elliptical orbits with sun at one focus.

    (ii) The arial velocity of the planet is constant.

    (iii)The square of the time period of revolution is directly proportional to the cube of semi-major axis of its elliptical orbit.

    Question 483
    CBSEENPH11017666
    Wired Faculty App

    State and prove Kepler's second law.

                             OR

    State Kepler’s second law and prove that it is based on the law of conservation of angular momentum.
    Solution
    The second law of planetary motion states that the arial velocity of a planet is constant. 
    We  know that,
              L with rightwards harpoon with barb upwards on top equals space 2 straight m space fraction numerator rightwards arrow for capital delta capital alpha of over denominator capital delta t end fraction  
     rightwards double arrow space space fraction numerator rightwards arrow for ΔΑ of over denominator Δt end fraction equals fraction numerator rightwards arrow for straight L of over denominator 2 straight m end fraction     
    The gravitational force exerted by the sun on planet produces zero torque, therefore, angular momentum straight L with rightwards harpoon with barb upwards on top of the planet is constant. 
    rightwards double arrow space space fraction numerator rightwards arrow for ΔA of over denominator ΔT end fraction equalsConstant. 
    Hence the result.

    Question 484
    CBSEENPH11017668
    Wired Faculty App

    A rigid body of moment of inertia I is rotating with uniform angular velocity ω about a given axis. Derive an expression for rotational kinetic energy of the body.

    Solution

    Consider a rigid body consisting of n particles rotating about an axis XY with uniform angular velocity straight omega.

    Let r1, r2...........,rn be the distances of particles from the axis of rotation.
    Different particles are at different distances from the axis of rotation, therefore, the magnitude of linear velocities of different particles will be different. 

                  

    Consider ith particle of mass mi at a distance ri from the axis of rotation.
    The linear speed of ith particle will be

                        Vi = ri ω

    Therefore, kinetic energy of ith particle due to rotation about XY-axis is, 
                <pre>uncaught exception: <b>mkdir(): Permission denied (errno: 2) in /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/util/sys/Store.class.php at line #56mkdir(): Permission denied</b><br /><br />in file: /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/util/sys/Store.class.php line 56<br />#0 [internal function]: _hx_error_handler(2, 'mkdir(): Permis...', '/home/config_ad...', 56, Array) #1 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/util/sys/Store.class.php(56): mkdir('/home/config_ad...', 493) #2 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/plugin/impl/FolderTreeStorageAndCache.class.php(110): com_wiris_util_sys_Store->mkdirs() #3 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/plugin/impl/RenderImpl.class.php(231): com_wiris_plugin_impl_FolderTreeStorageAndCache->codeDigest('mml=<math xmlns...') #4 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/plugin/impl/TextServiceImpl.class.php(59): com_wiris_plugin_impl_RenderImpl->computeDigest(NULL, Array) #5 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/service.php(19): com_wiris_plugin_impl_TextServiceImpl->service('mathml2accessib...', Array) #6 {main}</pre>

    Total kinetic energy of rotation of the rigid body about XY-axis is
    straight K space equals sum straight K subscript 1 space space space space equals sum 1 half straight m subscript 1 straight r subscript 1 squared straight ϖ squared space space space equals 1 half straight ϖ squared sum 1 half straight m subscript 1 straight r subscript 1 squared space space equals 1 half I straight ϖ squared T h e n comma space space straight K equals 1 half I straight ϖ squared 
    This is the required rotational K.E of the body. 

    Question 485
    CBSEENPH11017671
    Wired Faculty App

    A rigid body of moment of inertia I is rotating with uniform angular velocity ω about a given axis. Derive the expression for angular momentum associated with body.

    Solution

    Consider a rigid body consisting of particles of masses m1, m2........mn at the distances r1, r2...........rn respectively from the axis of rotation.
    Axis of rotation XY is revolving with angular velocity straight omega.
    The body is rigid, therefore the different particles revolve in a circular orbit of radii equal to their distances from the axis of rotation with their centers on XY axis, but all the particles complete one revolution in equal time.
    Therefore, the linear speed of different particles is different.
    Linear velocity of ith particle at a distance ri from the axis of rotation is,

                           vi =ri ω

    The linear momentum of ith particle is, 

                          pi =mi vi

    Linear momentum is along the tangent to the circular path followed by particle.
    Therefore, the linear momentum vector and radius vector are perpendicular to each other.
    Angular momentum is equal to product of linear momentum and perpendicular distance from axis of rotation to direction of motion, therefore angular momentum of a particle about XY-axis is,
    straight L subscript 1 equals straight p subscript 1 space straight r subscript 1 space space space space equals space straight m subscript 1 straight v subscript 1 straight r subscript 1 space space space space equals straight m subscript 1 left parenthesis straight r comma straight ϖ right parenthesis straight r subscript 1 space space space equals straight m subscript 1 straight r subscript 1 squared straight ϖ equals straight I subscript 1 space end subscript straight ϖ
    where,
    I1 is the moment of inertia of ith particle about XY-axis.
    Now, the total angular momentum of the body about XY is the vector sum of the angular momentum of all the constituent particles.
    All the particles have angular momentum in the same direction, therefore the magnitude of the angular momentum of body is equal to the sum of the magnitude of angular momentum of all the constituent particles of the body and is directed along the XY-axis. 
    therefore space space space space straight L equals sum straight L subscript 1 space space space space space space space space space space equals sum straight I subscript 1 straight ϖ space space space space space space space space space space equals straight ϖ sum straight I subscript 1 space space space space space space space space space space equals ϖI space rightwards double arrow space space straight L space equals space Iϖ

    Question 486
    CBSEENPH11017673
    Wired Faculty App

    Derive an expression for torque in terms of moment of inertia of body and its angular acceleration.

    Solution

    Consider a rigid body rotating about XY-axis under the action of external torque τ.
    As a result the body possesses the angular acceleration ‘α’ about the axis of rotation.
    All the constituent particles possess the same angular acceleration but the linear acceleration of different particles will be different.
    The acceleration of ith particle is,

                       ai = ri α

    The magnitude of external force on ith particle will be,

                  Fi =mi ai = mi ri α

    This force on the ith particle produces the turning effect.
    The magnitude of torque on this particle is equal to the product of force and the perpendicular distance from the axis of rotation to the line of action of force.
    T h a t space i s comma space space space space space space space straight tau subscript 1 equals straight r subscript 1 straight f subscript 1 space space space space space space space space space space equals straight r subscript 1 straight m subscript 1 straight r subscript 1 straight alpha space space space space space space space space space space equals space straight m subscript 1 straight r subscript 1 squared straight alpha equals straight I subscript 1 straight alpha 
    Therefore total torque acting on rigid body is, 
    therefore space space straight tau equals sum straight t subscript 1 space space space space space space space space equals sum straight I subscript 1 straight alpha space space space space space space space space space equals straight alpha sum straight I subscript 1 space space space space space space space space space equals αI rightwards double arrow space space space space space space straight tau equals straight I space straight alpha 

    Question 487
    CBSEENPH11017674
    Wired Faculty App

    If the polar ice caps of the earth melt, how will the length of day change?

    Solution

    When ice melts at poles, water will spread away from poles and hence moment of inertia of the earth will increase. Therefore, the length of the day changes. 

    In the absence of an external torque, angular momentum of the system remains conserved. 
    therefore space space space space space space space space space space space straight L space equals space I ϖ equals c o n s t a n t rightwards double arrow space space space space straight I fraction numerator 2 straight pi over denominator straight T end fraction equals c o n s t a n t rightwards double arrow space space space space space space space space space space space space straight I proportional to straight T
    If the moment of inertia increases, the duration of the day will increase.

    Question 488
    CBSEENPH11017675
    Wired Faculty App

    A planet rotates about the sun in an elliptical orbit whose eccentricity is e. What is the ratio of maximum to a minimum velocity of the planet?
    Solution
    Let a planet rotate about the sun in an elliptical orbit whose semi-major axis is a and eccentricity is e as shown. 

    Since, angular momentum of planet is constant
    i.e.,               mvr = constant,
    therefore v is maximum when planet is at the nearest point.
    i.e., at P1 and minimum when it is at the farthest point i.e. at P2.

    Distance of P1 from the sun =a - ae
                                               =a( 1- e)

    Distance of P2 from the sun =a + ae
                                              =a( 1 + e)

    According to the law of conservation of angular momentum, 
    mv subscript max space straight a open parentheses 1 minus straight e close parentheses equals m space v subscript min straight a left parenthesis 1 plus straight e right parenthesis space rightwards double arrow space space space space space space space space space straight v subscript max over straight v subscript min equals fraction numerator 1 plus straight e over denominator 1 minus straight e end fraction 

    Question 489
    CBSEENPH11017676
    Wired Faculty App

    A person exerts a force of 40N on a revolving door at a point 50cm from the axis. Find the torque produced by applied force.

    Solution

    Here,
    Force, F = 40N

    Distance from the axis, r = 50 cm = 0.5m

    The torque produced by force is

    τ = r F
      = 0.5 x 40
       = 20 Nm

    Question 490
    CBSEENPH11017677
    Wired Faculty App

    How large is a torque needed to accelerate a wheel of moment of inertia, I = 2kgm2 from rest to 30rps in 20s? 

    Solution

    Here, we have
     straight ϖ subscript 0 equals 0 space straight ϖ equals 30 space rps space equals space 60 space rad divided by straight s space straight t equals 20 straight s comma space straight I equals 2 space kgm squared
    Now, using the equation of rotational motion, we have
                     straight ϖ equals straight ϖ subscript 0 plus straight alpha space straight t
    therefore space space straight alpha space equals fraction numerator straight ϖ minus straight ϖ subscript 0 over denominator straight t end fraction space or space space straight alpha space equals space fraction numerator 60 straight pi minus 0 over denominator 15 end fraction equals 4 straight pi space rad divided by straight s squared Now comma space space space straight tau equals space Iα space space space space space equals 2 cross times 4 straight pi space space space space space equals 8 straight pi space Nm  

    Question 491
    CBSEENPH11017679
    Wired Faculty App

    A motor delivers 0.25hp when its shaft is running at the rate 1400 rpm. Find the torque acting on the shaft.

    Solution

    Here, we have
    Velocity, w=1400 rpm = 146.67 rad/s
    Power, P=0.25 hp = 0.25x746W=186.5W
    We know,  P= straight omega space straight tau
    therefore space space space space straight tau equals straight P over straight ϖ equals fraction numerator 186.5 over denominator 146.67 end fraction equals 1.27 Nm

    Question 492
    CBSEENPH11017683
    Wired Faculty App

    The angular velocity of a body of moment of inertia 12 kgm2 is given by 

    omega with rightwards harpoon with barb upwards on top equals negative 2 space straight i with hat on top plus 6 straight t space straight j with hat on top minus 8 straight t space straight k with hat on top space.Find the torque acting on the body.
    Solution

    Here, we have
    Moment of Inertia, I = 12 kgm2 
     omega with rightwards harpoon with barb upwards on top equals negative 2 space straight i with hat on top plus 6 straight t space straight j with hat on top minus 8 straight t space straight k with hat on top

    Therefore angular momentum of body is, 
    L with rightwards harpoon with barb upwards on top equals straight I space straight omega with rightwards harpoon with barb upwards on top space equals space 12 left parenthesis negative 2 space straight i with hat on top plus 6 straight t space straight j with hat on top minus 8 straight t space straight k with hat on top right parenthesis space space space space equals negative 24 space straight i with hat on top space plus space 72 straight t space straight j with hat on top space minus 96 straight t space straight k with hat on top 
    Now the torque on the body is,
    stack tau space with rightwards harpoon with barb upwards on top equals fraction numerator straight d space L with rightwards harpoon with barb upwards on top over denominator dt end fraction equals straight d over dt left parenthesis negative 24 space straight i with hat on top plus 72 straight t space straight j with hat on top minus 96 straight t space straight k with hat on top right parenthesis space space space space space space space space space space space space space space space space space equals space left parenthesis 72 space straight j with hat on top space minus 96 space straight k with hat on top right parenthesis
    Question 493
    CBSEENPH11017684
    Wired Faculty App

    A torque of 2.4Nm is applied on the flywheel of mass 8kg and radius 0.5m. Find the angular acceleration produced in the flywheel.

    Solution

    We have,
    <pre>uncaught exception: <b>mkdir(): Permission denied (errno: 2) in /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/util/sys/Store.class.php at line #56mkdir(): Permission denied</b><br /><br />in file: /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/util/sys/Store.class.php line 56<br />#0 [internal function]: _hx_error_handler(2, 'mkdir(): Permis...', '/home/config_ad...', 56, Array) #1 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/util/sys/Store.class.php(56): mkdir('/home/config_ad...', 493) #2 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/plugin/impl/FolderTreeStorageAndCache.class.php(110): com_wiris_util_sys_Store->mkdirs() #3 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/plugin/impl/RenderImpl.class.php(231): com_wiris_plugin_impl_FolderTreeStorageAndCache->codeDigest('mml=<math xmlns...') #4 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/plugin/impl/TextServiceImpl.class.php(59): com_wiris_plugin_impl_RenderImpl->computeDigest(NULL, Array) #5 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/service.php(19): com_wiris_plugin_impl_TextServiceImpl->service('mathml2accessib...', Array) #6 {main}</pre>,
    Mass, m=8kg
    Radius, r= 0.5 
    The moment of inertia  of flywheel is, 
    I = mr2 
      = 2 kgm2 
    Therefore angular acceleration of flywheel is,
     
    straight alpha equals straight tau over straight I equals fraction numerator 2.4 over denominator 2 end fraction equals 1.2 space rad divided by straight s squared

    Question 494
    CBSEENPH11017687
    Wired Faculty App

    A painter weighing 70kg paints the wall of a house, while standing on a uniform wooden plank of length 2m placed on a stand as shown in figure. The mass of the board is 12kg. Find whether the man at the position shown is safe or not.

    Solution
    The plank tends to turn about A in anticlockwise direction, due to the weight of the painter.
    The moment of weight of man about A is,
    straight tau with rightwards harpoon with barb upwards on top space equals space straight M space straight g space left parenthesis BA right parenthesis space space space space space equals 70 cross times 9.8 cross times 0.2 
         = 137.2 Nm in anti-clockwise direction.
    The plank is uniform, therefore centre of mass of the plank is at the center i.e. at C.
    Due to the weight of the plank, the plank tends to turn about A in clockwise direction.
    The moment of weight of plank about A is, 
    tau with rightwards harpoon with barb upwards on top space equals space mg space left parenthesis AC right parenthesis space space space space space equals space 12 cross times 9.8 cross times 0.5 
          = 58.8 Nm, in clockwise direction. 
    The torque which is acting anticlockwise about A is greater than that in the clockwise direction.
    Therefore the plank will turn about O and man will fall.
    Thus the man is unsafe.
    Question 495
    CBSEENPH11017689
    Wired Faculty App

    The velocity if a particle of mass 2 kg is 3 space straight i with hat on top space minus 4 space straight j with hat on top plus straight k with hat on top space when at negative 2 space straight i with hat on top space minus 4 space straight j with hat on top plus 3 space straight k with hat on top space Find the angular momentum of the particle.
    Solution

    Mass of particle, m = 2  kg
    Velocity of particle, straight v with rightwards harpoon with barb upwards on top = 3 i - 4 j + k
    Therefore, the  linear momentum of particle, 
    straight p with rightwards harpoon with barb upwards on top space equals space straight m space straight v with rightwards harpoon with barb upwards on top space equals space 2 space left parenthesis 3 space straight i with hat on top space minus space 4 space straight j with hat on top space plus space straight k with hat on top right parenthesis thin space space space space space equals space 6 space straight i with hat on top space minus space 8 space straight j with hat on top space plus space 2 straight k with hat on top Now comma space angular space momentum space of space particle space is comma space straight L with rightwards harpoon with barb upwards on top space equals space straight r with rightwards harpoon with barb upwards on top space straight x space straight p with rightwards harpoon with barb upwards on top space equals open vertical bar table row cell straight i with hat on top end cell cell straight j with hat on top end cell straight k row cell negative 2 end cell 4 3 row 6 cell negative 8 end cell 2 end table close vertical bar space space space space space space equals space 32 space straight i with hat on top space plus space 22 space straight j with hat on top space minus space 8 straight k with hat on top

    Question 496
    CBSEENPH11017695
    Wired Faculty App

    A particle of mass m is projected from the ground with velocity u at angle 0 with horizontal. Find the angular momentum of the particle about the point of projection at any instant. Deduce the result when a particle will be at the highest point. 
    Solution
    Consider the point of projection as origin, the horizontal direction as X-axis and vertically upward direction positive direction as Y-axis.

    The space velocity space of space particle space at space any space instant space straight t space is comma space straight v with rightwards harpoon with barb upwards on top space equals space straight u space cos space straight theta space straight i with hat on top space plus space left parenthesis space straight u space sinθ space minus space straight t right parenthesis thin space straight j with hat on top The space position space of space the space particle space at space any space instant space is comma space straight r with rightwards harpoon with barb upwards on top space equals space left parenthesis space straight u space cos space θt right parenthesis space straight i with hat on top space minus space left parenthesis space straight u space sinθt space minus space 1 half gt squared right parenthesis space straight j with hat on top space Now comma space the space angular space momentum space of space the space particle space is comma space straight L with rightwards harpoon with barb upwards on top space equals space straight r with rightwards harpoon with barb upwards on top space straight x space straight p with rightwards harpoon with barb upwards on top space equals space straight m space straight r with rightwards harpoon with barb upwards on top straight x space straight v with rightwards harpoon with barb upwards on top space space space equals space straight m open vertical bar table row cell straight i with hat on top end cell cell straight j with hat on top end cell cell straight k with hat on top end cell row ucosθt cell straight u space sinθt space minus space 1 half gt squared end cell cell space 0 space end cell row ucosθ cell usinθ space minus space gt end cell 0 end table close vertical bar space space space space equals space minus space 1 half mug space cosθ space straight t squared space straight k with hat on top space Time space at space which space the space particle space will space be space at space the space highest space point comma space straight t space equals space fraction numerator straight u space sin space straight theta over denominator straight g end fraction
    Therefore, angular momentum of particle when it will be at highest point is,
    L with rightwards harpoon with barb upwards on top space equals negative 1 half m u g space space c o s space 0 space straight t squared space space straight k with hat on top space space space equals space minus 1 half m u g space c o s space straight theta space open parentheses fraction numerator straight u space s i n space straight theta over denominator straight g end fraction close parentheses squared space straight k with hat on top space space space space equals space minus space fraction numerator m u cubed c o s space theta space s i n squared 0 over denominator 2 straight g end fraction straight k with hat on top
    Question 497
    CBSEENPH11017697
    Wired Faculty App

    An electron of mass 9.1 x 10-31 kg revolves in a circle of radius 0.529 Å around the nucleus with speed of 2.2x106 m/s. Show that its angular momentum is equal to h/2 π, where h is Planck's constant.

    Solution

    Mass of the electron = 9.1 cross times 10-31 kg
    Radius of the circle, r = 0.529 straight A with 0 on top 
    Speed = 2.2 x 106 m/s
    The angular momentum of electron is,
            L = mvr
               = 9.1 x 10-31 x 2.19 x 106 x 0.527 x 10-10 

               = 1.0502 x 10-34 J s
    Also space space space straight h space equals space fraction numerator straight h over denominator 2 straight pi end fraction equals fraction numerator 6.6 cross times 10 to the power of negative 34 end exponent over denominator 2 cross times 3.142 end fraction equals 1.0502 cross times 10 to the power of negative 34 end exponent space Js
    Thus,  L =fraction numerator straight h over denominator 2 straight pi end fraction 

    Question 498
    CBSEENPH11017698
    Wired Faculty App

    A particle of mass 2kg is moving with a velocity 4√10 m/s along the line y-3x+4=0. Find the angular momentum of the particle about the origin.

    Solution
    Here we have,
    Mass of the particle, m = 2kg
    Velocity, v = 4√10 m/s 
    Momentum of the particle, 
    straight p equals mv space equals 2 space cross times 4 square root of 10 space equals space 8 square root of 10 space Ns
    The particle moves along line y-3x + 4 = 0.
    Therefore, the perpendicular distance from origin to line =  fraction numerator 4 over denominator square root of left parenthesis 1 right parenthesis squared plus left parenthesis 3 right parenthesis squared end root end fraction space equals space fraction numerator 4 over denominator square root of 10 end fraction
    straight L space equals space pd space space space space equals space 8 square root of 10 cross times fraction numerator 4 over denominator square root of 10 end fraction space space space space equals space 32 space Nms 
    Question 499
    CBSEENPH11017699
    Wired Faculty App

    What will be the duration of the day if the earth suddenly shrinks to 1/64th of its original volume but mass remains unchanged? 

    Solution

    If the volume of earth suddenly shrinks to 1/64th of its original volume then the radius of the earth will decrease to 1/4th of its original radius.

    According to conservation of angular momentum,
    space space space space space space space space space space space space space space space space space space space space space space space space space space space space space straight I space straight ϖ space equals space constant rightwards double arrow space space open parentheses 2 over 5 MR squared close parentheses space fraction numerator 2 straight pi over denominator straight T end fraction equals 2 over 5 straight M open parentheses straight R over 4 close parentheses squared fraction numerator 2 straight pi over denominator straight T apostrophe end fraction space rightwards double arrow space space space space space space space space space space space space space space space space space space space space space space space 1 over straight T equals fraction numerator 1 over denominator 16 straight T apostrophe end fraction space rightwards double arrow space space space space space space space space space space space space space space space space space space space space space space space space space straight T apostrophe equals straight T over 15 space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space equals 24 over 16 hr equals 1.5 space hr space 
    So, duration of the day will be 1.5 hrs. 

    Question 500
    CBSEENPH11017701
    Wired Faculty App

    Two satellites of equal masses are orbiting the earth in different radii such that time period of revolutions are T and 2T. What is the ratio of moment of inertia of two satellites about the centre of the earth respectively?

    Solution
    Let R1 and R2 be the radii of orbits of satellites having time period T and 2T. 
                   

    According to Kepler's third law, 

    T2 ∝ R3 
    space space space therefore space space space space space space space open parentheses straight R subscript 1 over straight R subscript 2 close parentheses cubed equals open parentheses straight T subscript 1 over straight T subscript 2 close parentheses squared equals open parentheses fraction numerator straight T over denominator 2 straight T end fraction close parentheses squared equals 1 fourth therefore space space space space space space space space open parentheses straight R subscript 1 over straight R subscript 2 close parentheses squared equals open parentheses 1 fourth close parentheses to the power of 2 divided by 3 end exponent equals open parentheses 1 half close parentheses to the power of 4 divided by 3 end exponent 


    Ratio of the moment of inertia of first to second satellite is,
    straight I subscript 1 over straight I subscript 2 equals mR subscript 1 squared over mR subscript 2 squared equals open parentheses straight R subscript 1 over straight R subscript 2 close parentheses squared equals open parentheses 1 half close parentheses to the power of 4 divided by 3 end exponent

    Question 501
    CBSEENPH11017702
    Wired Faculty App

    A solid cylinder of mass 20 kg rotates about its axis with angular velocity 100 rad/s. The radius of cylinder is 0-25m. Find the magnitude of angular momentum and kinetic energy associated with cylinder.

    Solution

    Given,
    Mass of cylinder, m = 20 kg

    Radius of cylinder, r = 0.25m

    Angular velocity of cylinder, ω = 100 rad/s
    The moment of inertia of cylinder about its axis is,
          straight I equals 1 half mr squared equals 1 half cross times 20 cross times left parenthesis 0.25 right parenthesis squared
           = 0.625 kg m2 
    Now angular momentum of cylinder is, 
    L = I w
       = 0.625 x 100
       = 62.5 Js 
    Kinetic energy of cylinder is,
    straight K. straight E space equals 1 half 1 straight ϖ squared space space space space space space space space equals 1 half straight x 0.625 straight x open parentheses 100 close parentheses squared space space space space space space space space equals 3125 space straight J 

    Question 502
    CBSEENPH11017707
    Wired Faculty App

    A ring is rolling on a horizontal surface without slipping. Find the ratio of translation kinetic energy to rotational kinetic energy of ring.

    Solution

    Let m be the mass and r be the radius of ring.
    When the ring rolls without slipping, then v = rω,
    where v is the velocity of centre of mass of ring, and
    ω is its angular velocity.

    Now the translational kinetic energy of the ring is,
    straight K subscript straight t space equals space 1 half mv squared space space space space space equals 1 half mr squared straight ϖ squared space
    Rotational kinetic energy of the ring is,
    straight K subscript straight r space equals 1 half Iϖ to the power of 2 space end exponent space space space space space equals space 1 half mr squared straight ϖ squared Therefore comma space the space ratio space o f space translational space to space rotational straight K. straight E space is space given space by comma space therefore space space straight K subscript t over straight K subscript r equals 1

    Question 503
    CBSEENPH11017713
    Wired Faculty App

    A spherical solid ball rolls on a table. What fraction of its total kinetic energy is rotational kinetic energy?

    Solution

    When a body rolls without slipping, it possesses both rotational kinetic energy and translational kinetic energy.
    Total kinetic energy of solid ball is,
    K.E = Kt + Kr
    straight K. straight E space equals space 1 half space Iω squared space plus space 1 half mv squared space space space space space space space space equals 1 half open parentheses 2 over 5 MR squared close parentheses straight omega squared space plus space 1 half mv squared space space space space space space equals space 1 fifth MR squared space straight omega squared space plus space 1 half mv squared space space space space space space space equals space 1 fifth mv squared space plus space 1 half mv squared space straight K. straight E space equals space 7 over 10 space mv squared That space is comma space Rotational space straight K. straight E space equals space 1 fifth mv squared space So comma space Total space straight K. straight E space equals space 7 over 10 space mv squared space Fractional space rotational space straight K. straight E space equals space bevelled fraction numerator begin display style 1 fifth end style mv squared over denominator 7 over 10 space mv squared space end fraction space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space equals space 2 over 7 

    Question 504
    CBSEENPH11017730
    Wired Faculty App

    (a) A child stands at the centre of a turntable with his two arms outstretched. The turntable is set rotating with an angular speed of 40 rpm. How much is the angular speed of the child if he folds his hands back and thereby reduces his moment of inertia to 2/5 times the initial value? Assume that the turntable rotates without friction.

    (b) Show that the child's new kinetic energy of rotation is more than the initial kinetic energy of rotation. How do you account for this increase in kinetic energy? 
    Solution
    (a)
    Let I1 be the moment of inertia of child and ω1 be the angular velocity of the table when the arms are outstretched.
    Let, I2 be the moment of inertia of child and ω2 be the angular velocity of the table after the child folds his hands back. 
    According to the conservation of angular momentum,
    space space space space space space space space space space space straight I subscript 1 straight omega subscript 1 equals straight I subscript 2 straight omega subscript 2 space Here comma space straight omega subscript 1 equals 40 space rpm space and space straight I subscript 2 equals 2 over 5 straight I subscript 1 space therefore space space straight I subscript 1 space left parenthesis 40 right parenthesis space equals 2 over 5 straight I subscript 1 straight omega subscript 2 space rightwards double arrow space space space space space space space space straight omega subscript 2 equals 100 space rpm 
    (b)
    Initial kinetic energy of child is, 
                 <pre>uncaught exception: <b>mkdir(): Permission denied (errno: 2) in /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/util/sys/Store.class.php at line #56mkdir(): Permission denied</b><br /><br />in file: /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/util/sys/Store.class.php line 56<br />#0 [internal function]: _hx_error_handler(2, 'mkdir(): Permis...', '/home/config_ad...', 56, Array) #1 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/util/sys/Store.class.php(56): mkdir('/home/config_ad...', 493) #2 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/plugin/impl/FolderTreeStorageAndCache.class.php(110): com_wiris_util_sys_Store->mkdirs() #3 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/plugin/impl/RenderImpl.class.php(231): com_wiris_plugin_impl_FolderTreeStorageAndCache->codeDigest('mml=<math xmlns...') #4 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/plugin/impl/TextServiceImpl.class.php(59): com_wiris_plugin_impl_RenderImpl->computeDigest(NULL, Array) #5 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/service.php(19): com_wiris_plugin_impl_TextServiceImpl->service('mathml2accessib...', Array) #6 {main}</pre> 
    Kinetic energy of child after folding his hands is,
    straight K subscript 2 equals 1 half straight I subscript 1 straight omega subscript 1 squared 
    therefore space space straight K subscript 2 over straight K subscript 1 equals fraction numerator straight I subscript 2 straight omega subscript 2 squared over denominator straight I subscript 1 straight omega subscript 1 squared end fraction equals 2 over 5 open parentheses 100 over 40 close parentheses squared equals 5 over 2 rightwards double arrow space space space space straight K subscript 2 equals 5 over 2 straight K subscript 1
    That is, the new kinetic energy of child becomes 2.5 times the initial kinetic energy and this increase in kinetic energy is at the cost of internal muscular energy of the child.
    Question 505
    CBSEENPH11017735
    Wired Faculty App

    Torques of equal magnitude are applied to a hollow cylinder and a solid sphere, both having the same mass and radius. The cylinder is free to rotate about its standard axis of symmetry, and the sphere is free to rotate about an axis passing through its centre. Which of the two will acquire a greater angular speed after a given time? 

    Solution

    Torque is given by,
                       straight tau space equals space Iα
    therefore space space straight alpha space equals space straight tau over straight I 
    And using the equation of motion for rotational motion, 
    space space space space space space space space space space space space space space space space space space space space space space space space space space space straight omega equals straight omega subscript 0 space plus alpha t space If space straight omega subscript 0 equals 0 comma space Then space space space space space space space space space space space space space space space space space straight omega equals alpha t equals straight tau over straight I straight t

    So, on two different bodies, if the same torque is applied for the same time, the body, which has a smaller moment of inertia, acquires greater angular speed.
    We know that the moment of inertia of sphere is less than that of the hollow cylinder.
    Therefore, the solid sphere will acquire greater angular speed.

    Question 506
    CBSEENPH11017738
    Wired Faculty App

    A solid cylinder rolls down the inclined plane from a height 0.6m. Find the speed of cylinder at the bottom.

    Solution

    Let,
    m be the mass of the cylinder,
    r be the radius of the cylinder,
    v be the speed, and
    ω be the angular velocity of cylinder at the bottom.
    Therefore using the principle of conservation of energy,
     mgh space equals space 1 half space mv squared space plus space 1 half space Iω squared space space space space space space space space space space space equals space 1 half space mv squared space plus space 1 half open parentheses 1 half straight m space straight r squared space close parentheses space straight omega squared space space space space space space space space space space equals space space 1 half space mv squared space plus space 1 fourth straight m space straight r squared space space straight omega squared space space space space space space space space equals space 1 half space mv squared space space plus space 1 fourth straight m space straight v squared space space space space space space space space space equals space 3 over 4 space mv squared space rightwards double arrow space straight v space equals square root of 4 over 3 gh end root space We space have space here comma space straight h space equals space 0.6 space Therefore comma space straight v space equals space square root of 8 space equals space 2 square root of 2 space straight m divided by straight s

    Question 507
    CBSEENPH11017741
    Wired Faculty App

    By using law of conservation of energy show that acceleration of solid sphere rolling down the inclined plane without nslipping is 5/7 g sin θ.

    Solution
    Let, a solid sphere of mass m and radius R roll down the inclined plane of inclination θ and height h. 
                  
    Assume that, v is the velocity of the sphere when it reaches the ground.
    According to the law of conservation of energy,
    mgh space equals space 1 half space Iω squared space plus space 1 half space mv squared space space space space space space space space space equals space open parentheses 2 over 5 mR squared space straight omega squared close parentheses space plus space 1 half space mv squared space space space space space space space space equals space 1 fifth space mv squared space plus space 1 half space mv squared space space space space space space space space equals space 7 over 10 space mv squared 
    rightwards double arrow space space space space space straight v squared space equals space 10 over 7 space gh space space space space space space space space space space space space space space equals space 10 over 7 space gs space sin space straight theta space space space space space space space space space space space space space space space space space space space space space space... space left parenthesis 1 right parenthesis thin space
    where,
    s is the distance moved by the sphere along the inclined plane.
    If a is the acceleration of sphere with which it rolls down, then velocity at bottom will be,
                    straight v squared equals 2 as space space space space space space space space space space space space space space space space space space space space space space space space space space space space... left parenthesis 2 right parenthesis
    From (1) and (2), we have,
    10 over 7 gs space sin space straight theta space equals space 2 as rightwards double arrow space space space space straight alpha equals 5 over 7 straight g space sin space straight theta
    Question 508
    CBSEENPH11017750
    Wired Faculty App

    Is moment of inertia a scalar or a vector?

    Solution
    Moment of inertia is neither scalar nor vector, but it is a tensor quantity. 
    Question 509
    CBSEENPH11017751
    Wired Faculty App

    What is the moment of inertia of a point mass about an axis at a distance r?

    Solution
    Moment of Inertia of mass m about an axis at a distance r = mr2.
    Question 510
    CBSEENPH11017759
    Wired Faculty App

    A carpet of mass M made of inextensible material is rolled along its length in the form of a cylinder of radius R and kept on the floor. The carpet starts unrolling without sliding on the floor, when a negligible small push is given to it. Calculate the horizontal velocity of the axis of a cylindrical part of carpet, when its radius decreases to R/2.
    Solution
    The situation is illustrated as in figure below.
               
    Here, mass of the carpet is M.
    When the carpet unrolls and radius decreases to half, the mass of left unrolled carpet is, 
    straight m equals straight M over πR squared straight pi open parentheses straight R over 2 close parentheses squared equals straight M over 4 
    Initial  total energy of carpet is, 
    straight E subscript 1 equals straight K subscript rot plus straight K subscript tran plus straight P. straight E. space space space space equals 0 plus 0 plus Mg straight R over 2 space space space equals 1 half MgR 
    Let, v be the linear velocity, and
    straight omega be the angular velocity of unrolled carpet, when it becomes a cylinder of radius R/2.
    Now the total energy, 
    straight E subscript 2 space equals straight K subscript rot plus straight K subscript tran plus straight P. straight E space space space space space space equals 1 half Iω squared plus 1 half mv squared plus mg straight R over 2 space space space space space equals 1 half open square brackets 1 half straight m open parentheses straight R over 2 close parentheses squared close square brackets open parentheses fraction numerator straight v over denominator straight R divided by 2 end fraction close parentheses squared plus 1 half mv squared plus mg straight R over 2
    equals 3 over 4 mv squared plus 1 half mgR equals 3 over 16 Mv squared plus 1 over 8 MgR
    Using conservation of energy:
    3 over 16 Mv squared plus 1 over 8 MgR equals MgR rightwards double arrow space space space space space space space space space space space space space space space space space space space space space space space space space space straight v equals square root of 14 over 3 gR end root 
    This is the required horizontal velocity of the axis of a cylindrical part of carpet. 
    Question 511
    CBSEENPH11017761
    Wired Faculty App

    A thin circular ring of mass m and radius r is rotating about its axis with constant angular velocity ω. Three point masses each m are placed symmetrically on the ring gently. Find the angular velocity of ring after placing the masses.

    Solution

    Given, 
    Mass of the ring = m
    Radius of the ring = r
    Angular velocity = straight omega
    The moment of inertia of ring about its axis is;
                                I = mr2 
    Given that, three point masses  are placed symmetrically on the ring.
    The angular momentum of ring before masses are placed is,
                            straight L equals mr squared straight omega

    The moment of inertia of ring after three masses are placed on it is, 
                             straight I apostrophe equals straight I subscript ring plus straight I subscript masses space space space equals space mr squared plus 3 left parenthesis mr squared right parenthesis space space equals space 4 space mr squared 
    Let space space straight omega apostrophe be the angular velocity of the system after placing the masses.
    The angular momentum of system now is,
             <pre>uncaught exception: <b>mkdir(): Permission denied (errno: 2) in /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/util/sys/Store.class.php at line #56mkdir(): Permission denied</b><br /><br />in file: /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/util/sys/Store.class.php line 56<br />#0 [internal function]: _hx_error_handler(2, 'mkdir(): Permis...', '/home/config_ad...', 56, Array) #1 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/util/sys/Store.class.php(56): mkdir('/home/config_ad...', 493) #2 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/plugin/impl/FolderTreeStorageAndCache.class.php(110): com_wiris_util_sys_Store->mkdirs() #3 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/plugin/impl/RenderImpl.class.php(231): com_wiris_plugin_impl_FolderTreeStorageAndCache->codeDigest('mml=<math xmlns...') #4 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/plugin/impl/TextServiceImpl.class.php(59): com_wiris_plugin_impl_RenderImpl->computeDigest(NULL, Array) #5 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/service.php(19): com_wiris_plugin_impl_TextServiceImpl->service('mathml2accessib...', Array) #6 {main}</pre>
    Using conservation of angular momentum, 
        space space space space space space space space space space space space space space space space space 4 space mr squared straight omega apostrophe space equals space mr squared straight omega space space space space space space space space space space space space space space space space rightwards double arrow space space space space space space space space straight omega apostrophe equals straight omega divided by 4 space space space space space space space space space 
    Question 513
    CBSEENPH11017769
    Wired Faculty App

    A rod of mass M and length L is hinged at point O. A small bullet of mass m hits the rod with velocity u as shown in figure. The bullet gets embedded in the rod. Find the angular velocity of system after impact.

    Solution
    Before the collision, the rod is at rest and bullet is moving with velocity u at a distance 0.8L.
    Therefore angular momentum of system before impact is,

    straight L subscript 1 space equals 1 third ML squared left parenthesis 0 right parenthesis plus mu left parenthesis 0.8 straight L right parenthesis space space space space equals 0.8 space muL space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space... left parenthesis 1 right parenthesis

    Bullet after hitting gets embedded into the rod.
    Let straight omega be the angular velocity of the system after impact.
    The angular momentum of system after impact is,
    straight L subscript 2 space equals open square brackets 1 third ML squared plus straight m left parenthesis 0.8 straight L right parenthesis squared close square brackets straight omega space space space space space equals space open square brackets 1 third straight M plus 0.64 straight m close square brackets straight L squared straight omega space space space space space space space space space space space space space space space space space space space space space... left parenthesis 2 right parenthesis 
    Using the law of conservation of momentum
    space space space space space space space space space 0.8 space muL space equals space open square brackets 1 third straight M plus 0.64 straight m close square brackets straight L squared straight omega space rightwards double arrow space space space space space space space space space space space space space space space space space straight omega equals space fraction numerator 0.8 muL over denominator open square brackets begin display style 1 third straight M plus 0.64 straight m end style close square brackets straight L squared end fraction space space space space space space space space space space space space space space space space space space space space space space space space space equals fraction numerator 2.4 mu over denominator open square brackets straight M plus 1.92 straight m close square brackets straight L end fraction 
    This is the angular velocity of the system after the impact. 

    Question 514
    CBSEENPH11017775
    Wired Faculty App

    A cylinder of mass m, radius R and moment of inertia I is placed on a rough inclined plane. What should be the minimum coefficient of friction between cylinder and inclined plane so that cylinder rolls down the inclined plane without slipping?

    Solution

    Consider a cylinder placed on the inclined plane inclined at angle θ the with the horizontal.
    The various forces acting on the cylinder are:

    (i) Weight mg acting vertically downward.

    (ii) The friction F between the cylinder and surface of inclined plane.
                     
    On resolving the components of weight mg, along and perpendicular to the inclined plane we have mg sinθand mg cosθ respectively.
    Let a be the acceleration with which cylinder rolls down the inclined plane.

    Equation of linear motion down the inclined plane is,
    ma space equals space mg space sinθ space minus space straight F space space space space space space space space space space space space space space space space space space space... left parenthesis 1 right parenthesis
    Since the cylinder rolls due to the force of friction,
    therefore space space space space straight I space straight alpha equals FR space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space... left parenthesis 2 right parenthesis
    where alpha is angular acceleration of cylinder. 
    We know,
    straight alpha space equals space straight alpha over straight R space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space... left parenthesis 3 right parenthesis space therefore space space space space straight I straight alpha over straight R equals FR space rightwards double arrow space space space space space space space space straight F equals Iα over straight R squared space space space space space space space space space space space space space space space space space space space space... left parenthesis 4 right parenthesis 
    Substituting the value of F in equation (1), we get
    space space space space space space space space space space space space ma space equals space mg space sin space straight theta space minus Iα over straight R squared space rightwards double arrow space space space space space space space space space space space straight a space equals gsinθ minus Iα over mR squared space rightwards double arrow space space space space space space space space space space space space straight alpha equals fraction numerator straight g space sin space straight theta over denominator 1 plus begin display style straight I over mR squared end style end fraction 
    Substituting a  in (4),  we get,
    straight F space equals 1 over straight R squared fraction numerator straight g space sin space straight theta over denominator 1 plus begin display style straight I over mR squared end style end fraction space space space space equals fraction numerator mg space sin space straight theta over denominator 1 plus begin display style mR squared over straight I end style end fraction
    There will be no slipping if and only if
    Coefficient space of space friction comma space space straight mu greater or equal than straight F over straight R
    Therefore minimum coefficient of friction straight mu is, 
    space space space space space space space space straight mu space equals straight F over straight R equals fraction numerator mg space sinθ over denominator 1 plus begin display style mR squared over straight I end style end fraction cross times fraction numerator 1 over denominator mg space cosθ end fraction space space space space space space space space space space space equals fraction numerator tan space straight theta over denominator begin display style mR squared over straight I plus 1 end style end fraction space space rightwards double arrow space space space straight mu space equals space fraction numerator tan space straight theta over denominator begin display style mR squared over straight I plus 1 end style end fraction 

    Question 515
    CBSEENPH11017777
    Wired Faculty App

    A rope is wound round a hollow cylinder of mass 3 kg and radius 40 cm. What is the angular acceleration of the cylinder if the rope is pulled with a force of 30N? What is the linear acceleration of the rope?

    Solution

    We have,
    Mass of cylinder, m=3 kg
    Radius of the cylinder, r = 40 cm = 0.4 m

    Moment of inertia, I = mr2 = 3 (0.4)2 = 0.48 kg m2
    Since the rope leaves the cylinder tangential, therefore torque by tension in string is, 
    straight tau space equals space rF space equals space 0.4 space straight x space 30 space equals space 12 space Nm
    Also, 
    straight tau space equals thin space straight I space straight alpha
    Therefore, 
    straight alpha space equals space straight tau over straight I space equals space fraction numerator 12 over denominator 0.48 end fraction space equals space 25 space rad divided by sec squared
    Linear acceleration of rope is, 
    straight a space equals space straight r space straight alpha space equals space 0.4 space straight x space 25 space equals space 10 space straight m divided by straight s squared

    Question 516
    CBSEENPH11017779
    Wired Faculty App

    A flywheel of moment of inertia 0.04kgm2 and radius 20cm is mounted to rotate about a horizontal axis through its centre. A string of negligible mass is wrapped around the flywheel and mass of 0.2 kg is attached to the free end. Find how much the mass would fall in 3seconds?

    Solution

    Given,
    Moment of Inertia, I = 0.04 kgm2

    Radius, r=20cm=0.2m,  
    Mass, m=0.2kg
    The acceleration of mass is, 
    straight alpha equals fraction numerator straight g over denominator 1 plus begin display style straight I over mr squared end style end fraction equals fraction numerator 10 over denominator 1 plus begin display style fraction numerator 0.04 over denominator 0.2 cross times left parenthesis 0.2 right parenthesis squared end fraction end style end fraction equals 5 over 3 straight m divided by straight s squared
    Distance covered by the mass in 3s while falling is,
    <pre>uncaught exception: <b>mkdir(): Permission denied (errno: 2) in /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/util/sys/Store.class.php at line #56mkdir(): Permission denied</b><br /><br />in file: /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/util/sys/Store.class.php line 56<br />#0 [internal function]: _hx_error_handler(2, 'mkdir(): Permis...', '/home/config_ad...', 56, Array) #1 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/util/sys/Store.class.php(56): mkdir('/home/config_ad...', 493) #2 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/plugin/impl/FolderTreeStorageAndCache.class.php(110): com_wiris_util_sys_Store->mkdirs() #3 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/plugin/impl/RenderImpl.class.php(231): com_wiris_plugin_impl_FolderTreeStorageAndCache->codeDigest('mml=<math xmlns...') #4 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/plugin/impl/TextServiceImpl.class.php(59): com_wiris_plugin_impl_RenderImpl->computeDigest(NULL, Array) #5 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/service.php(19): com_wiris_plugin_impl_TextServiceImpl->service('mathml2accessib...', Array) #6 {main}</pre>

    Question 517
    CBSEENPH11017781
    Wired Faculty App

    A ring, disc and sphere all of same mass and radius roll down the inclined plane from the same height h. Which of the three reaches the bottom:

    (i) earliest (ii) last?
    Solution

    The body having maximum acceleration reaches the ground earliest and the body whose acceleration is minimum reaches the ground at last.

    We know acceleration of body rolling down the inclined plane without sliding is given by,
                    straight alpha equals fraction numerator straight g space sinθ over denominator 1 plus begin display style 1 over MR squared end style end fraction 
    Moment of inertia of ring = MR2 
    Moment of Inertia of the disc =<pre>uncaught exception: <b>mkdir(): Permission denied (errno: 2) in /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/util/sys/Store.class.php at line #56mkdir(): Permission denied</b><br /><br />in file: /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/util/sys/Store.class.php line 56<br />#0 [internal function]: _hx_error_handler(2, 'mkdir(): Permis...', '/home/config_ad...', 56, Array) #1 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/util/sys/Store.class.php(56): mkdir('/home/config_ad...', 493) #2 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/plugin/impl/FolderTreeStorageAndCache.class.php(110): com_wiris_util_sys_Store->mkdirs() #3 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/plugin/impl/RenderImpl.class.php(231): com_wiris_plugin_impl_FolderTreeStorageAndCache->codeDigest('mml=<math xmlns...') #4 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/plugin/impl/TextServiceImpl.class.php(59): com_wiris_plugin_impl_RenderImpl->computeDigest(NULL, Array) #5 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/service.php(19): com_wiris_plugin_impl_TextServiceImpl->service('mathml2accessib...', Array) #6 {main}</pre> 
    Moment on Inertia = <pre>uncaught exception: <b>mkdir(): Permission denied (errno: 2) in /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/util/sys/Store.class.php at line #56mkdir(): Permission denied</b><br /><br />in file: /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/util/sys/Store.class.php line 56<br />#0 [internal function]: _hx_error_handler(2, 'mkdir(): Permis...', '/home/config_ad...', 56, Array) #1 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/util/sys/Store.class.php(56): mkdir('/home/config_ad...', 493) #2 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/plugin/impl/FolderTreeStorageAndCache.class.php(110): com_wiris_util_sys_Store->mkdirs() #3 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/plugin/impl/RenderImpl.class.php(231): com_wiris_plugin_impl_FolderTreeStorageAndCache->codeDigest('mml=<math xmlns...') #4 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/plugin/impl/TextServiceImpl.class.php(59): com_wiris_plugin_impl_RenderImpl->computeDigest(NULL, Array) #5 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/service.php(19): com_wiris_plugin_impl_TextServiceImpl->service('mathml2accessib...', Array) #6 {main}</pre>
    therefore space space space space straight alpha subscript ring equals 1 half straight g space sinθ space space space space space space space straight alpha subscript disc equals 2 over 3 space straight g space sinθ space space space space space space space straight alpha subscript sphere equals 5 over 7 straight g space sin space straight theta
    Since the acceleration of sphere is maximum and that of the ring is minimum, therefore, the sphere will reach the ground the earliest and ring at last while starting from the same height. 

    Question 518
    CBSEENPH11017785
    Wired Faculty App

    A solid sphere rolls down an inclined plane with an acceleration g/2. What is the angle of inclination of inclined plane?

    Solution
    The acceleration of sphere rolling down an inclined plane is given by, 

                        straight alpha equals fraction numerator straight g space sin space straight theta over denominator 1 plus begin display style straight I over mr squared end style end fraction
    For solid sphere,space space space space straight I equals 2 over 5 mr squared
    therefore space space space space space space space space space space straight g over 2 equals fraction numerator straight g space s i n space straight theta over denominator 1 plus begin display style 2 over 5 end style end fraction equals 5 over 7 space straight g space s i n space straight theta rightwards double arrow space space space space s i n space straight theta space equals space 7 over 10 equals 0.7 rightwards double arrow space space space space space space space space space space straight theta space equals space space s i n to the power of negative 1 end exponent 0.7 
    This is the required angle of inclination of the inclined plane. 
    Question 519
    CBSEENPH11017787
    Wired Faculty App

    A disc rolls down the inclined plane of 30°. Find the acceleration of cylinder. Also find the angular acceleration of disc. The radius of disc is 10cm.

    Solution
    The acceleration of disc rolling down the inclined plane is given by,
    straight alpha equals fraction numerator straight g space sinθ over denominator 1 plus begin display style 1 over mr squared end style end fraction Here comma space space space space straight theta equals 30 degree comma space space space space space straight I equals 1 half mr squared therefore space space space straight alpha equals fraction numerator straight g space sin space 30 degree over denominator 1 plus begin display style 1 half end style end fraction equals straight g over 3 equals 3.27 space straight m divided by straight s squared
    The angular acceleration of disc is, 
    <pre>uncaught exception: <b>mkdir(): Permission denied (errno: 2) in /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/util/sys/Store.class.php at line #56mkdir(): Permission denied</b><br /><br />in file: /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/util/sys/Store.class.php line 56<br />#0 [internal function]: _hx_error_handler(2, 'mkdir(): Permis...', '/home/config_ad...', 56, Array) #1 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/util/sys/Store.class.php(56): mkdir('/home/config_ad...', 493) #2 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/plugin/impl/FolderTreeStorageAndCache.class.php(110): com_wiris_util_sys_Store->mkdirs() #3 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/plugin/impl/RenderImpl.class.php(231): com_wiris_plugin_impl_FolderTreeStorageAndCache->codeDigest('mml=<math xmlns...') #4 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/plugin/impl/TextServiceImpl.class.php(59): com_wiris_plugin_impl_RenderImpl->computeDigest(NULL, Array) #5 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/service.php(19): com_wiris_plugin_impl_TextServiceImpl->service('mathml2accessib...', Array) #6 {main}</pre> 
    Question 520
    CBSEENPH11017788
    Wired Faculty App

    A body of mass m, radius R, and radius of gyration K slides down a smooth inclined plane of inclination θ from height h and reach the bottom with speed v. Now the same inclined plane is made rough and same body rolls down the inclined plane without slipping and reach the ground with speed V. Find the ration v: V.

    Solution

    Acceleration of body when it slides down the inclined plane without rolling is,

                     a1 =gsinθ

    The speed of the body when it reaches the ground by slipping is,
                       straight v equals square root of 2 straight a subscript 1 straight I end root equals square root of 2 gIsin space straight theta end root 
    The acceleration of body when it rolls down the inclined plane without slipping is,
                 
    straight alpha subscript 2 space equals fraction numerator straight g space sin space straight theta over denominator 1 plus begin display style 1 over mR squared end style end fraction space space space space space space equals fraction numerator straight g space sin space straight theta over denominator 1 plus begin display style mK squared over mR squared end style end fraction space space space space space space equals fraction numerator straight g space sin space straight theta over denominator 1 plus begin display style straight K squared over straight R squared end style end fraction 
    The speed of the body when it reaches the ground by rolling is,
    straight V equals square root of 2 straight a subscript 1 straight I end root equals square root of fraction numerator 2 gI space sinθ over denominator 1 plus begin display style straight K squared over straight R squared end style end fraction end root space Now comma space the space ratio space is space given space by comma straight v over straight V equals square root of 1 plus straight K squared over straight R squared end root equals square root of fraction numerator straight K squared plus straight R squared over denominator straight R squared end fraction end root 

    Question 521
    CBSEENPH11019864
    Wired Faculty App

    Fill in the blanks: 

    a) The volume of a cube of side 1 cm is equal to .....m3
    Solution

    10-6 

    1 cm = 1 space cm space equals space 1 over 100 space straight m So comma space volume space of space the space cube space equals space 1 space cm cubed space straight i. straight e. comma space 1 space cm cubed space equals space open parentheses 1 over 100 close parentheses cross times space open parentheses 1 over 100 close parentheses cross times space open parentheses 1 over 100 close parentheses space straight m space therefore space 1 space cm cubed space equals space 10 to the power of negative 6 end exponent space straight m cubed So space volume space of space straight a space cube space of space side space 1 space cm is space equal space to space 10 to the power of negative 6 end exponent space cm cubed.

    Question 522
    CBSEENPH11019865
    Wired Faculty App

    The surface area of a solid cylinder of radius 2.0 cm and height 10.0 cm is equal to
    ...(mm)2

    Solution
    1.5 x 104 mm2

    The total surface area of a cylinder of radius r and height h is

    S = 2πr (r + h).
    Given space that comma space straight r space equals space 2 space cm space equals space 2 cross times 10 space mm space equals space 20 space mm space straight h space equals space 10 space cm space equals 10 space cross times 10 space mm space equals space 100 space mm space So comma space Surface space area comma space straight S space equals space 2 cross times 3.14 cross times 20 cross times left parenthesis 20 plus 100 right parenthesis space space space space space space space space space space space space space space space space space space space space space space space space space space space space space equals space 15072 space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space equals space 1.5 space cross times 10 to the power of 4 space mm squared

    Question 526
    CBSEENPH11019869
    Wired Faculty App

    Fill in the blanks by suitable conversion of unit:

    b) 1 m = ..... ly 

    Solution
     1.057 x 10-16
    Question 527
    CBSEENPH11019870
    Wired Faculty App

    Fill in the blanks by suitable conversion of units:

    c) 3.0 m s–2= .... km h –1

    Solution
    3.88 x 10-4 km h-2

    Tips: -

    1 space straight m space equals space 10 to the power of negative 3 end exponent space km 1 space straight s space equals space fraction numerator 1 over denominator 3600 space end fraction space straight h space rightwards double arrow space 1 space straight s to the power of negative 1 end exponent space equals space 3600 space straight h to the power of negative 1 end exponent space 1 space straight s to the power of negative 2 end exponent space equals left parenthesis 3600 right parenthesis squared space straight h to the power of negative 2 end exponent therefore space 3 space straight m space straight s to the power of negative 2 end exponent space equals left parenthesis 3 cross times 10 to the power of negative 3 end exponent space km right parenthesis space cross times left parenthesis 3600 right parenthesis squared space straight h to the power of negative 2 end exponent space space space space space space space space space space space space space space space space space space space space equals space 3.88 space cross times space 10 to the power of negative 4 end exponent space km space straight h to the power of negative 2 end exponent
    Question 529
    CBSEENPH11019872
    Wired Faculty App

    Explain this statement clearly : 

    “To call a dimensional quantity ‘large’ or ‘small’ is meaningless without specifying a
    standard for comparison”. In view of this, reframe the following statements wherever
    necessary : 

    (a) atoms are very small objects 

    (b) a jet plane moves with great speed 

    (c) the mass of Jupiter is very large 

    (d) the air inside this room contains a large number of molecules 

    (e) a proton is much more massive than an electron 

    (f) the speed of sound is much smaller than the speed of light

    Solution

    The given statement is true because a dimensionless quantity may be large or small in comparison to some standard reference. For example, the coefficient of friction is dimensionless. The coefficient of sliding friction is greater than the coefficient of rolling friction, but less than static friction. 

    a) An atom is a very small object in comparison to a soccer ball.

    b) A jet plane moves with a speed greater than that of a bicycle.

    c)  Mass of Jupiter is very large as compared to the mass of a cricket ball.

    d)  The air inside this room contains a large number of molecules as compared to that present in a geometry box.

    e) A proton is more massive than an electron.

    f) Speed of sound is less than the speed of light.

    Question 531
    CBSEENPH11019874
    Wired Faculty App

    Which of the following is the most precise device for measuring length : 

    (a) a vernier callipers with 20 divisions on the sliding scale

    b) a screw gauge of pitch 1 mm and 100 divisions on the circular scale

    (c) an optical instrument that can measure length to within a wavelength of light.

    Solution

    A device with minimum count is the most suitable to measure length.
    a) 
    straight L. straight C space of space vernier space calliper space equals space 1 space straight S. straight D space minus space 1 space straight V. straight D space space space space space space space space space space space space space space space space space equals space 1 space minus space 9 over 10 space equals space 1 over 10 space equals space 0.01 space cm  
    b) 
    straight L. straight C space of space screw space gauge space equals space fraction numerator Pitch over denominator Number space of space divisions end fraction space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space equals space 1 over 1000 space equals space 0.001 space cm space straight c right parenthesis straight L. straight C space of space an space optical space device space equals space Wavelength space of space light space tilde 10 to the power of negative 5 end exponent space cm space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space equals space 0.00001 space cm 

    Hence, it can be inferred that an optical instrument is the most suitable device to measure length.



    Question 532
    CBSEENPH11019875
    Wired Faculty App

    A student measures the thickness of a human hair by looking at it through a microscope of magnification 100. He makes 20 observations and finds that the average width of the hair in the field of view of the microscope is 3.5 mm. What is the estimate on the thickness of hair ?

    Solution

    Magnification of the microscope = 100 
    Average width of hair in the field of view of the microscope = 3.5 mm 
    Therefore, actual thinckness of the hair = 35 over 100 space equals space 0.035 space mm 

    Question 533
    CBSEENPH11019876
    Wired Faculty App

    Answer the following : 

    (a)You are given a thread and a metre scale. How will you estimate the diameter of
    the thread ?

    Solution

    The diameter of the thread is so small that it cannot be measured using a metre scale. 
    Wind a number of turns of threads on the metre scale so that the turns are closely touching one another.
    Measure the length (l) of the windings on the scale which contains n number of turns.
    Error converting from MathML to accessible text.

    Question 534
    CBSEENPH11019877
    Wired Faculty App

    Answer the following: 

    The mean diameter of a thin brass rod is to be measured by vernier callipers. Why is a set of 100 measurements of the diameter expected to yield a more reliable estimate than a set of 5 measurements only 

    Solution

    Random errors involved in a set of 100 measurements are very less as compared to the set of 5 measurements. Therefore, a set of 100 measurements is more reliable than a set of 5 measurements.

    Question 536
    CBSEENPH11019879
    Wired Faculty App

    State the number of significant figures in the following :

    a) 0.007 m2

    b) 2.64 1024kg

    (c) 0.2370 g cm–3

    (d) 6.320 J 

    (e) 6.032 N m     –2 

    (f) 0.0006032 m2

    Solution

    a) 1
    The given quantity is 0.007 m2

    If the number is less than one, then all zeros on the right of the decimal point (but left to the first non-zero) are insignificant. This means that here, two zeros after the decimal are not significant. Hence, only 7 is a significant figure in this quantity. 

    b) 3 

    Here, the power of 10 is irrelevant for the determination of significant figures. Hence, all the digits are significant figures.


    c) 4 
    d) 4
    e) 4
    f) 4

    Question 537
    CBSEENPH11019880
    Wired Faculty App

    The length, breadth and thickness of a rectangular sheet of metal are 4.234 m, 1.005 m, and 2.01 cm respectively. Give the area and volume of the sheet to correct significant figures.

    Solution
    Given, 
    Length, l = 4.234 m 
    Breadth,b = 1.005 m 
    Thickness, t = 2.01 cm = 2.01 × 10-2 m

    Area of the sheet = 2 (l × 0 + b × t + t × l)

                           = 2 (4.234 × 1.005 + 1.005 × 0.0201 + 0.0201 × 4.234)

                           = 2 (4.3604739)

                           = 8.7209478 m2

    Area can contain a maximum of three significant digits, therefore, rounding off, we get

    Area = 8.72 m2

    Also, volume = l × b × t

                   V = 4.234 × 1.005 × 0.0201

                      = 0.0855289
                      = 0.0855 m3 (Significant Figures = 3)

     

    Question 539
    CBSEENPH11019882
    Wired Faculty App

    A book with many printing errors contains four different formulas for the displacement y of a particle undergoing a certain periodic motion:

    left parenthesis straight a right parenthesis space straight y space equals space straight a space sin space 2 straight pi space straight t divided by straight T space left parenthesis straight b right parenthesis space straight y space equals space straight a space sin space vt space left parenthesis straight c right parenthesis space straight y space equals space left parenthesis straight a divided by straight T right parenthesis space sin space straight t divided by straight a straight d right parenthesis space straight y space equals space open parentheses straight a square root of 2 close parentheses space left parenthesis sin space 2 space πt divided by straight T space plus space cos space 2 πt space divided by straight T space right parenthesis 

    (a = maximum displacement of the particle, v = speed of the particle, T = time-period
    of motion).

    Rule out the wrong formulas on dimensional grounds.

    Solution
    The displacement y has the dimension of length.
    Therefore, the formula for it should also have the dimension of length.
    Trigonometric functions are dimensionless and their arguments are also dimensionless.
    Based on these considerations now checking each formula dimensionally, we have
    straight a right parenthesis thin space Dimensionally space correct straight y space equals space straight a space sin space fraction numerator 2 πt over denominator straight T end fraction Dimension space of space straight y space equals space straight M to the power of straight o straight L to the power of 1 straight T to the power of straight o space space Dimension space of space straight a space equals space straight M to the power of straight o straight L to the power of 1 straight T to the power of straight o space Dimsension space of space sin space fraction numerator 2 πt over denominator straight T end fraction space equals space straight M to the power of straight o straight L to the power of straight o straight T to the power of straight o space So comma space dimension space of space straight L. straight H. straight S space equals space Dimension space of space straight R. straight H. straight S rightwards double arrow space given space formula space is space dimensionally space correct. straight b right parenthesis thin space Incorrect space dimensionally. straight y space equals space straight a space sin space vt space Dimension space of space straight y space equals space straight M to the power of 0 space straight L to the power of 1 space end exponent straight T to the power of 0 space space Dimension space of space straight a space equals space straight M to the power of 0 space end exponent straight L to the power of 1 space straight T to the power of 0 space space Dimension space of space vt space equals space straight M to the power of 0 space straight L to the power of 1 space end exponent straight T to the power of – 1 end exponent space cross times space straight M to the power of 0 space end exponent straight L to the power of 0 space end exponent straight T to the power of 1 space end exponent equals space straight M to the power of 0 space straight L to the power of 1 space end exponent straight T to the power of 0 space The space argument space of space the space trigonometric space function space must space be space dimensionless comma space which space is space not space so space in space this space case. space Hence comma space given space formula space is space dimensionally space incorrect. straight c right parenthesis space Incorrect Dimension space of space straight y space equals space straight M to the power of 0 space space straight L to the power of 1 space space straight T to the power of 0 Dimension space of space begin inline style straight a over straight T end style equals space straight M to the power of 0 space end exponent space straight L to the power of 1 space space straight T to the power of – 1 end exponent space Dimension space of space straight t over straight a equals space equals space straight M 0 space straight L to the power of – 1 space end exponent straight T to the power of 1 space Argument space of space the space trignometric space function space is space dimensionless. space Hence comma space the space formula space is space dimensionally space incorrect. straight d right parenthesis thin space Correct Dimension space of space straight y space equals space straight M to the power of 0 space straight L to the power of 1 space end exponent straight T to the power of 0 Dimension space of space straight a space equals straight M to the power of 0 space straight L to the power of 1 space end exponent straight T to the power of 0 space Dimension space of space straight t over straight T space equals space straight M to the power of 0 space end exponent straight L to the power of 0 space end exponent straight T to the power of 0 space Since space the space srgument space of space the space trignometric space function space is space dimensionless comma space the space dimensions space of space straight y space and space straight a space are space the space same. space Hence comma space the given space formula space is space dimensionally space correct. space

    Question 540
    CBSEENPH11019883
    Wired Faculty App

    A famous relation in physics relates ‘moving mass’ m to the ‘rest mass’ moof a  particle in terms of its speed v and the speed of light, c. (This relation first arose as
    a consequence of special relativity due to Albert Einstein). A boy recalls the relation
    almost correctly but forgets where to put the constant c. He writes :

    straight m space equals space fraction numerator straight m subscript straight o over denominator left parenthesis square root of 1 minus straight v squared end root right parenthesis to the power of bevelled 1 half end exponent end fraction

    Guess where to put the missing c.

    Solution

    Given the relation, 
    straight m space equals space fraction numerator straight m subscript straight o over denominator left parenthesis square root of 1 minus straight v squared end root right parenthesis to the power of bevelled 1 half end exponent end fraction 
    Dimension of m = M1 L0 T0
    Dimension of mo = M1 L0 T0 
    Dimension of v = M0 L1 T–1
    Dimension of v2 = M0 L2 T–2
    Dimension of c = M0 L1 T–1

    The given formula will be dimensionally correct only when the dimension of L.H.S is the same as that of R.H.S.
    This is possible when the factor in the denominator is dimensionless. This is only possible if straight nu squared space is space divided space by space straight c squared. 
    Hence the correct relation is 
    straight m space equals space straight m subscript straight o over open parentheses 1 minus begin display style straight v squared over straight c squared end style close parentheses to the power of bevelled 1 half end exponent

    Question 543
    CBSEENPH11019886
    Wired Faculty App

    Explain this common observation clearly : If you look out of the window of a fast moving train, the nearby trees, houses etc. seem to move rapidly in a direction opposite to the train’s motion, but the distant objects (hill tops, the Moon, the stars etc.) seem to be stationary. (In fact, since you are aware that you are moving, these distant objects seem to move with you).

    Solution
    Line of sight is defined as an imaginary line joining an object and an observer's eye. When we observe nearby stationary objects such as trees, houses, etc. while sitting in a moving train, they appear to move rapidly in the opposite direction because the line of sight changes very rapidly.
    On the other hand, distant objects such as trees, stars, etc. appear stationary because of the large distance.
    As a result, the line of sight does not change its direction rapidly.
    Question 546
    CBSEENPH11019889
    Wired Faculty App

    Precise measurements of physical quantities are a need of science. For example, to ascertain the speed of an aircraft, one must have an accurate method to find its positions at closely separated instants of time. This was the actual motivation behind the discovery of radar in World War II. Think of different examples in modern science
    where precise measurements of length, time, mass etc. are needed. Also, wherever you can, give a quantitative idea of the precision needed.

    Solution
    It is indeed very true that precise measurements of physical quantities are essential for the development of science.
    For example, ultra-shot laser pulses (time interval ∼ 10–15 s) are used to measure time intervals in several physical and chemical processes.
    X-ray spectroscopy is used to determine the inter atomic separation or inter-planer spacing. 
    The development of mass spectrometer makes it possible to measure the mass of atoms precisely.
    Question 551
    CBSEENPH11019894
    Wired Faculty App

    Just as precise measurements are necessary in science, it is equally important to be able to make rough estimates of quantities using rudimentary ideas and common observations. Think of ways by which you can estimate the following (where an estimate is difficult to obtain, try to get an upper bound on the quantity):

    (e) the number of air molecules in your classroom.
    Solution

    Let, the volume of the classroom be V.
    One mole of air at NTP occupies 22.4 l.
    i.e., 22.4 × 10–3 m3volume.
    Number of molecules in one mole = 6.023 × 1023 
    ∴ Number of molecules in room of volume V,
    = 6.023 × 1023 × V / 22.4 × 10-3 
     =  134.915 × 1026 V 
    =  1.35 × 1028 V 

    Question 554
    CBSEENPH11019897
    Wired Faculty App

    It is claimed that two cesium clocks, if allowed to run for 100 years, free from any disturbance, may differ by only about 0.02 s. What does this imply for the accuracy of the standard cesium clock in measuring a time-interval of 1 s.

    Solution

    Total time for which the two watches run, T = 100 years
    Diffrence in the time = 0.02 sec 
    Error in 1 sec = fraction numerator 0.02 over denominator 3.1536 cross times 10 to the power of 9 end fraction space = 6.34 space cross times 10 to the power of negative 12 end exponent space straight s space almost equal to 10 to the power of negative 11 end exponent straight s
    Thus, accuracy of 1 part is 1011.

    Question 555
    CBSEENPH11019898
    Wired Faculty App

    Estimate the average mass density of a sodium atom assuming its size to be about 2.5 Å. (Use the known values of Avogadro’s number and the atomic mass of sodium). Compare it with the density of sodium in its crystalline phase : 970 kg m–3. Are the two densities of the same order of magnitude ? If so, why?

    Solution
    Diameter space of space sodium space atom space is space equal space to space the space size space of space sodium space atom. That space is space straight D space equals space 2.5 space straight Å space So comma space radius space of space sodium space atom comma space straight r space equals space left parenthesis 1 divided by 2 right parenthesis space cross times space 2.5 space straight Å space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space equals space 1.25 space straight Å space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space equals space 1.25 space cross times space 10 to the power of negative 10 end exponent space straight m space Volume space of space sodium space atom comma space straight V space equals space 4 over 3 πr cubed space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space equals space 4 over 3 cross times 3.14 cross times left parenthesis 1.25 cross times 10 to the power of negative 10 end exponent right parenthesis cubed space According space to space the space Avogadro space hypothesis comma one space mole space of space sodium space contains space 6.023 space cross times space 10 to the power of 23 space atoms space and space has space straight a space mass space of space 23 space straight g space or space 23 space cross times space 10 to the power of – 3 end exponent space kg. therefore space space Mass space of space one space atom space equals space fraction numerator space 23 space cross times space 10 to the power of – 3 end exponent space over denominator space 6.023 space cross times space 10 to the power of 23 end fraction space equals space straight m subscript 1 space Density space of space sodium space atom comma space straight rho space equals space straight m subscript 1 over straight V space Putting space values space in space this space equation comma space Density space of space sodium space atom comma space straight rho space equals 4.67 space cross times space 10 to the power of negative 3 end exponent space Kg space straight m to the power of negative 3 end exponent space Density space of space sodium space in space crystalline space phase space equals space 970 space kg space straight m to the power of negative 3 end exponent space
    Hence, the density of sodium atom and the density of sodium in its crystalline phase are not in the same order. This is because in solid phase, atoms are closely packed. Thus, the inter-atomic separation is very small in the crystalline phase. 
    Question 556
    CBSEENPH11019899
    Wired Faculty App

    The unit of length convenient on the nuclear scale is a fermi : 1 f = 10–15 m. Nuclear sizes obey roughly the following empirical relation :

    r = r0 A1/3 

    where,

     r is the radius of the nucleus, A its mass number, and
    ro is a constant equal to about, 1.2 f.

    Show that the rule implies that nuclear mass density is nearly constant for different nuclei.

    Estimate the mass density of sodium nucleus. Compare it with the average mass density of a
    sodium atom obtained in Exercise. 2.27.

    Solution
    Radius of nucleus r is given by the relation, 
                     r = r0 A1/3
    r0 = 1.2 f = 1.2 × 10-15 m 
    Volume of nucleus, V = (4 / 3) π r3 

                                    = (4 / 3) π (r0 A1/3)3 
                                    = (4 / 3) π r0 A       ... (i) 
    Now, the mass of a nuclei M is equal to its mass number.
    That is, 
    M = A amu = A × 1.66 × 10–27 kg
    Density of nucleus, ρ = Mass of nucleus / Volume of nucleus
                                = A X 1.66 × 10-27 / (4/3) π r03 A
                                = 3 X 1.66 × 10-27 / 4 π r03  Kg m-3
    Density of sodium nucleus is given by,
    ρSodium = 3 × 1.66 × 10-27 / 4 × 3.14 × (1.2 × 10-15)3

              = 4.98 × 1018 / 21.71
              = 2.29 × 1017 Kg m-3 
    Question 560
    CBSEENPH11019903
    Wired Faculty App

    It is a well known fact that during a total solar eclipse the disk of the moon almost completely covers the disk of the Sun. From this fact and from the information you can gather from examples 2.3 and 2.4, determine the approximate diameter of the moon.

    Solution
    Distance of the Moon from the Earth = 3.84 × 108 m 
    Distance of the Sun from the Earth = 1.496 × 1011 m 

    Diameter of the Sun = 1.39 × 109 m 
    During a lunar eclipse, the position of the sun, moon and earth is as shown below:

    Here comma space increment TRS space and space increment TPQ space are space similar. space So comma space space PQ over RS space equals space VT over UT space straight i. straight e. comma space fraction numerator 1.39 space cross times space 10 to the power of 9 over denominator RS end fraction space equals space fraction numerator 1.496 space cross times space 10 to the power of 11 over denominator 3.84 space cross times space 10 to the power of 8 end fraction space rightwards double arrow space RS space equals space left parenthesis 1.39 space 3.84 space cross times space 10 to the power of 8 right parenthesis space cross times space 10 to the power of 6 space space space space space space space space space space space space space equals space 3.57 space cross times space 10 to the power of 6 space Therefore comma space space space 3.57 space cross times space 10 to the power of 6 space space is space the space diameter space of space the space moon. space

    Question 562
    CBSEENPH11020118
    Wired Faculty App

    Give the location of the centre of mass of a

    (i) sphere,

    (ii) cylinder,

    (iii) ring, and

    (iv) cube, each of uniform mass density.

    Does the centre of mass of a body necessarily lie inside the body?
     
    Solution
    In all the four cases, as the mass density is uniform, the centre of mass is located at their respective geometrical centres.

    No, it is not necessary that the centre of mass of a body should lie on the body.
    For example, in the case of a circular ring, centre of mass is at the centre of the ring, where there is no mass. 
    Question 563
    CBSEENPH11020119
    Wired Faculty App

    In the HCl molecule, the separation between the nuclei of the two atoms is about 1.27 Å (1 Å = 10–10 m). Find the approximate location of the CM of the molecule, given that a chlorine atom is about 35.5 times as massive as a hydrogen atom and nearly all the mass of an atom is concentrated in its nucleus. 

    Solution

                             

    Distance between H and Cl atoms = 1.27 Å 
    Mass of H atom = m 

    Mass of Cl atom = 35.5m 

    Let the centre of mass of the system lie at a distance x from the Cl atom. 
    Distance of the centre of mass from the H atom = (1.27 – x)
    Let us assume that the centre of mass of the given molecule lies at the origin.
    Therefore, we can have
    fraction numerator left square bracket straight m space left parenthesis 1.27 space minus space straight x right parenthesis space plus space 35.5 space mx right square bracket over denominator left parenthesis space straight m space plus space 35.5 space straight m right parenthesis end fraction space equals space 0 
               m(1.27 – x) + 35.5mx =  0 
                                  1.27 - x = -35.5x 

    ∴                                       x = fraction numerator negative 1.27 over denominator left parenthesis 35.5 space minus space 1 right parenthesis end fraction
                                               =  -0.037 Å 
    Here, the negative sign indicates that the centre of mass lies at the left of the molecule.
    Hence, the centre of mass of the HCl molecule lies 0.037Å from the Cl atom. 
    Question 564
    CBSEENPH11020120
    Wired Faculty App

    A child sits stationary at one end of a long trolley moving uniformly with a speed V on a smooth horizontal floor. If the child gets up and runs about on the trolley in any manner, what is the speed of the CM of the (trolley + child) system?
     
    Solution
    The child is running arbitrarily on a trolley moving with velocity v. However, the running of the child will produce no effect on the velocity of the centre of mass of the trolley.

    This is because the force due to the boy’s motion is purely internal. Internal forces produce no effect on the motion of the bodies on which they act. Since no external force is involved in the boy–trolley system, the boy’s motion will produce no change in the velocity of the centre of mass of the trolley.
    Question 565
    CBSEENPH11020121
    Wired Faculty App

    Show that the area of the triangle contained between the vectors a and b is one half of the magnitude of a × b.

    Solution
    Consider two vectors OK = vector |a| and OM = vector |b|, inclined at an angle θ, as shown in the following figure.

                               
    In space increment space OMN comma space using space the space relation comma space sin space straight theta space equals space MN over OM space equals space fraction numerator MN over denominator open vertical bar straight b with rightwards harpoon with barb upwards on top close vertical bar end fraction space space MN space equals space open vertical bar straight b with rightwards harpoon with barb upwards on top close vertical bar space sin space straight theta open vertical bar straight a with rightwards harpoon with barb upwards on top space straight x space straight a with rightwards harpoon with barb upwards on top close vertical bar space equals space open vertical bar straight a with rightwards harpoon with barb upwards on top close vertical bar space open vertical bar straight b with rightwards harpoon with barb upwards on top close vertical bar space sin space straight theta space space space space space space space space space space space space space space space space equals space OK space straight x space MN space straight x space 2 over 2 space space space space space space space space space space space space space space space equals space 2 space straight x space Area space of space increment space OMK space therefore space space Area space of space increment space OMK space equals space 1 half space open vertical bar straight a with rightwards harpoon with barb upwards on top space straight x space straight b with rightwards harpoon with barb upwards on top close vertical bar 
    Area of the triangle contained between the vectors a and b is one-half of the magnitude of a x b. 

    Question 566
    CBSEENPH11020122
    Wired Faculty App

    Show that a. (b × c) is equal in magnitude to the volume of the parallelepiped formed on the three vectors, a, b and c. 

    Solution
    A parallelepiped with origin O and sides a, b, and c is shown in the following figure. 

                    
    Volume space of space the space given space parallelopiped space equals space abc space OC with rightwards harpoon with barb upwards on top space space equals space straight a with rightwards harpoon with barb upwards on top space stack OB space with rightwards harpoon with barb upwards on top space equals space straight b with rightwards harpoon with barb upwards on top OC with rightwards harpoon with barb upwards on top space space equals space straight c with rightwards harpoon with barb upwards on top Let space straight n with hat on top space be space straight a space unit space vector space perpendicular space to space both space straight b space and space straight c. space Hence comma space straight n with hat on top space and space straight a space have space the space same space direction. space therefore space straight b with rightwards harpoon with barb upwards on top space straight x space straight c with rightwards harpoon with barb upwards on top space equals space bc space Sin space straight theta space straight n with hat on top space space space space space space space space space space space space space space space space equals space bc space straight n with hat on top space space space space space space space space space space space space space space space space space equals space straight a with rightwards harpoon with barb upwards on top. space open parentheses straight b with rightwards harpoon with barb upwards on top space straight x space straight c with rightwards harpoon with barb upwards on top space close parentheses space space space space space space space space space space space space space space space space equals straight a. space left parenthesis bc space straight n with hat on top space right parenthesis space space space space space space space space space space space space space space space space equals space abc space Cosθ space straight n with hat on top space space space space space space space space space space space space space space space space equals space abc space cos space 0 space space space space space space space space space space space space space space space space space equals space abc space space space space space space space space space space space space space space space space equals space Volume space of space the space parallelopiped space
     Hence proved. 
    Question 568
    CBSEENPH11020124
    Wired Faculty App

    Two particles, each of mass m and speed v, travel in opposite directions along parallel lines separated by a distance d. Show that the vector angular momentum of the two particle system is the same whatever be the point about which the angular momentum is taken.

    Solution
    Let at a certain instant two particles be at points P and Q, as shown in the following figure.



    Angular momentum of the system about point P, 
    Lp = mv × 0 + mv × d 
        =  mvd                                   ...(i) 

    Angular momentum of the system about point Q, 
    LQ = mv × d + mv × 0  
        =  mvd                                   ...(ii) 

    Consider a point R, which is at a distance y from point Q.
    i.e.,               QR = y 

    ∴                   PR = d – y  
    Angular momentum of the system about point R, 
    LR = mv × (d - y) + mv × y 

        = mvd - mvy + mvy 

        = mvd                                   ...(iii) 

    Comparing equations (i), (ii), and (iii), we get
    LP = LQ = LR                              ...(iv)
    From equation (iv), we infer that that the angular momentum of a system does not depend on the point about which it is taken.
    Question 569
    CBSEENPH11020125
    Wired Faculty App

    A non-uniform bar of weight W is suspended at rest by two strings of negligible weight as shown in Fig.7.39. The angles made by the strings with the vertical are 36.9° and 53.1° respectively. The bar is 2 m long. Calculate the distance d of the centre of gravity of the bar from its left end.

    Solution
    The free body diagram of the bar is shown in the following figure.



    Length of the bar, l = 2 m 
    T1 and T2 are the tensions produced in the left and right strings respectively.
    At translational equilibrium, we have
    T1 Sin 36.90 = T2 Sin 53.10 

          straight T subscript 1 over straight T subscript 2 = 4 over 3 
    ⇒      T1 = open parentheses 4 over 3 close parentheses T2
    For rotational equilibrium, on taking the torque about the centre of gravity, we have
          T1 (Cos 36.9) × d = T2 Cos 53.1 (2 - d) 
                T1 × 0.800 d = T2 × 0.600 (2 - d) 
      open parentheses 4 over 3 close parentheses × T2 × 0.800d = T2 (0.600 × 2 - 0.600d) 
              1.067d + 0.6d = 1.2 
    ∴                         d = 1.2 / 1.67 
                                 = 0.72 m 
    Hence, the centre of gravity of the given bar lies 0.72 m from its left end. 
    Question 570
    CBSEENPH11020126
    Wired Faculty App

    A car weighs 1800 kg. The distance between its front and back axles is 1.8 m. Its centre of gravity is 1.05 m behind the front axle. Determine the force exerted by the level ground on each front wheel and each back wheel.

    Solution
    Mass of the car, m = 1800 kg 
    Distance between the front and back axles, d = 1.8 m
    Distance between the centre of gravity and the back axle = 1.05 m 
    The various forces acting on the car are shown in the following figure.

     

    Rf and Rb are the forces exerted by the level ground on the front and back wheels respectively. 
    At translational equilibrium, 
    Rf + Rb = mg 
                = 1800 × 9.8
                = 17640 N                               ... (i)
    For rotational equilibrium, on taking the torque about the C.G., we have
    Rf (1.05) = Rb (1.8 - 1.05)
            straight R subscript straight b over straight R subscript straight f space equals space 7 over 5
              Rb = 1.4 Rf                                 ...(ii) 
    Solving equations (i) and (ii), we get
    1.4Rf + Rf = 17640 
                Rf = 7350 N 
    ∴          Rb = 17640 – 7350
                    = 10290 N 
    Therefore, the force exerted on each front wheel = 7350 over 2  =  3675 N.
    The force exerted on each back wheel = 10290 over 2 =  5145 N 
    Question 571
    CBSEENPH11020127
    Wired Faculty App

    (a) Find the moment of inertia of a sphere about a tangent to the sphere, given the moment of inertia of the sphere about any of its diameters to be 2MR2/5, where M is the mass of the sphere and R is the radius of the sphere.

    (b) Given the moment of inertia of a disc of mass M and radius R about any of its diameters to be MR2/4, find its moment of inertia about an axis normal to the disc and passing through a point on its edge.

    Solution
    (a) The moment of inertia (M.I.) of a sphere about its diameter = fraction numerator 2 MR squared over denominator 5 end fraction

                                        
    According to the theorem of parallel axes, the moment of inertia of a body about any axis is equal to the sum of the moment of inertia of the body about a parallel axis passing through its centre of mass and the product of its mass and the square of the distance between the two parallel axes. 
    The M.I. about a tangent of the sphere = fraction numerator 2 MR squared over denominator 5 end fraction+ MR2  
                                                      = fraction numerator 7 MR squared over denominator 5 end fraction
    (b) The moment of inertia of a disc about its diameter = MR squared over 4 
    According to the theorem of perpendicular axis, the moment of inertia of a planar body (lamina) about an axis perpendicular to its plane is equal to the sum of its moments of inertia about two perpendicular axes concurrent with perpendicular axis and lying in the plane of the body. 
    The M.I. of the disc about its centre = MR squared over 4 + MR squared over 4  
                                                   = MR squared over 2
    The situation is shown in the given figure. 
                                       

    Applying the theorem of parallel axes, 
    The moment of inertia about an axis normal to the disc and passing through a point on its edge, 
     M.I = MR squared over 2 + MR2 = fraction numerator 3 MR squared over denominator 2 end fraction 
    Question 572
    CBSEENPH11020128
    Wired Faculty App

    Torques of equal magnitude are applied to a hollow cylinder and a solid sphere, both having the same mass and radius. The cylinder is free to rotate about its standard axis of symmetry, and the sphere is free to rotate about an axis passing through its centre. Which of the two will acquire a greater angular speed after a given time?

    Solution
    Let m and r be the respective masses of the hollow cylinder and the solid sphere. 
    The moment of inertia of the hollow cylinder about its standard axis, I1 = mr2 

    The moment of inertia of the solid sphere about an axis passing through its centre, I2 = open parentheses 2 over 5 close parenthesesmr2 

    We have the relation, 
    τ = Iα 

    where, 
    α = Angular acceleration, 
    τ = Torque, 
    I = Moment of inertia, 
    For the hollow cylinder, τ1 = I1 α1
    For the solid sphere, τn = In αn
    As an equal torque is applied to both the bodies, τ1 = τ2, 

    ∴      straight alpha subscript 2 over straight alpha subscript 1  = straight I subscript 1 over straight I subscript 2  = fraction numerator MR squared over denominator begin display style bevelled 2 over 5 end style space MR squared end fraction 
                α2 > α1                                      ...(i) 

    Now, using the relation, 
    ω = ω0 + αt 

    where,
    ω0 = Initial angular velocity 
      t = Time of rotation 
     ω = Final angular velocity 
    For equal ω0 and t, we have
        ω ∝ α                                                 … (ii) 

    From equations (i) and (ii), 
    ω2 > ω1 

    Hence, the angular velocity of the solid sphere will be greater than that of the hollow cylinder.
    Question 573
    CBSEENPH11020129
    Wired Faculty App

    A solid cylinder of mass 20 kg rotates about its axis with angular speed 100 rad s–1. The radius of the cylinder is 0.25 m. What is the kinetic energy associated with the rotation of the cylinder? What is the magnitude of angular momentum of the cylinder about its axis?

    Solution
    Mass of the cylinder, m = 20 kg 
    Angular speed, ω = 100 rad s–1 

    Radius of the cylinder, r = 0.25 m 
    The moment of inertia of the solid cylinder, 
       I = mr2 / 2 
         = (1/2) × 20 × (0.25)2 

         = 0.625 kg m2 

    ∴ Kinetic energy = (1/2) I ω2 

                          = (1/2) × 6.25 × (100)2 
                          = 3125 J 
    Therefore,
    Angular momentum, L = Iω 

                                   = 6.25 × 100 
                                   = 62.5 Js 
    That is, there is a magnitude of 62.5 Js angular momentum of the cylinder about it's axis. 
    Question 574
    CBSEENPH11020130
    Wired Faculty App

    (a) A child stands at the centre of a turntable with his two arms outstretched. The turntable is set rotating with an angular speed of 40 rev/min. How much is the angular speed of the child if he folds his hands back and thereby reduces his moment of inertia to 2/5 times the initial value? Assume that the turntable rotates without friction.

    (b) Show that the child’s new kinetic energy of rotation is more than the initial kinetic energy of rotation. How do you account for this increase in kinetic energy?
     
    Solution
    (a) 
    Initial angular velocity, ω1= 40 rev/min 
    Final angular velocity = ω2 

    The moment of inertia of the boy with stretched hands = I1 

    The moment of inertia of the boy with folded hands = I2 

    The two moments of inertia are related as, 
                             I2 = open parentheses 2 over 5 close parentheses I1
    Since no external force acts on the boy, the angular momentum L is a constant. 
    Hence, for the two situations, 
                I2ω2  =  I1 ω1 

                    ω2 = open parentheses straight I subscript 1 over straight I subscript 2 close parentheses ω1 

                         equals space open square brackets fraction numerator straight I subscript 1 over denominator begin display style bevelled 2 over 5 end style space straight I subscript 1 end fraction close square brackets space straight x space 40 space equals space open parentheses 5 over 2 close parentheses space straight x space 40 space equals space 100 space rev divided by min 

    (b) 
    Final K.E. = 2.5 Initial K.E.
    Final kinetic rotation, EF = (1/2) I2 ω22 

    Initial kinetic rotation, EI =  (1/2) I1 ω12 
        straight E subscript straight f over straight E subscript straight i =open parentheses 1 half close parentheses I2 ω22 / open parentheses 1 half close parentheses I1 ω12 

             =open parentheses 2 over 5 close parentheses I1 (100)2 / I1 (40)2 

             = 2.5 
    Therefore, 
    EF = 2.5 E1 

    The increase in the rotational kinetic energy is attributed to the internal energy of the boy. 
    Question 575
    CBSEENPH11020131
    Wired Faculty App

    A rope of negligible mass is wound round a hollow cylinder of mass 3 kg and radius 40 cm. What is the angular acceleration of the cylinder if the rope is pulled with a force of 30 N? What is the linear acceleration of the rope? Assume that there is no slipping.

    Solution
    Mass of the hollow cylinder, m = 3 kg 
    Radius of the hollow cylinder, r = 40 cm = 0.4 m
    Applied force, F = 30 N 
    The moment of inertia of the hollow cylinder about its geometric axis, 
    I = mr2 

      = 3 × (0.4)2 
      = 0.48 kg m2 

    Torque, τ = F × r 
                 =  30 × 0.4 
                 = 12 Nm 
    For angular acceleration α, torque is also given by the relation, 
       τ = Iα 
       α = τ / I 
         =  12 / 0.48
         = 25 rad s-2 

    Linear acceleration = τα 
                             = 0.4 × 25
                             = 10 m s–2 
    This is the required linear acceleration of the rope. 
    Question 576
    CBSEENPH11020132
    Wired Faculty App

    To maintain a rotor at a uniform angular speed of 200 rad s–1, an engine needs to transmit a torque of 180 Nm. What is the power required by the engine?
    (Note: uniform angular velocity in the absence of friction implies zero torque. In practice, applied torque is needed to counter frictional torque). Assume that the engine is 100 % efficient. 

    Solution
    Given,
    Angular speed of the rotor, ω = 200 rad/s
    Torque required, τ = 180 Nm
    The power of the rotor (P) is related to torque and angular speed by the relation, 
    P = τω 
      = 180 × 200
      = 36 × 103 

      = 36 kW 
    Hence, the power required by the engine is 36 kW. 
    Question 577
    CBSEENPH11020133
    Wired Faculty App

    From a uniform disk of radius R, a circular hole of radius R/2 is cut out. The centre of the hole is at R/2 from the centre of the original disc. Locate the centre of gravity of the resulting flat body.

    Solution
    Mass per unit area of the original disc = σ 
    Radius of the original disc = R 

    Mass of the original disc, M = πR2σ 
    The disc with the cut portion is shown in the following figure,
                 

    Radius of the smaller disc = R/2 
    Mass of the smaller disc, M’ = π (R/2)2σ
                                              = π R2σ / 4 
                                               =  M / 4 
    Let O and O′ be the respective centres of the original disc and the disc cut off from the original.
    As per the definition of the centre of mass, the centre of mass of the original disc is supposed to be concentrated at O, while that of the smaller disc is supposed to be concentrated at O′.
    It is given that, 
    OO′= begin inline style straight R over 2 end style
    After the smaller disc has been cut from the original, the remaining portion is considered to be a system of two masses.
    The two masses are, 
    M (concentrated at O), and –M′ equals straight M over 4concentrated at O′.
    (The negative sign indicates that this portion has been removed from the original disc.)
    Let x be the distance through which the centre of mass of the remaining portion shifts from point O. 
    The relation between the centres of masses of two masses is given as, 
    x =open parentheses fraction numerator straight m subscript 1 straight r subscript 1 space plus space straight m subscript 2 straight r subscript 2 over denominator straight m subscript 1 space plus straight m subscript 2 end fraction close parentheses
    For the given system,
    x =fraction numerator left square bracket straight M space straight x space 0 space minus space straight M apostrophe space straight x space open parentheses begin display style straight R over 2 end style close parentheses over denominator straight M space plus space left parenthesis negative straight M apostrophe right parenthesis end fraction
       = negative straight R over 6
    (The negative sign indicates that the centre of mass gets shifted toward the left of point O.) 

    The centre of gravity of the resulting flat body is R/6; from the original centre of the body and opposite to the centre of the cut portion.
    Question 578
    CBSEENPH11020134
    Wired Faculty App

    A metre stick is balanced on a knife edge at its centre. When two coins, each of mass 5 g are put one on top of the other at the 12.0 cm mark, the stick is found to be balanced at 45.0 cm. What is the mass of the metre stick?

    Solution
    Let W and W′ be the respective weights of the metre stick and the coin.



    The mass of the metre stick is concentrated at its mid-point.
     i.e., at the 50 cm mark. 
    Mass of the meter stick = m’

    Mass of each coin, m = 5 g 
    When the coins are placed 12 cm away from the end P, the centre of mass gets shifted by 5 cm from point R toward the end P.
    The centre of mass is located at a distance of 45 cm from point P. 
    The net torque will be conserved for rotational equilibrium about point R. 
    10 × g(45 - 12) - m'g(50 - 45) = 0 
    ∴                                  m' = 66 g 

    Hence, the mass of the metre stick is 66 g.
    Question 579
    CBSEENPH11020135
    Wired Faculty App

    A solid sphere rolls down two different inclined planes of the same heights but different angles of inclination.

    (a) Will it reach the bottom with the same speed in each case?

    (b) Will it take longer to roll down one plane than the other?

    (c) If so, which one and why?

    Solution
    (a) Mass of the sphere = m

    Height of the plane = h

    Velocity of the sphere at the bottom of the plane = v

    At the top of the plane, the total energy of the sphere = Potential energy = mgh

    At the bottom of the plane, the sphere has both translational and rotational kinetic energies. 
    Hence,
    Total energy = 1 halfmv2 + 1 half I ω2

    Using the law of conservation of energy, 
    1 halfmv2 + 1 half I ω2 = mgh 
    For a solid sphere, the moment of inertia about its centre, I = open parentheses 2 over 5 close parenthesesmr2 
    Hence, equation (i) becomes
    1 halfmr2 +1 half [open parentheses 2 over 5 close parenthesesmr2] ω2  =  mgh 

                 1 halfv2 + open parentheses 1 fifth close parenthesesr2ω2  =  gh 
    But we have the relation, v = rω 
    ∴      1 halfv2 + open parentheses 1 fifth close parenthesesv2  =  gh 

                      v2 (7 over 10) = gh 

                                 v =square root of 10 over 7 end rootgh 

    Hence, the velocity of the sphere at the bottom depends only on height (h) and acceleration due to gravity (g).
    Both these values are constants.
    Therefore, the velocity at the bottom remains the same from whichever inclined plane the sphere is rolled.
    (b) Consider two inclined planes with inclinations θ1 and θ2, related as, 
                            θ1 < θ2 

    The acceleration produced in the sphere when it rolls down the plane inclined at θ1 is, 
                          a1 = g sin θ1 

    The various forces acting on the sphere are shown in the following figure.


     
    R1 is the normal reaction to the sphere.
    Similarly, the acceleration produced in the sphere when it rolls down the plane inclined at θ2 is, 
                           a2 = g sin θ2 
    The various forces acting on the sphere are shown in the following figure.

     
    R2 is the normal reaction to the sphere. 
    θ2 > θ1; sin θ2 > sin θ1                          ... (i) 
    ∴ a2 > a1                                                … (ii) 
    Initial velocity, u = 0 
    Final velocity, v = Constant 
    Using the first equation of motion, we can obtain the time of roll as
    v = u + at 

    ∴ t ∝ (1/α) 
    For inclination θ1 : 
                          t1 ∝ (1/α1) 
    For inclination θ2 : 
                           t2 ∝ (1/α2) 
    From above equations, we get, 
                               t2 < t1 

    Hence, the sphere will take a longer time to reach the bottom of the inclined plane having the smaller inclination. 

    Question 580
    CBSEENPH11020136
    Wired Faculty App

    A hoop of radius 2 m weighs 100 kg. It rolls along a horizontal floor so that its centre of mass has a speed of 20 cm/s. How much work has to be done to stop it?

    Solution
    Given,
    Radius of the hoop, r = 2 m
    Mass of the hoop, m = 100 kg
    Velocity of the hoop, v = 20 cm/s = 0.2 m/s
    Total energy of the hoop = Translational K.E. + Rotational K.E.
                                ET = 1 halfmv2 + 1 halfI ω2
    Moment of inertia of the hoop about its centre, I = mr2 

                               ET = 1 halfmv2 +1 half (mr2)ω2 
    Using the relation, v = rω 
    ∴                          ET = 1 halfmv2 + 1 halfmr2ω2 

                                   = 1 halfmv2 +1 halfmv2 
                                   = mv2 

    The work required to be done for stopping the hoop is equal to the total energy of the hoop. 
    ∴ Required work to be done, W = mv2 
                                            = 100 × (0.2)2 
                                            = 4 J.
    Question 581
    CBSEENPH11020137
    Wired Faculty App

    The oxygen molecule has a mass of 5.30 × 10–26 kg and a moment of inertia of 1.94×10–46 kg m2 about an axis through its centre perpendicular to the lines joining the two atoms. Suppose the mean speed of such a molecule in a gas is 500 m/s and that its kinetic energy of rotation is two thirds of its kinetic energy of translation. Find the average angular velocity of the molecule.
     
    Solution
    Mass of an oxygen molecule, m = 5.30 × 10–26 kg
    Moment of inertia, I = 1.94 × 10–46 kg m2 

    Velocity of the oxygen molecule, v = 500 m/s 
    The separation between the two atoms of the oxygen molecule = 2r  
    Mass of each oxygen atom = straight m over 2
    Hence, moment of inertia I, is calculated as
    (straight m over 2)r2 + (straight m over 2)r2 = mr2 

                         r = open parentheses straight I over straight m close parentheses to the power of begin inline style bevelled 1 half end style end exponent 
    open parentheses fraction numerator 1.94 space straight x space 10 to the power of negative 46 end exponent over denominator 5.36 space straight x space 10 to the power of negative 26 end exponent end fraction close parentheses to the power of begin inline style bevelled 1 half end style end exponent =  0.60 × 10-10 m 
    It is given that, 
           KErot = open parentheses 2 over 3 close parenthesesKEtrans 

    open parentheses 1 half close parentheses I ω2 = open parentheses 2 over 3 close parentheses ×open parentheses 1 half close parentheses × mv2 

         mr2ω2 =open parentheses 2 over 3 close parenthesesmv2 

               ω =open parentheses 2 over 3 close parentheses to the power of begin inline style bevelled 1 half end style end exponent open parentheses straight v over straight r close parentheses 
                  =open parentheses 2 over 3 close parentheses to the power of begin inline style bevelled 1 half end style end exponentopen parentheses fraction numerator 500 over denominator 0.6 space straight x space 10 to the power of negative 10 end exponent end fraction close parentheses
                  = 6.80 × 1012 rad/s.
    Question 582
    CBSEENPH11020138
    Wired Faculty App

    A solid cylinder rolls up an inclined plane of angle of inclination 30°. At the bottom of the inclined plane the centre of mass of the cylinder has a speed of 5 m/s. 

    (a) How far will the cylinder go up the plane? 

    (b) How long will it take to return to the bottom?
    Solution
    Initial velocity of the solid cylinder, v = 5 m/s 
    Angle of inclination, θ = 30° 
    Let the cylinder go up the plane upto a height h. 
    Using the relation,
    1 half space mv squared space plus space 1 half space straight I space straight omega squared space equals space mgh
    1 half space mv squared space plus space 1 half open parentheses 1 half space mr squared close parentheses straight omega squared space equals space mgh space 3 over 4 mv squared space equals space mgh space straight h space equals 3 fraction numerator straight v squared over denominator 4 straight g end fraction space equals space 3 space straight x space fraction numerator 52 over denominator 4 space straight x space 9.8 end fraction space space space equals space 1.913 space straight m space
    If s is the distance up the inclined plane, then as
          sin θ =open parentheses straight h over straight s close parentheses 
              s =open parentheses fraction numerator straight h over denominator sin space straight theta end fraction close parentheses  
                =open parentheses fraction numerator 1.913 over denominator sin space 30 to the power of straight o end fraction close parentheses space equals space 3.826 space straight m
    Therefore, time taken to return to the bottom of the inclined plane,  

      straight t space equals space square root of fraction numerator 2 space straight s space open parentheses 1 space plus space bevelled straight k squared over straight r squared close parentheses over denominator straight g space sin space straight theta end fraction end root space space space equals space fraction numerator 2 space straight x space 3.826 space open parentheses 1 space plus space bevelled 1 half close parentheses over denominator 9.8 space sin space 30 to the power of straight o end fraction space space space equals space 1.53 space sec
    1.53 sec is required for the balloon to return to the bottom. 


     
    Question 583
    CBSEENPH11020139
    Wired Faculty App

    As shown in Fig.7.40, the two sides of a step ladder BA and CA are 1.6 m long and hinged at A. A rope DE, 0.5 m is tied half way up. A weight 40 kg is suspended from a point F, 1.2 m from B along the ladder BA. Assuming the floor to be frictionless and neglecting the weight of the ladder, find the tension in the rope and forces exerted by the floor on the ladder. (Take g = 9.8 m/s2)




    (Hint: Consider the equilibrium of each side of the ladder separately.)
    Solution

    The given question is illustrated in the figure below: 

    NB = Force exerted on the ladder by the floor point B 
    NC = Force exerted on the ladder by the floor point C 
    T = Tension in the rope 
    BA = CA = 1.6 m 
    DE = 0. 5 m 
    BF = 1.2 m 
    Mass of the weight, m = 40 kg 
    Draw a perpendicular from A on the floor BC. This intersects DE at mid-point H. 
    ΔABI and ΔAIC are similar triangles. 
    ∴       BI = IC 
    Hence, I is the mid-point of BC. 
    DE || BC 
     BC = 2 × DE = 1 m 
    AF = BA – BF = 0.4 m                          … (i) 

    D is the mid-point of AB.  
    Hence, we can write, 
    AD = open parentheses 1 half close parentheses × BA  =  0.8 m                    ...(ii) 

    Using equations (i) and (ii), we get
    FE = 0.4 m 
    Hence, F is the mid-point of AD. 
    FG||DH and F is the mid-point of AD.
    Hence, G will also be the mid-point of AH. 
    ΔAFG and ΔADH are similar 
        FG over DH space equals space AF over AD FG over DH space equals space fraction numerator 0.4 over denominator 0.8 end fraction space equals space 1 half space FG space equals space 1 half space DH space space space space space space space equals space open parentheses 1 half close parentheses space straight x space 0.25 space space space space space space space space equals space 0.125 space straight m space 

    In ΔADH, 
    AH = (AD2 - DH2)1/2 

         = (0.82 - 0.252)1/2 
         =  0.76 m
    For translational equilibrium of the ladder, the upward force should be equal to the downward force. 
    Nc + NB  = mg  = 392                      … (iii) 

    For rotational equilibrium of the ladder, the net moment about A is 
    -NB × BI + mg × FG + NC × CI + T × AG - T × AG  =  0 
    -NB × 0.5 + 40 × 9.8 × 0.125 + NC × 0.5  =  0 
    (NC - NB) × 0.5 = 49 
    NC - NB = 98                                      ...(iv) 

    Adding equations (iii) and (iv), we get
    NC = 245 N 
    NB = 147 N 
    For rotational equilibrium of the side AB, consider the moment about A, 
                  -NB × BI + mg × FG + T × AG  =  0 
    -245 × 0.5 + 40 X 9.8 × 0.125 + T × 0.76  =  0  
    ∴                                                    T = 96.7 N.
    T is the required tension in the rope. 

    Question 584
    CBSEENPH11020140
    Wired Faculty App

    A man stands on a rotating platform, with his arms stretched horizontally holding a 5 kg weight in each hand. The angular speed of the platform is 30 revolutions per minute. The man then brings his arms close to his body with the distance of each weight from the axis changing from 90cm to 20cm. The moment of inertia of the man together with the platform may be taken to be constant and equal to 7.6 kg m2. 

    (a) What is his new angular speed? (Neglect friction.) 

    (b) Is kinetic energy conserved in the process? If not, from where does the change come about?

    Solution
    (a) 
    Moment of inertia of the man-platform system = 7.6 kg m2

    Moment of inertia when the man stretches his hands to a distance of 90 cm,
    2 × m r2 = 2 × 5 × (0.9)2 

                = 8.1 kg m2 

    Initial moment of inertia of the system, Ii = 7.6 + 8.1
                                                                   = 15.7 kg m2 

    Angular speed, ωi = 300 rev/min 
    Angular momentum, Li = Iiωi  =  15.7 × 30                         ...(i) 
    Moment of inertia when the man folds his hands to a distance of 20 cm
    2 × mr2 = 2 × 5 (0.2)2 
               = 0.4 kg m2 

    Final moment of inertia, If = 7.6 + 0.4 = 8 kg m2  
    Final angular speed = ωf 

    Final angular momentum, Lf = Ifωf = 0.79 ωf                   .... (ii) 
    From the conservation of angular momentum, we have
                           Iiωi  =  Ifωf 

    Therefore,
                          ωf =fraction numerator 15.7 space straight x space 30 over denominator 8 end fraction space equals space 58.88 space rev divided by min 
    (b) Kinetic energy is not conserved in the given process. But, the kinetic energy increases as there is a decrease in the moment of inertia. 
    The additional kinetic energy comes from the work done by the man to fold his hands toward himself.
    Question 585
    CBSEENPH11020141
    Wired Faculty App

    A bullet of mass 10 g and speed 500 m/s is fired into a door and gets embedded exactly at the centre of the door. The door is 1.0 m wide and weighs 12 kg. It is hinged at one end and rotates about a vertical axis practically without friction. Find the angular speed of the door just after the bullet embeds into it.

    (Hint: The moment of inertia of the door about the vertical axis at one end is ML2/3.)

    Solution
    Given, 
    Mass of the bullet, m = 10 g = 10 × 10–3 kg 
    Velocity of the bullet, v = 500 m/s 
    Thickness of the door, L = 1 m 
    Radius of the door, r = m / 2 
    Mass of the door, M = 12 kg 
    Angular momentum imparted by the bullet on the door, 
        α = mvr 

          = (10 × 10-3 ) × (500) × 1 half
          =  2.5 kg m2 s-1                                   ...(i) 
    Moment of inertia of the door is given by, 
    I thin space equals space ML squared over 3 space space equals space open parentheses 1 third close parentheses space x space 12 space x space 1 squared space space space equals space 4 space k g space m squared space B u t comma space alpha space equals space I space omega space omega equals space alpha over I space space space equals space fraction numerator 2.5 over denominator 4 end fraction space space space space equals space 0.625 space r a d divided by s e c  
    This is the required angular speed of the door just after the bullet embeds into it. 
    Question 586
    CBSEENPH11020142
    Wired Faculty App

    Two discs of moments of inertia I1 and I2 about their respective axes (normal to the disc and passing through the centre), and rotating with angular speeds ω1 and ω2 are brought into contact face to face with their axes of rotation coincident. (a) What is the angular speed of the two-disc system? (b) Show that the kinetic energy of the combined system is less than the sum of the initial kinetic energies of the two discs. How do you account for this loss in energy? Take ω1 ≠ ω2.

    Solution
    (a) Moment of inertia of disc I= I1 
    Angular speed of disc I = ω1 
    Moment of inertia of disc II = I2 

    Angular speed of disc II = ω2 

    Angular momentum of disc I, L1 = I1ω1 

    Angular momentum of disc II, L2 = I2ω2 

    Total initial angular momentum Li = I1ω1 + I2ω2 

    When the two discs are joined together, their moments of inertia get added up. 
    Moment of inertia of the system of two discs, I = I1 + I2 

    Let ω be the angular speed of the system. 
    Total final angular momentum, LT = (I1 + I2) ω 
    Using the law of conservation of angular momentum, we have
                   Li = LT 

    I1ω1 + I2ω2 = (I1 + I2)ω 
    ∴              ω =fraction numerator straight I subscript 1 straight omega subscript 1 plus space straight I subscript 2 straight omega subscript 2 over denominator straight I subscript 1 space plus space straight I subscript 2 end fraction
    (b) Kinetic energy of disc I, E1 = open parentheses 1 half close parentheses I1ω12 
    Kinetic energy of disc II, E2= open parentheses 1 half close parentheses I2ω22 
    Total initial kinetic energy, Ei =open parentheses 1 half close parentheses ( I1ω12 + I2ω22) 
    When the discs are joined, their moments of inertia get added up. 
    Moment of inertia of the system, I = I1 + I2 

    Angular speed of the system = ω 
    Final kinetic energy is given by,
          Ef = open parentheses 1 half close parentheses( I1 + I2) ω2

              = open parentheses 1 half close parentheses ( I1 + I2) open square brackets fraction numerator open parentheses straight I subscript 1 space straight omega subscript 1 space plus space straight I subscript 2 straight omega subscript 2 close parentheses over denominator open parentheses straight I subscript 1 space plus space straight I subscript 2 close parentheses end fraction close square brackets squared 
              =open parentheses 1 half close parentheses  fraction numerator open parentheses straight I subscript 1 space straight omega subscript 1 space plus space straight I subscript 2 straight omega subscript 2 close parentheses squared over denominator open parentheses straight I subscript 1 space plus space straight I subscript 2 close parentheses end fraction 
    ∴  Ei - Ef = 1 half open square brackets fraction numerator straight I subscript 1 straight omega subscript 1 squared space plus straight I subscript 2 straight omega subscript 2 squared space space minus space open parentheses begin display style bevelled 1 half end style close parentheses space open parentheses straight I subscript 1 straight omega subscript 1 space plus space straight I subscript 2 straight omega subscript 2 close parentheses space squared over denominator straight I subscript 1 space plus space straight I subscript 2 end fraction close square brackets
    Solving the equation, we get 
              = fraction numerator straight I subscript 1 space straight I subscript 2 space open parentheses straight omega subscript 1 space minus space straight omega subscript 2 close parentheses squared over denominator 2 space left parenthesis straight I subscript 1 space plus space straight I subscript 2 right parenthesis end fraction 
    All the quantities on RHS are positive 
    Therefore,
     Ei - Ef > 0 
    i.e.,       Ei > Ef 

    When the two discs come in contact with each other, there is a frictional force between the two. Hence there would be a loss of Kinetic Energy.
    Question 587
    CBSEENPH11020143
    Wired Faculty App

    (a) Prove the theorem of perpendicular axes. 

    (Hint: Square of the distance of a point (x, y) in the x–y plane from an axis through the origin perpendicular to the plane is (x2 + y2). 

    Solution
    The theorem of perpendicular axes states that the moment of inertia of a planar body (lamina) about an axis perpendicular to its plane is equal to the sum of its moments of inertia about two perpendicular axes concurrent with the perpendicular axis and lying in the plane of the body. 
    A physical body with centre O and a point mass m,in the x–yplane at (x, y) is shown in the following figure below. 

                             
    Moment of inertia about x-axis, Ix = mx2 
    Moment of inertia about y-axis, Iy = my2 

    Moment of inertia about z-axis, Iz = m(x2 + y2)1/2 

    Therefore, 
    Ix + Iy = mx2 + my2 

             = m(x2 + y2) 
             = straight m space open square brackets square root of straight x squared plus straight y squared end root close square brackets to the power of begin inline style bevelled 1 half end style end exponent 
    I    x + Iy = Iz 

    Hence, the theorem is proved.
     
    Question 588
    CBSEENPH11020144
    Wired Faculty App

    26. (b) Prove the theorem of parallel axes. 

    (Hint: If the centre of mass is chosen to be the origin ∑ miri = 0).

    Solution
    The theorem of parallel axes states that the moment of inertia of a body about any axis is equal to the sum of the moment of inertia of the body about a parallel axis passing through its centre of mass and the product of its mass and the square of the distance between the two parallel axes. 

                              
    Suppose a rigid body is made up of n particles, having masses m1, m2, m3, … , mn, at perpendicular distances r1,r2, r3, … , rn respectively from the centre of mass O of the rigid body. 
    The moment of inertia about axis RS passing through the point O, 
    straight I space subscript RS space equals space sum from straight i space equals space 1 to straight n of space straight m subscript straight i space straight r subscript straight i squared space The space perpendicular space distance space of space mass space straight m comma space from space the space axis space QP space equals space straight a space plus space straight r subscript straight i space Hence comma space the space moment space of space Inertia space about space axis space QP thin space is comma space straight I subscript QP space equals space sum from straight i space equals 1 to straight n of space straight m subscript straight i space left parenthesis thin space straight a space plus space straight r subscript straight i right parenthesis squared space space space space space space space equals space sum from straight i space equals 1 to straight n of space straight m subscript straight i space left parenthesis thin space straight a squared space plus space straight r subscript straight i squared space plus space 2 ar subscript straight i space right parenthesis space squared space space space space space space space equals space sum from straight i space equals 1 to straight n of space straight m subscript straight i space straight a squared space space plus space sum from straight i space equals 1 to straight n of space space straight m subscript straight i space straight r subscript straight i squared space plus space sum from straight i space equals 1 to straight n of space space straight m subscript straight i space space 2 ar subscript straight i space space space space space space space equals space space sum from straight i space equals 1 to straight n of space straight m subscript straight i space straight a squared space plus space 2 space sum from straight i space equals 1 to straight n of space space straight m subscript straight i space straight r subscript straight i squared

    The moment of Inertia of all the particles about the axis passing through the centre of mass is zero, at the centre of mass. 
    That is, 
    2 space sum from straight i space equals space 1 to straight n of space straight m subscript straight i space straight a space straight r subscript straight i space equals space 0 space left square bracket space because space straight a space not equal to space 0 space right square bracket therefore space sum space straight m subscript straight i space straight r subscript straight i space equals space 0 space Also comma space sum from straight i space equals space 1 to straight n of space straight m subscript straight i space equals space straight M thin space where comma space straight M space is space the space total space mass space of space the space rigid space body. space Therefore comma space straight I subscript QP space equals space straight I subscript RS space plus space Ma squared
    Thus the theorem is proved. 


    Question 589
    CBSEENPH11020145
    Wired Faculty App

    Prove the result that the velocity v of translation of a rolling body (like a ring, disc, cylinder or sphere) at the bottom of an inclined plane of a height h is given by v2 = 2gh/ [1 + (k2/R2) ].

    Using dynamical consideration (i.e. by consideration of forces and torques). Note k is the radius of gyration of the body about its symmetry axis, and R is the radius of the body. The body starts from rest at the top of the plane. 

    Solution
    A body rolling on an inclined plane of height h,is shown in figure below:


    Let,
    m     = Mass of the body 
    R = Radius of the body 
    K = Radius of gyration of the body 
    v = Translational velocity of the body 
    h =Height of the inclined plane 
    g = Acceleration due to gravity 
    Total energy at the top of the plane, E­1= mgh 

    Total energy at the bottom of the plane, Eb = KErot + KEtrans 

                                                            = open parentheses 1 half close parentheses I ω2 +open parentheses 1 half close parentheses mv2
    But I = mk2 and ω = v / R 

    ∴ Eb = open parentheses 1 half close parentheses (mk2)open parentheses straight v squared over straight R squared close parentheses + open parentheses 1 half close parenthesesmv2
          = open parentheses 1 half close parenthesesmv2 (1 +straight k squared over straight R squared)
    From the law of conservation of energy, we have
                     ET = Eb 

                  mgh = open parentheses 1 half close parenthesesmv2 (1 +straight k squared over straight R squared)
    ∴                v = open parentheses fraction numerator 2 gh over denominator 1 space plus space begin display style straight k squared over straight R squared end style end fraction close parentheses 
    Hence, the result. 
    Question 590
    CBSEENPH11020146
    Wired Faculty App

    A disc rotating about its axis with angular speed ωois placed lightly (without any translational push) on a perfectly frictionless table. The radius of the disc is R. What are the linear velocities of the points A, B and C on the disc shown in Fig. 7.41? Will the disc roll in the direction indicated?

    Solution
    Angular speed of the disc = ωo 

    Radius of the disc = R 

    Using the relation for linear velocity, v = ωoR 

    For point A: 
    vA = Rωo, in the direction tangential to the right.
    For point B: 
    vB = Rωo, in the direction tangential to the left. 
    For point C:
    vc = open parentheses straight R over 2 close parenthesesωo, in the direction same as that of vA.
    vA = Rωo; 
    vB = Rω0, 
    vc = open parentheses straight R over 2 close parenthesesω0
    The disc will not roll.

    The directions of motion of points A, B, and C on the disc are shown in the following figure 

    Since the disc is placed on a frictionless table, it will not roll. This is because the presence of friction is essential for the rolling of a body.
    Question 591
    CBSEENPH11020147
    Wired Faculty App

    Explain why friction is necessary to make the disc in Fig. 7.41 roll in the direction indicated.

    (a) Give the direction of frictional force at B, and the sense of frictional torque, before perfect rolling begins.

    (b) What is the force of friction after perfect rolling begins?

    Solution
    A torque is required to roll the given disc. According to the definition of torque, the rotating force should be tangential to the disc. Since the frictional force at point B is along the tangential force at point A, a frictional force is required for making the disc roll.

    (a) Force of friction acts opposite to the direction of velocity at point B. The direction of linear velocity at point B is tangentially leftward. Hence, frictional force will act tangentially rightward. The sense of frictional torque before the start of perfect rolling is perpendicular to the plane of the disc in the outward direction.
    (b) Frictional force acts opposite to the direction of velocity at point B. Therefore, perfect rolling will begin when the velocity at that point becomes equal to zero. This will make the frictional force acting on the disc zero.
    Question 592
    CBSEENPH11020148
    Wired Faculty App

    A solid disc and a ring, both of radius 10 cm are placed on a horizontal table simultaneously, with initial angular speed equal to 10 π rad s-1. Which of the two will start to roll earlier? The co-efficient of kinetic friction is μk = 0.2.

    Solution
    Radii of the ring and the disc, r = 10 cm = 0.1 m
    Initial angular speed, ω0 =10 π rad s–1 

    Coefficient of kinetic friction, μk = 0.2 
    Initial velocity of both the objects, u = 0 
    The frictional force causes motion between two objects. 
    As per Newton’s second law of motion,
    Frictional force, f =ma 
                      μkmg= ma 

    where, 
     a = Acceleration produced in the objects 
    m = Mass 
    Therefore,
        a = μkg                                           … (i)

    As per the first equation of motion,
    Final velocity of the objects can be obtained as,
    v = u + at 

       = 0 + μkgt 

       = μkgt                                             … (ii)

    The torque applied by the frictional force will act in perpendicularly outward direction and cause reduction in the initial angular speed. 
    Torque, τ= – I α 
    where, 
    α = Angular acceleration 
    μkmgr = –I α  
    ∴
     α = fraction numerator negative straight mu subscript straight k space end subscript mgr over denominator straight I end fraction                                   ...(iii) 

    Using the first equation of rotational motion to obtain the final angular speed, 
    ω = ω0 + αt 

       = ω0 + (-μkmgr / I)t                          ...(iv) 

    Rolling starts when linear velocity, 
                v = rω 
    Therefore, 
      v = r (ω0 - μkmgrt / I)                         ...(v) 

    Equating equations (ii) and (v), we get
    μkgt = r (fraction numerator straight omega subscript straight o space minus space straight mu subscript straight k mgrt over denominator straight I end fraction) 
          = rω0 - fraction numerator straight mu subscript straight k mgr squared straight t over denominator straight I end fraction                          ...(vi)

    For the ring, 
           I = mr2 

    ∴ μkgt = rω0 -fraction numerator straight mu subscript straight k space mgr squared straight t over denominator mr squared end fraction
            = rω0 - μkgt 
    2μkgt = rω0 

    ∴    t = fraction numerator rω subscript straight o over denominator 2 straight mu subscript straight k straight g end fraction 
           = fraction numerator 0.1 space straight x space 10 space straight x space 3.14 over denominator 2 space straight x space 0.2 space straight x space 9.8 end fraction
           =0.80 s                                       ...(vii)

    For the disc:
              I = open parentheses 1 half close parenthesesmr2
    ∴    μkgt = rω0 - μkmgr2t / 1 halfmr2 

                = rω0 - 2μkgt 

        3μkgt = rω0 

    Therefore,
     t = fraction numerator rω subscript straight o over denominator 3 space straight mu subscript straight k straight g end fraction

        = fraction numerator 0.1 space straight x space 10 space straight x space 3.14 space over denominator 3 space straight x space 0.2 space straight x space 9.8 end fraction space equals space 0.53 space sec      ...(viii) 

    Since td > tr, the disc will start rolling before the ring. 
    Question 593
    CBSEENPH11020149
    Wired Faculty App

    A cylinder of mass 10 kg and radius 15 cm is rolling perfectly on a plane of inclination 30°. The coefficient of static friction µs = 0.25. 

    (a) How much is the force of friction acting on the cylinder? 

    (b) What is the work done against friction during rolling? 

    (c) If the inclination θ of the plane is increased, at what value of θ does the cylinder begin to skid, and not roll perfectly?

    Solution
    Mass of the cylinder, m = 10 kg 
    Radius of the cylinder, r = 15 cm = 0.15 m 
    Co-efficient of kinetic friction, µk = 0.25 
    Angle of inclination, θ = 30° 
    Moment of inertia of a solid cylinder about its geometric axis, I =open parentheses 1 half close parentheses mr2
    The various forces acting on the cylinder are shown in the following figure, 


    The acceleration of the cylinder is given as, 
        a = mg Sinθ / [m + (I/r2) ] 
          = mg Sinθ / [m + {(1/2)mr2/ r2} ] 
          = open parentheses 2 over 3 close parentheses g Sin 30° 
          =open parentheses 2 over 3 close parentheses × 9.8 × 0.5 
          =  3.27 ms-2
    (a) Using Newton’s second law of motion,
      Net force, fnet = ma 
     
     mg Sin 30° - f = ma 

                         f = mg Sin 30° - ma 

                           = 10 × 9.8 × 0.5 - 10 × 3.27 
           49 - 32.7 = 16.3 N
    (b) During rolling, the instantaneous point of contact with the plane comes to rest.
    Hence, the work done against frictional force is zero.
    (c) For rolling without skid,
    Using the relation, 
                        μ = (1/3) tan θ 
                  tan θ = 3μ
                           = 3 × 0.25 
    ∴ θ = tan-1 (0.75)
          = 36.87°.
    Question 594
    CBSEENPH11020150
    Wired Faculty App

    Read each statement below carefully, and state, with reasons, if it is true or false;

    (a) During rolling, the force of friction acts in the same direction as the direction of motion of the CM of the body.

    Solution

    a) False
    Frictional force acts opposite to the direction of motion of the centre of mass of a body. In the case of rolling, the direction of motion of the centre of mass is backward. Hence, frictional force acts in the forward direction.

    Tips: -

    a) Frictional force acts opposite to the direction of motion of the centre of mass of a body. In the case of rolling, the direction of motion of the centre of mass is backward. Hence, frictional force acts in the forward direction.
    Question 595
    CBSEENPH11020151
    Wired Faculty App

    Read each statement below carefully, and state, with reasons, if it is true or false;

    (b) The instantaneous speed of the point of contact during rolling is zero. 

    Solution
    True
    Rolling can be considered as the rotation of a body about an axis passing through the point of contact of the body with the ground. Hence, its instantaneous speed is zero. 

    Question 596
    CBSEENPH11020152
    Wired Faculty App

    Read each statement below carefully, and state, with reasons, if it is true or false

    The instantaneous acceleration of the point of contact during rolling is zero.

    Solution
    False
    This is because when a body is rolling, its instantaneous acceleration is not equal to zero. Instantaneous acceleration will have some value. 
    Question 597
    CBSEENPH11020153
    Wired Faculty App

     Read each statement below carefully, and state, with reasons, if it is true or false;

    (d) For perfect rolling motion, work done against friction is zero.

    Solution
    True
    The perfect rolling begins because the force of friction becomes zero. Hence work done against friction is zero.
    Question 598
    CBSEENPH11020154
    Wired Faculty App

    Read each statement below carefully, and state, with reasons, if it is true or false;

    A wheel moving down a perfectly frictionless     inclined plane will undergo slipping (not rolling) motion.

    Solution
    True
    The rolling occurs only on account of friction which is a tangential force capable of providing torque. When the inclined plane is perfectly smooth, the wheel will simply slip under the effect of its own weight. 
    Question 601
    CBSEENPH11020282
    Wired Faculty App

    A student measured the length of a rod and wrote it as 3.50 cm. Which instrument did he use to measure it?

    • A meter scale.

    • A vernier calliper where the 10 divisions in vernier scale match with 9 division in the main scale and main scale have 10 divisions in 1 cm.

    • A screw gauge having 100 divisions in the circular scale and pitch as 1 mm.

    • A screw gauge having 100 divisions in the circular scale and pitch as 1 mm.
    Solution

    B.

    A vernier calliper where the 10 divisions in vernier scale match with 9 division in the main scale and main scale have 10 divisions in 1 cm.

    If student measures 3.50 cm, it means that there is an uncertainty of order 0.01 cm
    For vernier scale with 1 MSD = 1mm
    and 9 MSD = 10 VSD
    LC of vernier scale with 1 MSD - 1 VSD
    1 over 10 open parentheses 1 minus 9 over 10 close parentheses space equals space 1 over 100 cm

    Question 608
    CBSEENPH11020298
    Wired Faculty App

    A cylindrical tube, open at both ends, has a fundamental frequency, f, in the air. The tube is dipped vertically in water so that half of it is in water. The fundamental frequency of the air-column is now

    • m1r1:m2r2

    • m1 :m2

    • r1 :r2

    • 1:1
    Solution

    C.

    r1 :r2

    As their period of revolution is same, so its angular speed is also same. Centripetal acceleration is circular path,
    a= ω2r
    Thus, 
    straight a subscript 1 over straight a subscript 2 space equals space fraction numerator straight omega squared straight r subscript 1 over denominator straight omega squared straight r subscript 2 end fraction space equals space straight r subscript 1 over straight r subscript 2

    Question 609
    CBSEENPH11020310
    Wired Faculty App

    A screw gauge gives the following reading when used to measure the diameter of a wire. Main scale reading : 0 mm
    Circular scale reading: 52 division
    Given that 1 mm on the main scale corresponds to 100 divisions of the circular scale. The diameter of wire from the above data is

    • 0.052 cm

    • 0.026 cm

    • 0.005

    • 0.52
    Solution

    A.

    0.052 cm

    Least count of screw gauge =1/100 mm = 0.01 mn
    Diameter - Divisions on cirular scale × least count + main scale reading = 52 × 1/100 + 0
    100 = 0.52 mm
    diameter = 0.052 cm

    Question 610
    CBSEENPH11020313
    Wired Faculty App

    A thin horizontal circular disc is rotating about a vertical axis passing through its centre. An insect is at rest at a point near the rim of the disc. The insect now moves along a diameter of the disc to reach its other end. During the journey of the insect, the angular speed of the disc 

    • remains unchanged

    • continuously decreases

    • continuously increases 

    • first increases and then decreases
    Solution

    D.

    first increases and then decreases


    From angular momentum conservation about the vertical axis passing through centre. When the insect is coming from circumference to center. Moment of inertia first decreases then increase. So angular velocity increase than decrease. 
    Question 615
    CBSEENPH11020342
    Wired Faculty App

    The moment of inertia of a uniform cylinder of length

    • 1

    • fraction numerator 3 over denominator square root of 2 end fraction
    • square root of 3 over 2 end root
    • fraction numerator square root of 3 over denominator 2 end fraction

    Solution

    C.

    square root of 3 over 2 end root straight I space equals space fraction numerator straight m calligraphic l squared over denominator straight I 2 end fraction space plus mR squared over 4 or space straight I space equals space straight m over 4 open parentheses calligraphic l squared over 3 plus straight R squared close parentheses space space.... space left parenthesis 1 right parenthesis Also comma space straight m space equals space πR squared calligraphic l rho rightwards double arrow space R squared space equals space fraction numerator m over denominator pi </div> </div> <div class=
    Question 616
    CBSEENPH11020346
    Wired Faculty App

    A slender uniform rod of mass M and length

    • fraction numerator 3 straight g over denominator 2 calligraphic l end fraction space cos space straight theta
    • fraction numerator 2 space straight g over denominator 3 space calligraphic l end fraction space cos space straight theta
    • fraction numerator 3 straight g over denominator 2 space calligraphic l end fraction space sin space straight theta
    • fraction numerator 2 space straight g over denominator 3 space calligraphic l end fraction space sin space straight theta

    Solution

    C.

    fraction numerator 3 straight g over denominator 2 space calligraphic l end fraction space sin space straight theta
    Taking torque about pivot τ = Iα
    mg space sin space straight theta space calligraphic l over 2 space equals space fraction numerator straight m calligraphic l squared over denominator 3 end fraction space straight alpha straight alpha space equals space fraction numerator 3 straight g over denominator 2 space calligraphic l end fraction space sinθ
    Question 618
    CBSEENPH11020374
    Wired Faculty App

    The dimension of magnetic field in M, L, T and C (Coulomb) is given as

    • MLT−1C−1

    • MT2C−2

    • MT−1C−1

    • MT−2C−1
    Solution

    C.

    MT−1C−1

    F = qvB
    B = F/qv
    = MC−1 T−1

    Question 622
    CBSEENPH11020515
    Wired Faculty App

    From a disc of radius R and mass M, a circular hole of diameter R, whose rim passes through the centre is cut. What is the moment of inertia of the remaining part of the disc, about a perpendicular axis, passing through the centre?

    • 13 MR2/32

    • 11MR2/32

    • 9MR2/32

    • 15MR2/32
    Solution

    A.

    13 MR2/32

    Illustrating the above figure,

    As given in the above fig.,
    Moment of Inertia of the disc, I = Iremain + I(R/2)
    rightwards double arrow space straight I subscript remain space equals space straight I space minus space straight I subscript straight R divided by 2 end subscript

    Now, using the values, we get
    equals space MR squared over 2 space minus space open square brackets fraction numerator begin display style straight M over 4 end style open parentheses begin display style straight R over 2 end style close parentheses squared over denominator 2 end fraction plus straight M over 4 open parentheses straight R over 2 close parentheses squared close square brackets equals space MR squared over 2 space minus open square brackets MR squared over 32 plus MR squared over 16 close square brackets equals MR squared over 2 minus open square brackets fraction numerator MR squared space plus space 2 MR squared over denominator 32 end fraction close square brackets equals space MR squared over 2 minus fraction numerator 3 MR squared over denominator 32 end fraction equals space fraction numerator 16 space MR squared minus 3 MR squared over denominator 32 end fraction
    That is,
    Iremain = fraction numerator 13 space MR squared over denominator 32 end fraction
    Question 623
    CBSEENPH11020521
    Wired Faculty App

    A disc and a sphere of the same radius but differnt masses roll off two inclined planes of the same altitude and length. which one of the two objects gets to the bottom of the plane first?

    • Sphere

    • Both reach at the same time

    • Depends on their masses

    • Disc
    Solution

    A.

    Sphere

    a) Acceleration of an object rolling down an inclined plane is given by,
    straight a space equals space fraction numerator straight g space sin space straight theta over denominator 1 plus begin display style straight I over mr squared end style end fraction where comma space straight theta space equals space angle space of space inclination space of the space inclined space plane
    m = mass of the object,
    I = moment of Inertia about the axis through the centre of mass.
    For space disc comma
    straight I over mr squared space equals space fraction numerator bevelled 1 half space m r squared over denominator m r squared end fraction equals 1 half
    For solid sphere,
    straight I over mr squared equals fraction numerator bevelled 2 over 5 space m r squared over denominator m r squared end fraction space equals space 2 over 5
    For hollow sphere,
    straight I over mr squared equals fraction numerator bevelled 2 over 3 space m r squared over denominator m r squared end fraction equals 2 over 3
    therefore space straight a subscript disc space equals space fraction numerator straight g space sin space straight theta over denominator 1 plus begin display style 1 half end style end fraction space space space space space space space space space space space space space equals 2 over 3 space straight g space sinθ space space space space space space space space space space space space space equals space 0.66 space straight g space sin space straight theta straight a subscript solid space sphere end subscript space equals space fraction numerator straight g space sin space straight theta over denominator 1 plus begin display style 2 over 5 end style end fraction space space space space space space space space space space space space space space space space space space space equals space 5 over 7 space straight g space sin space straight theta straight a subscript hollow space sphere end subscript space equals space fraction numerator straight g space sin space straight theta over denominator 1 plus begin display style 2 over 3 end style end fraction space space space space space space space space space space space space space space space space space space space space space equals space 3 over 5 space straight g space sin space straight theta space space space space space space space space space space space space space space space space space space space space equals space 0.6 space straight g space sin space straight theta Clearly comma space straight a subscript solid space sphere end subscript space greater than space straight a subscript disk greater than straight a subscript hollow space sphere end subscript 
    Therefore, the given sphere is a solid sphere.
    asolid sphere = ahollow sphere > adisk

    Question 625
    CBSEENPH11020540
    Wired Faculty App

    A rod of weight w is supported by two parallel knife edges A and B and is in equilibrium in a horizontal position. The knives are at distance d from each other. The centre of mass of the rod is at distance x from A. The normal reaction on A is

    • wx over straight d
    • wd over straight x
    • fraction numerator straight w left parenthesis straight d minus straight x right parenthesis over denominator straight x end fraction
    • fraction numerator straight w left parenthesis straight d minus straight x right parenthesis over denominator straight d end fraction

    Solution

    D.

    fraction numerator straight w left parenthesis straight d minus straight x right parenthesis over denominator straight d end fraction

    As the weight w balances the normal reactions.

    So, 
    w= N1 +N2   ----- (i) 
    Now balancing torque about the COM,
    i.e. anti -clockwise momentum= clockwise momentum

    rightwards double arrow space straight N subscript 1 straight x space equals space straight N subscript 2 space left parenthesis straight d minus straight x right parenthesis Putting space the space value space of space straight N subscript 2 space from space equation space left parenthesis straight i right parenthesis comma space we space get straight N subscript 1 straight x space equals space left parenthesis straight W minus straight N subscript 1 right parenthesis left parenthesis straight d minus straight x right parenthesis rightwards double arrow space straight N subscript 1 straight x space equals space wd minus space wx minus straight N subscript 1 straight d space plus straight N subscript 1 straight x straight N subscript 1 straight d space equals negative space straight w space left parenthesis straight d minus straight x right parenthesis straight N subscript 1 space equals space fraction numerator straight w left parenthesis straight d minus straight x right parenthesis over denominator straight d end fraction
    Question 628
    CBSEENPH11020547
    Wired Faculty App

    The dimensions of left parenthesis straight mu subscript straight o straight epsilon subscript straight o right parenthesis to the power of negative 1 divided by 2 end exponent are

    • [L1/2 T-1/2]

    • [L-1T]

    • [LT-1]

    • [L1/2 T1/2]
    Solution

    C.

    [LT-1]

    left parenthesis straight mu subscript straight o straight epsilon subscript straight o right parenthesis to the power of negative 1 divided by 2 end exponent space is space velocity space of space light Hence space straight c space equals space fraction numerator 1 over denominator square root of straight mu subscript straight o straight epsilon subscript straight o end root end fraction space equals space left square bracket LT to the power of negative 1 end exponent right square bracket
    Question 630
    CBSEENPH11020552
    Wired Faculty App

    The moment of inertia of a uniform circular disc is maximum about an axis perpendicular to the disc and passing through

    • B

    • C

    • D

    • A
    Solution

    A.

    B

    Moment of inertia of circular disc = 1/2mR2
    Thus, as the distance between the centre and the point increases, a moment of inertia increases.

    Question 631
    CBSEENPH11020562
    Wired Faculty App

    If force (F), velocity (v) and time (T) are taken as fundamental units, then the dimensions of mass are

    • [FvT-1]

    • [FvT-2]

    • [Fv-1T-1]

    • [Fv-1T]
    Solution

    D.

    [Fv-1T]

    We know that,
    F = ma
    That is,
    F = mv/t
    straight m space equals space Ft over straight v
    So, left square bracket straight M right square bracket space equals space fraction numerator left square bracket straight F right square bracket left square bracket straight T right square bracket over denominator left square bracket straight V right square bracket end fraction space equals space left square bracket space Fv to the power of negative 1 end exponent straight T right square bracket

    Question 633
    CBSEENPH11020571
    Wired Faculty App

    When a mass is rotating in a plane about a fixed point, its angular momentum is directed along

    • a line perpendicular to the plane of rotation

    • the line making an angle of 45o to the plane of rotation

    • the radius

    • the tangent to the orbit
    Solution

    A.

    a line perpendicular to the plane of rotation

    We knows  L = m (r x v)
    So, here, angular momentum is directed along a line perpendicular to the plane of rotation.

    Question 636
    CBSEENPH11020601
    Wired Faculty App

    In an experiment, four quantities a,b, c and d are measured with percentage error 1%, 2%, 3% and 4% respectively. Quantity P is calculated as follows:

    P = fraction numerator straight a cubed straight b squared over denominator cd end fraction space percent sign, error in P is

    • 14 %

    • 10 %

    • 7 %

    • 4 %
    Solution

    A.

    14 %

    Here, P = fraction numerator straight a cubed straight b squared over denominator cd end fraction
    Therefore,
    fraction numerator increment straight P over denominator straight P end fraction space x space 100
    = open parentheses fraction numerator 3 increment straight a over denominator straight a end fraction plus fraction numerator 2 increment b over denominator b end fraction plus fraction numerator increment c over denominator c end fraction plus fraction numerator increment d over denominator d end fraction close parentheses x 100
    = 3 fraction numerator increment straight a over denominator straight a end fraction straight x 100 space plus space 2 space fraction numerator increment straight b over denominator straight b end fraction straight x 100 space plus space fraction numerator increment straight c over denominator straight c end fraction straight x 100 space plus space fraction numerator increment straight d over denominator straight d end fraction straight x 100
    = 3 x1 + 2 x 2 + 3 +4 
    = 3 + 4 + 3 + 4
    = 14 %

    Question 637
    CBSEENPH11020608
    Wired Faculty App

    A rod PQ of mass M and length L is hinged at end P. The rod is kept horizontal by a massless string tied to point Q as shown in the figure. When a string is cut, the initial angular acceleration of the rod is,

    • 3g / 2L

    • g/L

    • 2g/L

    • 2g/3L
    Solution

    A.

    3g / 2L

    Torque on the rod = moment of weight of the rod about P

    straight tau space equals space mg space straight L over 2              ... (i)
    Moment of inertia of rod about,
    straight P space equals space ML squared over 3              ... (ii)
    As straight tau = I straight alpha
    From equations (i) and (ii), we get
    Mg space straight L over 2 space equals space ML squared over 3 straight alpha therefore space space space space straight alpha space equals space fraction numerator 3 straight g over denominator 2 straight L end fraction

    Question 642
    CBSEENPH11020636
    Wired Faculty App

    The dimensions of left parenthesis straight mu subscript straight o straight epsilon subscript straight o right parenthesis to the power of negative 1 divided by 2 end exponent are

    • [L-1T]

    • [LT-]

    • [L-1/2 T1/2]

    • [L1/2 T-1/2]
    Solution

    B.

    [LT-]

    This expression for speed of light and the dimensions of speed of light are [LT-1].

    Question 645
    CBSEENPH11020654
    Wired Faculty App

    straight A space force space straight F space equals space straight alpha stack space straight i with hat on top space plus space 3 space straight j with hat on top space plus space 6 space straight k with hat on top space is space acting space at space straight a space point space straight r equals space 2 straight i with hat on top space minus space 6 space straight j with hat on top space minus 12 space straight k with hat on top. The value of α for which angular momentum about origin is conserved is

    • -1

    • 2

    • zero

    • 1
    Solution

    A.

    -1

    When the resultant external torque acting on a system is zero, the total angular momentum of a system is zero, the total angular momentum of a system remains constant.This is the principle of the conservation of angular momentum.
    Given,
    straight F space equals space straight alpha space straight i with hat on top space plus 3 space straight j with hat on top space plus space 6 space straight k with hat on top space is space acting space at space straight a space point space straight r equals space 2 space straight i with hat on top space minus 6 space straight j with hat on top space minus 12 space straight k with hat on top As comma space angular space momentum space about space origin space is space consverved straight i. straight e. comma Torque space equals space constant rightwards double arrow space Torque space equals space 0 space rightwards double arrow space straight r space xF space equals space 0 open vertical bar table row cell straight i with hat on top end cell cell straight j with hat on top end cell cell straight k with hat on top end cell row 2 cell negative 6 end cell cell negative 12 end cell row straight alpha 3 6 end table close vertical bar space equals space 0 rightwards double arrow space left parenthesis negative 36 space plus 36 right parenthesis straight i with hat on top space minus space left parenthesis 12 space plus space 12 straight alpha right parenthesis space straight j with hat on top space plus left parenthesis 6 space plus 6 straight alpha right parenthesis space straight k with hat on top space equals 0 rightwards double arrow space 0 space straight i with hat on top space minus space 12 space straight i with hat on top space left parenthesis space 1 plus straight alpha right parenthesis straight i with hat on top space plus space 6 space left parenthesis 1 space plus space straight alpha right parenthesis straight k with hat on top space equals 0 6 space left parenthesis 1 space plus straight alpha right parenthesis space equals 0 space rightwards double arrow space straight alpha space equals negative 1

    Question 646
    CBSEENPH11020656
    Wired Faculty App

    The cylindrical tube of a spray pump has radius R, one end of which has n fine holes, each of radius r. if the speed of the liquid in the tube is v, the speed of the ejection of the liquid through the holes is

    • vR2/n2r2

    • vR2/nr2

    • vR2/n3r2

    • v2R/nr
    Solution

    B.

    vR2/nr2

    During the streamline of viscous and incompressible fluid through a pipe varying cross -section, the product of the area of cross section and normal fluid velocity (Av) remains constant throughout the flow.
    Consider a cylindrical tube of a spray pump has radius R, one end having n fine holes, each of radius r and speed of liquid in the tube is v as shown in the figure.

    According to equation of continuity
    Av = constant
    where A is a cylindrical tube and v is the velocity of the liquid in a tube. 
    Volume in flow rate = volume an outflow rate
    πR2v = nπr2v'
    v' = R2v/nr2
    Thus, speed of the ejection of the liquid through the holes is R2v/nr2

    Question 647
    CBSEENPH11020657
    Wired Faculty App

    Point masses m1 and m2 are placed at the opposite ends of a rigid rod of length L and negligible mass. The rod of length L and negligible mass. The rod is to be set rotating about an axis perpendicular to it. The position of point P on this rod through which the axis should pass so that the work required to set the rod rotating with angular velocity straight omega subscript straight o  is minimum is given by 

    • straight x space equals space fraction numerator straight m subscript 1 straight L over denominator straight m subscript 1 plus straight m subscript 2 end fraction
    • straight x space equals space straight m subscript 1 over straight m subscript 2 straight L
    • straight x space equals space straight m subscript 2 over straight m subscript 1 straight L
    • straight x space equals space fraction numerator straight m subscript 2 straight L over denominator straight m subscript 1 space plus straight m subscript 2 end fraction

    Solution

    D.

    straight x space equals space fraction numerator straight m subscript 2 straight L over denominator straight m subscript 1 space plus straight m subscript 2 end fraction

    As two point masses m1 and m2 are placed at opposite ends of a rigid rod of length L and negligible as shown in the figure. 

    Total moment of inertia of the rod
     I = m1x2  + m2(L-x)2
    I = m1x2 + m2L2 +m2x2 - 2m2Lx
    As I is minimum i.e 
    dI over dx space equals space 2 straight m subscript 1 space straight x space plus space space 0 space plus space 2 xm subscript 2 space minus space 2 straight m subscript 2 straight L space equals space 0 rightwards double arrow space straight x space left parenthesis 2 straight m subscript 1 space plus space 2 straight m subscript 2 right parenthesis space equals space 2 straight m subscript 2 straight L rightwards double arrow space straight x space equals space fraction numerator straight m subscript 2 straight L over denominator straight m subscript 1 plus straight m subscript 2 end fraction
    When I is minimum, the work done by rotating a rod will1 divided by 2 straight omega squared space with space angular space velocity space straight omega subscript straight o be minimum.

    Question 653
    CBSEENPH11020696
    Wired Faculty App

    A gramophone record is revolving with an angular velocity ω. A coin is placed at a distance r from the centre of the record. The static coefficient of friction is μ. The coin will revolve with the record if

    • r = μgω2

    • straight r space less than space straight omega squared over μg
    • straight r space less or equal than space μg over straight omega squared
    • straight r space greater or equal than space μg over straight omega squared

    Solution

    C.

    straight r space less or equal than space μg over straight omega squared

    When the disc spins the frictional force between the gramophone record and coin is μ mg
    The coin will revolve with record, if 

    straight F subscript frictional space greater or equal than space straight F space centripetal straight mu space mg space greater or equal than space straight m space straight omega squared straight r μg over straight r space greater or equal than space straight omega squared space

    Question 655
    CBSEENPH11020705
    Wired Faculty App

    From a circular disc of radius R and mass 9M, a small disc of mass M and radius R/3 is removed concentrically. The moment of inertia of the remaining disc about an axis perpendicular to the plane of the disc and passing through its centre is

    • 40MR2/9

    • MR2

    • 4 MR2

    • 4MR2/9
    Solution

    A.

    40MR2/9

    The moment of inertia of the remaining disc about axis perpendicular to the plane of the disc and passing through its centre.

    straight I space equals space straight I subscript 1 minus straight I subscript italic 2 space equals fraction numerator 9 MR squared over denominator 2 end fraction minus MR squared over 18 equals space fraction numerator 81 space MR squared minus MR squared over denominator 18 end fraction equals space fraction numerator 40 space MR squared over denominator 9 end fraction

    Question 658
    CBSEENPH11020717
    Wired Faculty App

    If the dimensions of a physical quantity are given by Ma Lb Tc, the physical quantity will be 

    • pressure if a = 1, b = -1, c= 2

    • velocity if a = 1, b = 1, c = -2

    • acceleration if a = 1, b = 1, c =-2

    • force if a = 0, b = -1, c =- 2
    Solution

    A.

    pressure if a = 1, b = -1, c= 2

    Dimensions of velocity = [ M0L1T-1]
    Here, a = 0, b =1, c = -1
    Dimension of acceleration = [ M0L1T-2]
    Dimension of force = [ M1L1T-2]
    Dimension of pressure = [M1L-1T-2
    The physical quantity is pressure.

    Question 659
    CBSEENPH11020718
    Wired Faculty App

    Under the influence of a uniform magnetic field, a charged particle moves with constant speed v in a circle of radius R. The time period of rotation of the particle

    • depends on v and not on R

    • depends on R and not on v

    • is independent of both v and R

    • depends on both v and R
    Solution

    C.

    is independent of both v and R

    The time period of circular motion of their charged particle is given by 

    straight T space equals space fraction numerator 2 πr over denominator straight v end fraction space equals space fraction numerator 2 straight pi over denominator straight v end fraction space straight x mv over Bq straight T space equals space fraction numerator 2 πm over denominator Bq end fraction
    Hence, time period of rotation of the charged particle in uniform magnetic field is independent of both v and R.

    Question 661
    CBSEENPH11020731
    Wired Faculty App

    Four identical thin rods each of mass M and length l , inertial form a square frame. moment of inertia of this frame about an axis through the centre of the square and perpendicular to this plane is 

    • 4Ml2/3

    • 2Ml2/3

    • 13Ml2/3

    • Ml2/3
    Solution

    A.

    4Ml2/3

    Apply theorem of the parallel axis and the total moment of inertia will be the sum of the moment of inertia of each rod.
    Moment of inertia of rod about an axis through its centre of mass and perpendicular to rod + (mass) of rod) x (perpendicular distance between two axes)

    = MI squared over 12 space plus space M open parentheses l over 2 close parentheses squared space equals space open parentheses fraction numerator M I squared over denominator 3 end fraction close parentheses M o m e n t space o f space i n e r t i a space o f space t h e space s y s t e m space space equals space fraction numerator M l cubed space over denominator 3 end fraction x space 4 space equals space 4 over 3 space M l squared

    Question 662
    CBSEENPH11020737
    Wired Faculty App

    A bar magnet having a magnetic moment of 2 x 104 JT-1 is free to rotate in a horizontal plane. A horizontal magnetic field B = 6 x 10-4 T exists in the space. The work  is done in taking the magnet slowly from a direction parallel to the field to a direction 60o from the field is

    • 0.6 J

    • 12 J

    • 6 J

    • 2 J
    Solution

    C.

    6 J

    The work done in rotating a magnetic dipole against the torque acting on it, when placed in a magnetic field is stored inside it in the form of potential energy. 
    When magnetic dipole is rotated from initial position θ = θ1 to final position θ = θ2, then work done = MB (cos θ1 = cos θ2)
    space equals space MB space open parentheses 1 minus 1 half close parentheses space equals space fraction numerator 2 space straight x space 10 to the power of 4 space straight x space 6 space straight x space 10 to the power of negative 4 end exponent over denominator 2 end fraction space equals space 6 space straight J

    Question 664
    CBSEENPH11020743
    Wired Faculty App

    The ratio of the radii of gyration of a circular disc to that of a circular ring, each of same mass and radius, around their respective axes is 

    • square root of 3 colon square root of 2
    • 1 colon square root of 2
    • square root of 2 colon 1
    • square root of 2 colon space square root of 3

    Solution

    B.

    1 colon square root of 2

    The square root of the ratio of the moment of inertia of rigid body nad its mass is called radius of gyration.
    As in key idea, radius of gyration is given by
    straight K space equals space square root of straight I over straight M end root For space given space problem straight K subscript disc over straight K subscript ring space equals space square root of straight I subscript disc over straight I subscript ring end root But space straight I subscript disc space left parenthesis about space its space axis right parenthesis space equals space 1 half MR squared and space straight I subscript ring space left parenthesis about space its space axis right parenthesis space equals MR squared Where space straight R space is space the space the space radius space of space both space bodies. Therefore comma space eq left parenthesis straight i right parenthesis space becomes straight K straight K subscript disc over straight K subscript ring space equals space square root of fraction numerator begin display style 1 half end style MR squared over denominator MR squared end fraction end root space equals space 1 colon square root of 2

    Question 665
    CBSEENPH11020754
    Wired Faculty App

    A thin conducting ring and radius R is given charge +Q. The electric field at the centre O of the ring due to the charge on the part AKB of the ring is E. The electric field at the centre due to the charge on the part ACDB of the ring is 

    • 3E along KO

    • E along OK

    • E along KO

    • 3 E along OK
    Solution

    B.

    E along OK


    The fields at O due to AC and BD cancel each other. The field due to CD is acting in the direction OK and equal magnitude to E due to AKB.
    Question 666
    CBSEENPH11020757
    Wired Faculty App

    A thin rod of length L and Mass M is bent at its midpoint into two halves so that the angle between them is 90o. The moment of inertia of the bent rod about an axis passing through the bending point and perpendicular to the plane defined by the two halves of the rod is 

    • ML2/24

    • ML2/12

    • ML2/6

    • fraction numerator square root of 2 space ML squared over denominator 24 end fraction

    Solution

    B.

    ML2/12

    Since rod is bent at the middle, so each part of it will have same length (L/2) and mass (M/2) as shown,

    Moment of inertia of each part through its one end
     equals space 1 third open parentheses straight M over 2 close parentheses open parentheses straight L over 2 close parentheses squared
    Hence, net moment of inertia through its middle point O is
    straight I space equals space 1 third open parentheses straight M over 2 close parentheses open parentheses straight L over 2 close parentheses squared space plus 1 third open parentheses straight M over 2 close parentheses open parentheses straight L over 2 close parentheses squared equals 1 third open square brackets ML squared over 8 plus ML squared over 8 close square brackets space equals space ML squared over 12

    Question 668
    CBSEENPH11020768
    Wired Faculty App

    A Wheel has an angular acceleration of 3.0 rad/s2 and an initial angular speed of 2.00 rad/s. In a time of 2s it has rotated thorough an angle (in radian) of:

    • 6

    • 10

    • 12

    • 4
    Solution

    B.

    10

    Angular acceleration is time derivative of angular speed and angular speed is time derivative of angular displacement.
    By definition,
    straight alpha space equals space dω over dt ie comma space dω space equals space αdt So comma space if space in space time space straight t space the space angular space speed space of space straight a space body space changes space from space straight omega subscript straight o space to space straight omega integral subscript straight omega subscript straight o end subscript superscript straight omega space dω space equals space integral subscript 0 superscript straight r space straight alpha space dt If space straight alpha space is space constant straight omega space minus straight omega subscript straight o space equals space αt straight omega space equals space straight omega subscript straight o space plus αt Now comma space as space by space definition straight omega space equals space dθ over dt Eq space left parenthesis straight i right parenthesis space becomes dθ over dt space equals space space straight omega subscript straight o space plus αt dθ space equals space left parenthesis straight omega subscript straight o space plus αt right parenthesis dt So space straight m space if space in space time space straight t space angular space displacement space is space straight theta integral subscript 0 superscript straight theta dθ space equals space integral subscript 0 superscript straight t space left parenthesis straight omega subscript straight o space plus αt right parenthesis space dt Given comma space straight alpha space equals space 3.0 space rad divided by straight s squared comma space space equals space 2.0 space rad divided by straight s comma space straight t space equals space 2 straight s Hence comma space straight theta space equals space 2 space straight x space 2 space plus space 1 half space straight x space 3 space left parenthesis 2 right parenthesis squared or space straight theta space equals space 4 plus 6 space equals space 10 space rad

    Question 669
    CBSEENPH11020769
    Wired Faculty App

    A uniform rod AB of length l and mass m is free to rotate about point A. The rod is released from rest in the horizontal position. Given that the moment of inertia of the rod about A  is ml2/3, the initial angular acceleration of the rod will be:

    • 2g/3l

    • mgl/2

    • 3gl/2

    • 3g/2l
    Solution

    D.

    3g/2l

    The moment of inertia of the uniform rod about an axis through one end and perpendicular to its length is 
    l = ml3/3
    where m is a mass of rod and l is the
    straight tau space equals space mg straight iota over 2 Iα space equals space mg straight l over 2 ml squared over 3 straight alpha space equals space mg straight l over 2 straight alpha space equals space fraction numerator 3 straight g over denominator 2 straight l end fractionlength.
    Torque acting on centre of gravity of rod is given by

    Question 670
    CBSEENPH11020773
    Wired Faculty App

    A particle of mass m moves in the XY plane with a velocity v along the straight line AB. If the angular momentum of the particle with respect to origin O is LA when it is at A and LB when it is B, then: 

    • LA > LB

    • LA = LB

    • the relationship between LA and LB depends upon the slope of the line AB

    • LA < LB
    Solution

    B.

    LA = LB

    From the definition of angular momentum,
    bold L with bold rightwards arrow on top bold space bold equals bold space bold r with bold rightwards arrow on top bold space bold italic x bold P with bold rightwards arrow on top bold space equals space r m v space s i n space ϕ space left parenthesis negative bold k with bold rightwards arrow on top right parenthesis

    Therefore, the magnitude of L is 
    L = mvr sin Φ = mvd
    where d = r sin Φ is the distance of closest approach of the particle so the origin. As d is same for both the particles, hence, LA = LB

    Question 672
    CBSEENPH11020794
    Wired Faculty App

    The velocity v of a particle at time t is given by straight v equals at plus fraction numerator straight b over denominator straight t plus straight c end fraction comma where a, b and c are constants, The dimensions of a, b and c are respectively:

    • open square brackets LT to the power of negative 2 end exponent close square brackets comma space open square brackets straight L close square brackets space and space open square brackets straight T close square brackets
    • open square brackets straight L squared close square brackets space open square brackets straight T close square brackets space and space open square brackets LT squared close square brackets
    • open square brackets LT squared close square brackets comma space open square brackets LT close square brackets space and space open square brackets straight L close square brackets
    • open square brackets straight L close square brackets comma space open square brackets LT close square brackets space and space open square brackets straight T squared close square brackets

    Solution

    A.

    open square brackets LT to the power of negative 2 end exponent close square brackets comma space open square brackets straight L close square brackets space and space open square brackets straight T close square brackets

    According to principle of homogeneity of dimensions, the dimensions of all the terms in a physical expression should be same.
       The given expression is 
                          straight v equals at plus fraction numerator straight b over denominator straight t plus straight c end fraction
    From principle of homogeneity
         open square brackets straight a close square brackets space open square brackets straight t close square brackets space equals space open square brackets straight v close square brackets space space space space space open square brackets straight a close square brackets space equals space fraction numerator open square brackets straight v close square brackets over denominator open square brackets straight t close square brackets end fraction space equals space fraction numerator open square brackets LT to the power of negative 1 end exponent close square brackets over denominator open square brackets straight T close square brackets end fraction space equals space open square brackets LT to the power of negative 2 end exponent close square brackets Similarly comma space space open square brackets straight c close square brackets space equals space open square brackets straight t close square brackets space equals space open square brackets straight T close square brackets Further comma space space space space space fraction numerator open square brackets straight b close square brackets over denominator open square brackets straight t plus straight c close square brackets end fraction space equals open square brackets straight v close square brackets or space space space space space space space space space space space space open square brackets straight b close square brackets space equals space open square brackets straight v close square brackets space open square brackets straight t plus straight c close square brackets or space space space space space space space space space space space space open square brackets straight b close square brackets space equals space open square brackets LT to the power of negative 1 end exponent close square brackets space open square brackets straight T close square brackets space equals space open square brackets straight L close square brackets

    Question 675
    CBSEENPH11024911
    Wired Faculty App

    Calculate the degrees of freedom for monoatomic, diatomic and triatomic gas
    Solution

    D.O.F can be easily calculated using this equation
    D. O. F=3N
    Where N represents no. Of molecules
    Mono atomic (N =1)
    D.O.F=3 +0+0
    It has only 3 translation
    So Mono atomic gas molecule has 3 D. O. F
    Diatomic (N=2)
    D. O. F=3+2+1
    It has 3 translation, 2 rotation and 1 vibration
    So Diatomic gas molecule has 6 D. O. F
    Triatomic (N=3)
    D. O. F=3+3+(3N-6)
    It has 3 translation, 3 rotation and 3 vibration
    So Triatomic gas molecule has 9 D. O. F

     

    Question 676
    CBSEENPH11026114
    Wired Faculty App

    The length, breadth and thickness of a block are given by l = 12 cm, b = 6 cm and t = 2.45 cm. The volume of the block according to the idea of significant figures should be

    • 1 × 102 cm3

    • 2 × 102 cm3

    • 1.76 × 102 cm3

    • None of these
    Solution

    B.

    2 × 102 cm3

    Using relation for volume

    Given:- length of a block  =  12 cm

              breadth of a block  =  6 cm

            thickness of a block  =  2.45cm

    V =length  × breadth  × thickness

      =12  × 6  × 2.45  = 176.4

     =1.764 ×102 cm3

    The minimum number of significant figures is 1 in thickness, hence the volume will contain only one significant figure.

    Therefore V= 2 ×102 cm3 

    Question 677
    CBSEENPH11026133
    Wired Faculty App

    In a system of units if force (F), acceleration (A), and time (T) are taken as fundamental units, then the dimensional formula of energy is:

    • FA2T

    • FAT2

    • F2AT

    • FAT
    Solution

    B.

    FAT2

    Let energy E = Fx Ay T3

    Writing the dimensions on both sides

    M L2 T -2 =M  L T -2 x L T -2 y T 2M L2 T -2 =Mx Lx+y T -2x-2y+zOn comparing  both sides x =1x + y=1⇒  y = 2-x = 2 = 2-1 =1and   -2x - 2y +z = -2⇒  z = -2×1-2×1+z∴  Dimensional formula of energy     =F A T2

    Question 678
    CBSEENPH11026149
    Wired Faculty App

    A force F is given by F =at + bt2 ,where, t is time. What are the dimensions of a and b ?

    • M L T -1 and M L T0

    •  M L T -3 and M L2 T4

    • M L T -4 and M L T1

    • M L T -3 and M L T -4
    Solution

    D.

    M L T -3 and M L T -4

    Force F = at + bt2

    From the principle of homogeneity 

    Dimension of at = dimension of F

    ∴Dimension of a = Ft                                 =M L T -2T                                = M L T -3Dimension of b = dimension of Bb = Ft2= M L T-2T2                 = M L T -4

    Question 679
    CBSEENPH11026184
    Wired Faculty App

    Dimensions of resistance in an electrical circuit, in terms of dimension of mass M, of length L, of time T and of current I, would be

    • [ M L2 T-3 I-1]

    • [M L2 T-1]

    • [M L2 T-1 I-1]

    • [M L2 T-3 I-2 ]
    Solution

    D.

    [M L2 T-3 I-2 ]

    Resistance

    R = Dimensions of workDimensions of charge  Dimension of current  

        = M L2T -2I T I

    R = [ M L2 T-3 I-2 ]

    Question 680
    CBSEENPH11026228
    Wired Faculty App

    The physical quantity having the dimensions [ M-1 L-3 T3 A2 ] is

    • resistance

    • resistivity

    • electrical conductivity

    • electromotive force
    Solution

    C.

    electrical conductivity

    Resistivity ρ =mne2τ

    ∵   ρ = ML-3 AT2 T

              = [ M L3 A-2 T-3 ]

    So, electrical conductivity

           σ = 1ρ

           [ σ ] = 1σ

                   = [M-1 L-3 A2 T3 ]

    Question 681
    CBSEENPH11026395
    Wired Faculty App

    A physical quantity X is given by X = 2 K3 l2m n The percentage error in the measurement of k, l,  m  and n are 1 %, 2%, 3% and 4% respectively. The value of X is uncertain by

    • 8%

    • 10%

    • 12%

    • None of these
    Solution

    C.

    12%

    Given:-

               X = 2 k3 l2m n

    ∴       ∆XX = 3 ∆kk + 2 ∆ll + ∆mm + 12 ∆nn

    Percentage error in X

         ∆XX × 100 = 3∆kk + 2∆ll + ∆mm + 12∆nn

                          = 3 × 1% + 2 × 2% + 3% + 12 × 4%

                           = 3% + 4% + 3% + 2%

    ⇒     ∆XX × 100 = 12%

    Mock Test Series

    Sponsor Area

    Sponsor Area

    NCERT Book Store

    NCERT Sample Papers

    Entrance Exams Preparation

    3