Science Chapter 1 Matter In Our Surroundings
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    NCERT Solution For Class 9 About 2.html

    Matter In Our Surroundings Here is the CBSE About 2.html Chapter 1 for Class 9 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 9 About 2.html Matter In Our Surroundings Chapter 1 NCERT Solutions for Class 9 About 2.html Matter In Our Surroundings Chapter 1 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 9 About 2.html.

    Question 1
    CBSEENSC9004630

    Convert the following temperature to Celsius scale.

    300K

    Solution

    To change a temperature on the kelvin scale to the celsius scale. Subtract 273 form given temperature.

    300 K = 300 - 273 = 27°C

    Question 2
    CBSEENSC9004631

    Convert the following temperature to Celsius scale.

    573 K

    Solution
    To change a temperature on the kelvin to celsius. Subtract 273 from given temperature.

    573 K = 573 – 273 = 300°C
    Question 3
    CBSEENSC9004632

    Convert the following temperatures to Kelvin scale

    25°C

    Solution

    To change a temperature on the celsius to kelvin. Add 273 to given temperature. i.e.,
    25°C = 25 + 273 = 298K

    Question 4
    CBSEENSC9004633

    What is the physical state of water at

    100 °C

    Solution

    At 1000C water is in vapoues phase. 

    Question 5
    CBSEENSC9004634

    For any substance, why does the temperature remain constant during the change of state ?

    Solution

    During the change of state temperature remains constant because the heat given to the matter is used up in changing the state of matter. This is called latent heat.

    Question 6
    CBSEENSC9004635

    Suggest a method to liquify atmospheric gases.

    Solution

    Liquefaction of gases is the process by which substances in their gaseous state are converted to the liquid state. When pressure on a gas is increased, its molecules closer together, and its temperature is reduced, which removes enough energy to make it change from the gaseous to the liquid state.


    Question 7
    CBSEENSC9004636

    Convert the following temperatures to Kelvin scale.

    35°C

    Solution

    To change a temperature on the celsius scale to kelvin scale. Add 273 to given temperature.

    35°C = 35 + 273 = 308K

    Question 8
    CBSEENSC9004637

    Give reason for the following observations.

    Naphthalene balls disappear with time without leaving any solid.

    Solution

    Naphthalene shows the property of sublimation. Evaporation of naphthalene takes place easily and so it disappears during course of time without leaving any solid.

    Question 9
    CBSEENSC9004638

    Give reason for the following observations.

    We can get the smell of perfume sitting several metres away.

    Solution

    Perfume has high degree of vaporisation and its vapours diffuse into air easily. That is why we can smell perfume sitting several metres away.

    Question 10
    CBSEENSC9004639

    What is the physical state of water at

    a) 25°C
    b)00C
    c)1000C

    Solution
    Physical state of water at different temperature:
    (a)Liquid
    (b) Solid and Liquid
    (c) Liquid and Vapours
    Question 11
    CBSEENSC9004640

    What is the physical state of water at
    0°C

    Solution

    Water is in solid state at 0°C.

    Question 12
    CBSEENSC9004641

    What is the physical state of water at
    100°C

    Solution

    Water is in gas phase at 100°C.

    Question 13
    CBSEENSC9004642

    Give two reasons to justify
    Water at room temperature is a liquid.

    Solution

    At room temperature water is a liquid because of the following characteristic of liquid:

    1) At room temperature, water has no shape but a fixed volume that is, it occupies the shape of the container in which it is kept.

    2) Water at room temperature is a liquid because it has fluidity.

    Question 14
    CBSEENSC9004643

    Give two reasons to justify
    An iron almirah is a solid at room temperature.

    Solution

    An iron almirah is a solid at room temperature because:

    1) It has a definite shape and volume.

    2) It is rigid at room temperature. 

    Question 15
    CBSEENSC9004644

    Why is ice at 273K more effective in cooling than water at the same temperature ?

    Solution

    Ice at 273 K is less energetic than water. It is because of the difference in the latent heat of fusion which is present in water at the same temperature in the form of extra energy.

    Question 16
    CBSEENSC9004645

    What produces more severe burns, boiling water or steam ?

    Solution

    Steam produces more severe burns than boiling water. This is because steam has more energy than boiling water,present in it in the form of latent heat of vaporization.

    Question 17
    CBSEENSC9004646

    8. Name A, B, C, D, E and F in the following diagram showing state change :

    Solution

    A — Melting

    B — Evaporation

    C — Liquefaction

    D — Fusion

    E — Sublimation

    F — Solidification

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    Question 18
    CBSEENSC9004647

    Both boiling and evaporation convert a liquid into vapour. What is the difference between the two process

    Solution

    Evaporation is phase transition from liquid to vapour phase. It occur at temperatures below the boiling temperature at given temperature. Evaporation is a surface phenomenon.

    Boiling is phase transition from liquid state to gas phase that occurs at or above the boiling point. Boiling is bulk phenomenon. 

    Question 19
    CBSEENSC9004648

    Define Evaporation

    Solution

    The phenomenon of change of a liquid into vapours at any temperature below its boiling point is called evaporation. For example if Water left in open vessel at room temperature disappears after some time due to evaporation.

    Question 20
    CBSEENSC9004649

    Discuss the factors which affect evaporation

    Solution

    There are four factors which affect evaporation.

    (a) Surface area. Evaporation is a surface phenomenon. Thus escaping of particles from liquid state to vapour state depends on surface area. Therefore, the rate of evaporation increases with surface area.

    (b) Temperature of the system. With increase of surface temperature, the number of particles with larger kinetic energy increases and there is greater chance of escape of particles from liquid to vapour state. Therefore, rate of evaporation increases with temperature.

    (c) Wind. More particles of liquid vapour would be carried away from the surface of the liquid with increasing speed of wind. Thus rate of evaporation increases with speed of wind.

    (d) Humidity. Humidity is the amount of vapour present in the air. At a given temperature, air cannot hold more than a fixed amount of water vapour. Therefore, the rate of evaporation decreases with increase in the humidity of air.

    Question 21
    CBSEENSC9004650

    Why does evaporation causes cooling of a liquid system

    Solution

    Due to evaporation, only high energy particles leave the surface. This results in decrease of average kinetic energy of the particles in the system. Kinetic energy is related to temperature. As a result there is a drop in temperature of the liquid. Thus evaporation causes cooling.

    Question 22
    CBSEENSC9004651

    Why do wet clothes kept in shade dry slowly on a rainy day even in summer ?

    Solution

    During rainy day, the humidity of air increases, i.e., air is quite saturated with water vapour or air cannot hold more water vapour. Therefore, rate of evaporation of water from wet clothes decreases. So wet clothes dry slowly in rainy day.

    Question 23
    CBSEENSC9004652

    Why do we see water droplets on the outer surface of a glass containing ice-cold water

    Solution

    The air around ice-cold water container contains water vapours which have higher energy than ice-cold water. Therefore, water vapour present in the air loses energy when come in contact with cold water. These are then converted into liquid state, which we see as water droplets.

    Question 24
    CBSEENSC9004653

    Why should we wear cotton clothes in summer

    Solution

    In summer we feel hot. The body temperature should remain constant. In order to keep body cool, the water from the body gets evaporated and gets deposited on the body surface as sweat.
     
    Cotton allows better air circulation, which helps in absorbing and removing body moisture caused by sweat. It therefore minimizes fungal breeding by keeping your body dry and cool. Therefore we should wear cotton clothes in summer.

    Question 25
    CBSEENSC9004654

    How does our body maintain its temperature during summer

    Solution

    During summer, we perspire more because of the mechanism of our body which keeps us cool. During evaporation, the particle at the surface of the liquid gain energy from body surface and change into vapour phase. In this process the heat energy equal to latent heat of vaporization is absorbed from the body leaving the body cool. 

    Question 26
    CBSEENSC9004655

    Define (i) Latent heat of fusion and (ii) Latent heat of vaporisation

    Solution

    Latent heat of fusion is defined as the amount of energy required to convert of solid to liquid at 1 atmospheric pressure at constant temperature.


    Latent heat of vaporsiation is defined as the amount of heat energy required to convert water to gas phase at 100°C and 1 atmosheric pressure.



    Question 27
    CBSEENSC9004656

    Benzene is a liquid. At 80°C, liquid benzene is in equilibrium with its vapors. It is found that particles of benzene vapors are more energetic than particles of liquid benzene. Explain the observation.

    Solution

    Particles of benzene vapors are more energetic because of benzene vapours have gained extra energy in the form of latent heat of vaporisation. Hence it is more energetic.

     

    Question 28
    CBSEENSC9004657

    What is the relation between boiling point of a liquid and the molecular forces of attraction between the particles of a liquid?

    Solution

    Boiling point of liquid depend on the intermolecular force, Stronger the intermolecular forces of attraction, higher would be the boiling point.  For example, particles of alcohol have weaker intermolecular forces of attraction than of water particles. This is in conformity with the fact that alcohol boils at lower temperature, of 78°C as compared to boiling point of water which is 100°C.

    Question 29
    CBSEENSC9004658

    Change the following temperatures to Kelvin scale.

    1) 40°C 

    2) 80°C 

    3) 213°C

    Solution
    To change a temperature on the celsius scale to the kelvin scale. Add 273to given temperature.

    1) 40°C = 40 +273 = 313K

    2) 80°C = 80 +273 =353K

    3) 213°C = 213 +273=486K
    Question 30
    CBSEENSC9004659

    Why is ice at 273K is less energetic than water at the same temperature

    Solution

    Cooling takes place when heat is removed from a system, in case of ice at 273 K, it will take latent heat from the medium to convert itself into water at 273K there will be a change in physical state, whereas in case of water at 273 K there will be no change in state. Hence lesser energy will be taken from the medium. 

    Question 31
    CBSEENSC9004660

    What is sublimation ? Give an example with illustration.

    Solution

    Generally matter changes its state on heating from solid to liquid and from liquid to gas. However, there are some substances that change directly from solid state to gaseous state and vice versa. This change of state from solid to gas without changing into liquid state or vice versa is called sublimation. For example, camphor on heating directly changes into vapors. Another example of Sublimation is ammonium chloride.
    Illustration:


    1) Take some ammonium chloride. Powder it and put in a China dish. Put an inverted funnel over the China dish. Put some cotton plug on the item of the funnel. Now heat the China dish .
    2) You will notice fine crystal particles depositing on the inner walls of the funnel. Thus, on heating ammonium chloride is directly converted into ammonium chloride vapours which condense on the inner walls of the China dish.

    Fig. Sublimation of ammonium chloride.

    Question 32
    CBSEENSC9004661

    Select the substances which follow the process of sublimation : Ice, ammonium chloride, salt, naphthalene, ghee, coconut oil, camphor.

    Solution

    Sublimation is the transition of a substance directly from the solid to the gas phase without passing through the intermediate liquid phase. 

    Ammonium chloride, naphthalene and camphor follow the process of sublimation.

    Question 33
    CBSEENSC9004662

    Ghee freezes at room temperature and mustard oil does not (in winter). Which of these has a higher melting point and lower intermolecular forces

    Solution

    Ghee freezes at room temperature and mustard oil does not, it shows that intermolecular forces between particles of ghee are stronger compared to those of mustard oil. Higher the intermolecular forces, higher is the melting point. Therefore, ghee has a higher melting point than that of mustard oil.

    Question 34
    CBSEENSC9004663

    Which of the following shows phenomenon of sublimation ? Solid water, solid carbon dioxide, solid alcohol, solid oxygen.

    Solution

    Sublimation is phenomenon in which solid phase is directly converted to liquid phase or vice versa.

    Solid carbon dioxide.

    Question 35
    CBSEENSC9004664

    What is the effect of temperature and pressure on a gas ?

    Solution

    At a given pressure, reduction of temperature liquefies a gas at some point. At a given temperature, increase of pressure also results in liquifying a gas. 

    Question 36
    CBSEENSC9004665

    How do you differentiate between a gas and a vapour ?

    Solution

    Vapour phase is not a state of matter. Vapour is a substance in the gas phase at a temperature lower than its critical point. it means that the vapour can be condensed to a liquid by increasing the pressure on it without reducing the temperature.

    Gas is the state of matter distinguished from the solid and liquid state. Gas has relatively low density and viscosity, having great expansion and contraction which change with pressure and temperature.

    Question 37
    CBSEENSC9004666

    How are particles of matter affected with increasing or reducing pressure on the matter at a given temperature ?

    Solution

    On increasing pressure, particles of a matter come closer and these move apart on reducing pressure at a given temperature. Thus if pressure is increasingly applied on a gas, particles of the gas come closer and closer and eventually the gas may change into liquid and then solid form. Fig. shows three cylinders in which a matter is present as gas, liquid and solid.



    Changing state of matter on application of pressure : (a) solid; (b) liquid; (c) gas.

    Question 38
    CBSEENSC9004667

    How do you define pressure ?

    Solution

    The pressure is the force per unit area exerted by a gas on the walls of the container. It is measured in atmospheres (atm) or Pascals (Pa) 1 atm = 1.01 x 105 Pa

    Question 39
    CBSEENSC9004668

    What is atmospheric pressure ?

    Solution

    The pressure of air in atmosphere is called atmospheric pressure. It is 1 atmosphere at sea level and is also called normal atmospheric pressure.

    The unit of pressure is Pascal (Pa). 1 atmosphere = 1.01 × 105 Pa.

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    Question 40
    CBSEENSC9004669

    Define Evaporation.

    Solution

    The phenomenon of change of a liquid into vapours at any temperature below its boiling point is called evaporation. For example if water left in open vessel at room temperature disappears after some time due to evaporation.

    Question 41
    CBSEENSC9004670

    Why does evaporation causes cooling of a liquid system ?

    Solution
    Due to evaporation, only high energy particles leave the surface. This results in decrease of average kinetic energy of the particles in the system. The particle of liquid absorb energy from the surrounding to regain the energy lost during evapotation. This absorption of energy from the surroundings make the liquid cool.
    Question 42
    CBSEENSC9004671

    Why does a desert cooler cool better on a hot dry day.

    Solution

    On a hot dry day, the rate of evaporation is direcrtly proportional to the temperature and inversely proportional to humidity. Therefore, On the hot dry day the rate of evaporation increase thus causing a cooling effect.

    Question 43
    CBSEENSC9004672

    How does the water kept in an earthen pot (matka) become cool during summers ?

    Solution

    Earthen pots have small pores through which the water seeps and  evaporation leads to cooling of the water.

     

    Question 44
    CBSEENSC9004673

    Why does our palm feel cold when we put some acetone or petrol or perfume on it ?

    Solution

    Acetone, perfume and petrol are volatile liquids, i.e., they have low boiling points. When kept on palm, they can absorb enough energy from the palm or surroundings and evaporate and leave the palm cool.


    Question 45
    CBSEENSC9004674

    Why are we able to sip hot tea or milk faster from a saucer rather than a cup ?

    Solution

    Hot tea or milk cover larger surface area on a saucer and resulting in increase of evaporation. The tea or milk thus cool faster on a saucer and we can sip it more comfortably.


    Question 46
    CBSEENSC9004675

    What types of clothes should we wear during summer season ?

    Solution

    We should wear cotton clothes during summer season. Because during summer we perspire more to maintain the temperature of our body. Cotton is a good absorber of water (sweat). It then exposes the water to atmosphere for easy evaporation leaving us dry and cool.

    Question 47
    CBSEENSC9004676

    What is plasma ?

    Solution

    Plasma is a state of matter in which the state consists of super energetic and super excited particles. These particles are in the form of ionised gases. 

    Question 48
    CBSEENSC9004677

    Is it true to say that fluorescent tube contains only plasma ? Explain.

    Solution

    It is not correct to say that fluorescent tube contains only plasma. Inside a fluorescent tube there is helium gas or some other gas. The gas gets ionised, that is, gets charged when electrical energy flows through it. 

    Question 49
    CBSEENSC9004678

    Why do the sun or the stars glow ?

    Solution

    Sun or the stars glow because at their core the pressure is so high that hydrogen fuses to create helium. This process releases a great amount of energy, which result in growing the sun or stars.

     

    Question 50
    CBSEENSC9004679

    Do all particles or plasma glow similarly ?

    Solution

    The plasma glows with different colours depending on the nature of the containing gas. Neon sign bulb (containing neon gas) glows differently than a fluorescent tube (containing helium gas).

    Question 51
    CBSEENSC9004680

    What is the difference between a gas and plasma ?

    Solution

    Gas consists of particles which are neutral and are associated with energies comparable with atmospheric temperature.

    Plasma are super energenic particles in the form of ionized gases.

    Question 52
    CBSEENSC9004681

    How can matter be classified on the basis of their physical nature ?

    Solution

    Matter can classified in three different states.
    solid, liquid and gas.

    These states of matter arise due to the variation in the characteristics of the particles of matter.

    Question 53
    CBSEENSC9004682

    Which of the following substances is most compressible ?

    Carbon dioxide, water, sodium chloride salt.

    Solution

    Carbon dioxide is most compressible becuase of the larger space between the molecule. 

    Question 54
    CBSEENSC9004683

    Which property of a gas results in study pressure of the gas ?

    Solution

    The constant bombardment of gas particles with the walls of the container of the ?as gives pressure of the gas.

    Question 56
    CBSEENSC9004685
    Question 57
    CBSEENSC9004686

    Diffusion takes place fastest when two gases are brought in contact with each other ? Which characteristic of gases is responsible for this behaviour ?

    Solution

    Force of attraction between the gases molecule is very low. Therefore gases are in a state of constant rapid motion in all direction. This characteristic behaviour of gases responsible for diffusion.

    Question 58
    CBSEENSC9004687

    Which state of matter has no open surface ?

    Solution

    Gases.

    Question 59
    CBSEENSC9004688
    Question 60
    CBSEENSC9004689

    Which states of matter cannot form heaps ?

    Solution

    Liquids and gases cannot form heaps.

    Question 61
    CBSEENSC9004690

    Which states of matter take the shape of the containing vessel ?

    Solution

    Liquids and gases states of matter take the shape of the containing vessel depending upon condition.

    Question 65
    CBSEENSC9004694

    What name is given to the energy possessed by a particle due to its motion ?

    Solution

    The energy posses  by a partcle due to its motion known as Kinetic energy.

    Question 66
    CBSEENSC9004695
    Question 67
    CBSEENSC9004696

    Which of the following matter has lowest density ?

    Water, sugar, salt, ice.

    Solution

    Liquids generally have lower density as compared to solids. Density of water is the maximum at 4°C. The density of ice is less than that of the water. Thus, ice floats on water. Hence ice has lowest density.

    Question 68
    CBSEENSC9004697

    State the property of matter that enables aquatic plants and animals survive in water

    Solution

    Diffusion of gases into water helps aquatic plants and animals to survive in water.

    Question 69
    CBSEENSC9004698

    How does the rate of diffusion change with temperature ?

    Solution

    Kinetic energy of gas molecule increase with increasing temperature. Hence rate of difussion also increases with temperature.

    Question 71
    CBSEENSC9004700
    Question 73
    CBSEENSC9004702
    Question 75
    CBSEENSC9004704

    Name the two parameters which determine the physical state of matter.

    Solution

    Temperature and pressure can determine the physical state of matter.

    Question 78
    CBSEENSC9004707

    What is normal atmospheric pressure ?

    Solution

    Atmospheric pressure at sea level is 1 atmosphere, and is taken as the normal atmospheric pressure. 1 atmosphere = 1.01 × 105 Pa. 

    Question 79
    CBSEENSC9004708

    All materials occupy ______ and have ______.

    Solution

    space

    ,

    mass

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    Question 80
    CBSEENSC9004709
    Question 81
    CBSEENSC9004710

    A material in gaseous state can fill any ______________.

    Solution

    container/vessel

    Question 83
    CBSEENSC9004712
    Question 89
    CBSEENSC9004718
    Question 92
    CBSEENSC9004721

    Evaporation results in ___________.

    Solution

    cooling

    Question 93
    CBSEENSC9004722

    Evaporation is  ___________  phenomenon.

    Solution

    surface

    Question 94
    CBSEENSC9004723
    Question 95
    CBSEENSC9004724
    Question 103
    CBSEENSC9004732
    Question 108
    CBSEENSC9004737
    Question 110
    CBSEENSC9004739
    Question 112
    CBSEENSC9004741

    Arrange the following substances in increasing order of forces of attraction between the particles — water, sugar, oxygen.

    Solution

    The intermolecular force of attraction is least in gas, followed by liquid and maximum in gas. Therefore it least in oxygen followed by water and maximum in sugar. Therefore the increasing order :

    Oxygen<water<sugar

    Question 113
    CBSEENSC9004742

    What do you mean by latent heat?

    Solution

    Latent heat define as the quantity of heat absorbed or released by a substance undergoing a change of state, such as At 0°C, when the ice melts, the temperature of the system does not change though the heat is continuously supplied. The given heat is used up in overcoming the force of intermolecular attraction. This results in the conversion of ice into water. As this heat energy is absorbed by ice without any rise in temperature of the system, this heat is considered as hidden in the system and is known as latent heat.

    Question 114
    CBSEENSC9005651

    An athlete completes one round of a circular track of diameter 200 m in 40 s. What will be the distance covered and the displacement at the end of 2 minutes 20 s?

    Solution

    Diameter of circular track (D) = 200 m
    Radius of circular track (r) = 200/2=100 m
    Time taken by the athlete for one round (t) = 40 s
    Distance covered by athlete in one round (s) = 2π r
    = 2 × ( 22 / 7 )×100
    Speed of the athlete (v) = Distance / Time
    = (2 × 2200)/(7 × 40)
    = 4400 / 7 × 40
    ∴ Distance covered in 2 minutes 20
    seconds (s) or 140 s
    = Speed (s) × Time(t)
    = 4400 / (7 × 40) × (2×60 + 20)
    = 4400 /( 7 × 40) × 140
    = 4400 × 140 /7 × 40
    =2200 m
    Number of round in 40 s =1 round
    Number of round in 140 s =140/40
    =3.5 rounds

    Question 115
    CBSEENSC9005652

    Joseph jogs from one end A to the other end B of a straight 300 m road in 2 minutes and 30 seconds and then turns around and jogs 100 m back to point C in another 1 minute. What are Joseph’s average speeds and velocities in jogging (a) from A to B and (b) from A to C?

    Solution
    i) For motion from A to B, 

    Disatnce covered = 300 m
    Time taken = 2 minutes 30 seconds = 2.50 × 60 = 150 s

    Average speed = Distance coveredTime taken

                          = 300150= 2 m/s

    Displacement = 300 m

    So, average velocity = Displacement time taken
                                  = 300150= 2 m/s

    ii) In the second case, there is motion from A to B to C, 

    Distance covered = 300+ 100 = 400 m
    Displacement = AB - CB = 300 - 100 = 200 m
    Time taken = 2.50 + 1.00 = 3.50 min = 3.50×60 = 210 sec

    So, Average speed = distance travelledTime taken

                                =400210 = 1.90 m/s

    Average velocity = Displacementtime taken 

                             = 200210 = 0.952 m/s








    Question 118
    CBSEENSC9005655

    A driver of a car travelling at 52 km h–1 applies the brakes and accelerates uniformly in the opposite direction. The car stops in 5 s. Another driver going at 34 km h–1 in another car applies his brakes slowly and stops in 10 s. On the same graph paper, plot the speed versus time graphs for the two cars. Which of the two cars travelled farther after the brakes were applied?

    Solution

    Initial speed of car A = 52 km/hr
    Time at which the car A comes to a stop = 5 s
    Initial speed of car B = 34 km/h
    Time at which the car B comes to rest = 10 s
    Speed versus time graph for two cars:

    In Fig. AB and CD are the speed-time graphs for the two cars whose initial speeds are 52 km/h and , respectively. 
    Distance covered by first car before coming to rest is equal to Are of traingle AOB. 
    
space equals space 1 half cross times AO space cross times space BO
space equals space 1 half cross times space 52 space km divided by straight h space cross times space 5 space straight S
equals space 1 half cross times fraction numerator 52 space cross times space 5 over denominator 18 end fraction straight m divided by straight s space cross times space 5 space straight s space
equals space 36.1 space straight m
    Distance covered by the second car before coming to rest is, 
    space equals space Area space of space triangle space COD space equals space 1 half cross times CO cross times DO
equals space 1 half cross times space 34 space km divided by straight h space cross times space 10 space straight s
equals space 1 half cross times fraction numerator 34 space cross times space 5 over denominator 18 end fraction straight m divided by straight s space cross times space 10 space straight s
equals space 47.2 space straight m. 
    Thus, the second car travels farther than the first car after the brakes are applied.


    Question 119
    CBSEENSC9005656

    Fig. shows the distance-time graph of three objects A, B and C. Study the graph and answer the following questions:

    (a) Which of the three is travelling the fastest?

    (b) Are all three ever at the same point on the road?

    (c) How far has C travelled when B passes A?

    (d) How far has B travelled by the time it passes C?

    Solution
    left parenthesis straight i right parenthesis
Speed space of space straight A space equals space Slope space of space PN 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 10 minus 6 over denominator 1.1 space minus 0 end fraction

space space space space space space space space space space space space space space space space space space space space space equals 40 over 11 km divided by straight h

space space space space space space space space space space space space space space space space space space space space space equals 3.63 space km divided by straight h

Speed space of space straight B space equals space Slope space of space OM 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 6 minus 0 over denominator 0.7 minus 0 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 60 over 7 km divided by straight h space
space space space space space space space space space space space space space space space space space space space space equals space 8.57 space km divided by straight h

Speed space of space straight C space equals space Slope space of space QM 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 6 minus 2 over denominator 0.7 minus 0 end fraction space

space space space space space space space space space space space space space space space space space space space space equals 40 over 7 km divided by straight h

space space space space space space space space space space space space space space space space space space space equals space 5.71 space km divided by straight h

    Thus, object B is travelling the fastest. 
    (ii) No, all three objects do not meet at any point on the road. 
    (iii) When B passes A at point N (at 1.2 hours), C is at a distance of 8 km from the origin O. 
    (iv) B passes C at 0.7 hours. During this time B covers distance = 6 km. 



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    Question 122
    CBSEENSC9005659

    State which of the following situations are possible and give an example of each of these:

    (a) a body moving with constant acceleration but with zero velocity.

    (b) a body moving in a certain direction with an acceleration in the perpendicular direction.

    Solution

    (a) Even when a body's velocity is zero a body can have acceleration. 

    For example, when a body is thrown up, at it's highest point veocity is zero. But, the acceleration of the body is equal to acceleration due to gravity.

    (b) It is possible to have acceleration in the perpendicular direction for a body moving in a certain direction. 

    For example, an aeroplane moving horizontally is acted upon by acceleration due to gravity that acts vertically downwards. 

    Question 124
    CBSEENSC9005661

    Define the terms rest and motion. Give one example for each.

    Solution
    Rest: If a body does not change its position with respect to its surroundings, the body is said to be at rest.
    E.g. A table lying in a room is at rest with respect to the walls of the room. 

    Motion: A body is said to be in motion if it changes its position with respect to its surroundings.
    E.g., A car running on the road is in motion with respect to the lamp posts, trees or bus stop on the roadside
    Question 125
    CBSEENSC9005662

    Give one example each for directly and indirectly perceivable motions.

    Solution

    (i) Directly perceivable motion: Motion of vehicles moving on a road.

    (ii) Indirectly perceivable motion: We perceive the motion of air indirectly by observing the movement of dust, leaves and branches of trees.

    Question 126
    CBSEENSC9005663

    Show that rest and motion are relative terms.

    Solution

    Rest and motion are relative terms. this can be illustrated below using the following explanation.

    Let's consider for example, a passenger sitting in a moving train. The passenger is at rest relative to other passenger of his compartment. But, the passenger is in motion relative to the outside trees, lamp posts, railway stations, etc. 
    So, we can see that rest and motion are in respect to relativity. 

    Question 127
    CBSEENSC9005664

    Describe the various types of motions observed in bodies.

    Solution

    Various types of motion as observed in bodies are: 

    1. Translatory motion: When a body moves as a whole along a straight or curved path, it is said to be in translatory motion.
    Translatory motion is again of two types:

    (i) Rectilinear motion: Here a body moves as a whole along a straight path.
    For example, a train moving on straight rails has translatory rectilinear motion.

    (ii) Curvilinear motion: In this case a body moves as a whole along a curved path.
    For example, motion of a bicycle taking a turn along a curved path.

    2. Rotatory motion: When a body rotates about a fixed point or axis, it exhibits a rotatory motion.
    For example, motion of a flywheel about a shaft.

    3. Vibratory or oscillatory motion: When a body moves to and fro about a mean position, the motion is said to be vibratory or oscillatory motion.
    For example, the motion of the pendulum of a wall-clock.

    4. Complex motion: When the motion of a body may be a combination of more than one types of motion, it is said to be a complex motion. 
    For example, a ball rolling down an inclined plane has both translatory and rotatory motions.

    Question 128
    CBSEENSC9005665

    What is meant by a point object? Give some examples.

    Solution

    Whenever the size of the object is much smaller than the distance it moves in a given time interval, the size of the object can be neglected. The objet can be regraded as point object in such cases.

    Examples:

    (i) A car covering a distance of 10 km can be treated as a point object.

    (ii) Earth can be regarded as a point object for studying its motion around the Sun.

    Question 129
    CBSEENSC9005666

    How can we specify the position of an object?

    Solution

    The position of an object can be specified by choosing:
    (i) a fixed point called ‘origin’ or reference point, and
    (ii) a fixed line passing through the origin, called reference axis.
    So the position of an object can be fully described by knowing:
    (i) its distance from origin O and
    (ii) the angle θ which the line joining the origin O and the object makes with the reference axis.
    In Fig.  the position of an object located at point P is 6 m from the origin and 30° north of east.

    Fig. To specify the position of an object (Scale: 1 cm = 1m).

    Question 130
    CBSEENSC9005667

    What are scalar quantities? Give examples.

    Solution

    Physical quantities which require only magnitude and not the direction for their complete description of motion are called ‘scalars’ or ‘scalar quantities’.

    For example, distance, speed, time, area, etc. are some of the scalar quantities.

    Question 131
    CBSEENSC9005668

    What are vector quantities? Give examples.

    Solution

    Physical quantities which need both magnitude and direction for their complete description of motion are called ‘vectors’ or ‘vector quantities’.

    Some examples of vector quantities are displacement, velocity, force, etc.

    Question 132
    CBSEENSC9005669

    Classify the following as scalar and vector quantities:

    Temperature, mass, volume, speed, displacement, time, distance, velocity, electric charge, acceleration, force, momentum, density, work, weight, energy.

    Solution

    Scalar quantities: Temperature, mass, volume, speed, time, distance, electric charge, density, work and energy.

    Vector quantities: Displacement, velocity, acceleration, force, momentum and weight.

    Question 133
    CBSEENSC9005670

    What is meant by motion in a straight line ? Give some examples of such a motion.

    Solution

    When the position of a body changes with time, the body is said to be moving in a straight line. 

    Examples of motion in a straight line or rectilinear motion are:

    (i) A bus moving on a straight road,

    (ii) A train moving on a straight track,

    (iii) A runner running along a straight track,

    (iv) A ball moving along a straight path, and

    (v) An object falling vertically downwards towards the surface of the earth.

    Question 134
    CBSEENSC9005671

    With the help of a suitable example, explain the terms distance and displacement.

    Solution

    In Fig. 8.2, suppose a body moves from position A to B through C.
    Then, distance travelled = AC + BC

                     Fig. 8.2

    Distance:
    i) It is the length of the actual path travelled by a body between its initial and final positions. 
    ii) Distance is a scalar quantity because it has only magnitude and no direction.

    iii) Distance covered is always positive or zero.
    iv) Distance describes the total distance moved by an object. 

    Displacement:
    i) The change in the position of an object in a given direction is known as displacement.
    ii) It is the shortest distance measured in the direction from the initial to the final position of the body.
    iii) Displacement has both magnitude and direction, so it is a vector quantity. 
    iv) Displacement may be positive, negative or zero. 
    v) Displacement is used to locate the final position of an object with reference to its initial position at a given time.
    vi) The magnitude of displacement is always less than or equal to the distance travelled. 


    Question 135
    CBSEENSC9005672

    Give some points of differences between distance and displacement.

    Solution

    Differences between distance and displacement are: 

    Distance

    Displacement

    1. Distance is the length of the actual path traversed by a body, irrespective of its direction of motion.

    1. Displacement is the shortest distance between the initial and final positions of a body in a given direction.

    2. Distance between two given points may be same or different for different paths chosen.

    2. Displacement between two given points is always same.

    3. It is a scalar quantity.

    3. It is a vector quantity.

    4. Distance covered is always positive.

    4. Displacement covered may be positive, negative or zero.

    Question 136
    CBSEENSC9005673

    An object has moved through a distance, can it have zero displacement? If yes, support your answer with an example.

    Solution

    Yes. Suppose a person throws a ball upwards through height h and catches back
    the ball.
    Then,
    Distance covered by the ball = h + h  = 2h
    Displacement of the ball = 0.

    Question 137
    CBSEENSC9005674

    A farmer moves along the boundary of a square field of side 10 m in 40 s. What will be the magnitude of displacement of the farmer at the end of 2 minutes 20 seconds?

    Solution

    If the farmer starts from point A, then at the end of 2 minutes and 20 seconds (= 140 seconds), he will reach the diagonally opposite corner C.
    The magnitude of displacement of the farmer is:
    AC space equals space square root of AB squared plus BC squared end root space

space space space space space space space equals space square root of 10 squared plus 10 squared end root space

space space space space space space space equals 10 square root of 2 space
space
space space space space space space space equals space 10 space cross times space 1.414 space

space space space space space space space equals space 14.14 space straight m. space 

    Question 138
    CBSEENSC9005675

    Which of the following is true for displacement?

    (a) It can not be zero.

    (b) Its magnitude is greater than the distance travelled by the object.

    Solution

    a) Statement is false. Displacement can be zero. If a body starts from a point and then comes back to the same point through the same path, displacement is zero. 

    b) The magnitude of the displacement is always less than or equal to the distance travelled by the object. 

    Question 139
    CBSEENSC9005676

    A particle moves from a point P directly towards another point Q which is at a distance of 10 cm from P. It then comes back to P directly. Calculate:

    (i) the total distance travelled; and

    (ii) the displacement of the particle.

    Solution

    Given, 
    Distance between points P and Q = 10 cm

    (i) Total distance travelled = PQ + QP
                                           = (10 + 10) cm
                                           = 20 cm. 

    (ii) Displacement of the particle will be zero, because the particle comes back to the initial position. That is, the initial and final position is same. 

    Question 140
    CBSEENSC9005677

    A body thrown in the vertically upward direction rises upto a height h and comes back to the position of start.

    Calculate:

    (i) the total distance travelled by the body; and

    (ii) the displacement of the body.

    Solution

    Given, a body thrown vertically upward comes back to it's position of start. The body rise upto a height 'h' when thrown vertically up.

    (i) The total distance travelled = h + h = 2h.

    (ii) The displacement is zero, because the initial and final positions coincide. 

    Question 141
    CBSEENSC9005678

    In a long distance race, the athletes were expected to take four rounds of the track such that the line of finish was same as the line of start. Suppose the length of the track was 200 m.

    (a) What is the total distance to be covered by the athletes ?

    (b) What is the displacement of the athletes when they touch the finish line ?

    (c) Is the motion of the athletes uniform or non-uniform ?

    (d) Is the displacement of an athlete and the distance moved by him at the end of the race equal ?

    Solution

    Given, 
    Length of the track = 200 m
    Expected rounds athlete are supposed to take = 4

    (a) Total distance covered by each athlete = 4 × 200 = 800 m.

    (b) As the line of finish is the same as the line of start, so displacement of each athlete = 0.

    (c) Motion of the athletes is non-uniform as the speed of the athlete will be changing at each point.

    (d) From parts (a) and (b), the distance and displacement of an athlete are not equal at the end of the race. 

    Question 142
    CBSEENSC9005679
    Question 143
    CBSEENSC9005680

    A person travels a distance of 1.5 m towards east, then 2.0 m towards south and finally 4.5 m towards east. Find:

    (i) the total distance travelled, and

    (ii) the total displacement.

    Solution

    Given, a person travels 1.5 m towards east, then 2.0 m towards south and then 4.5 m towards east. 


    The displacement of 1 m by 1 cm length of a line is rep
    The various displacements can be represented as follows:
    Distance space travelled space equals space OA space plus space AB space plus space BC

space space space space space space space space space 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 plus space 2.0 space plus space 4.5
space
space space space space space space space space space 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 8.0 space straight m

Displacement space space equals OC with rightwards arrow on top space equals space 6.3 space m space comma space along space OC.



    Question 144
    CBSEENSC9005681

    A person moves on a semicircular track of radius 42 m during a morning walk. He starts at one end of the track and reaches the other end. Find the
    i) distance covered and
    ii) the displacement of the person.

    Solution

    Given, 
    Radius of the semicircular track, r = 42 m

    i) Distance covered is equal to the circumference of the semicircular track = pi * r
                                     = 3.14 *42
                                     = 131.8 m
                                     = 132 m

    ii) Displacement of the person = diameter of the semicircle = (42+ 42) = 84 m



    Question 145
    CBSEENSC9005682

    A body travels a distance of 8 m from A to B, then moves a distance of 6 m at right angles to AB. What is the total distance travelled ? Find the magnitude of resultant displacement.

    Solution

    Distance travelled from A to B = 8 m
    Distance travelled at right angles to AB = 6m

    So, total distance covered = 8+6 = 14 m

    Displacement = shortest distance covered by the body in any given time. 

    Therefore, Displacement = 82+62 =64+36 =100 = 10 m


    Question 146
    CBSEENSC9005683

    A body moves towards east by a distance of 3 km and turns towards north and moves a distance of 4 km. What is the magnitude of displacement of the body?

    Solution

    Given, 
    Distance travelled in east direction = 3 km
    Towards north = 4 km

    The magnitude of the displacement would be the resultant BC. 

    Therefore, using Pythagoras theorem, 

    BC = CA2+AB2      = 32+42      =9+16      =25       = 5 km

    That is, 5 km is the magnitude of displacement.

    Question 147
    CBSEENSC9005684

    What is meant by uniform motion ? Give an example.

    Solution

    When an object covers equal distances in equal intervals of time, however small the time interval may be, then the motion of the object is said to be uniform motion.

    Consider for example, a bus moves 10 km in the first 15 minutes, 10 km in second 15 minutes, 10 km in third 15 minutes and so on.
    Then one can say that the bus is in uniform motion.

    Question 148
    CBSEENSC9005685

    What is non-uniform motion? Give some examples.

    Solution

    When an object covers unequal distances in equal intervals of time, the object is said to be in non-uniform motion.
    the motions that we see in our daily life are usually examples of non- uniform motion.

    For example, if we drop a ball from the roof of a building, we will note that the ball covers 4.9 m in the 1st second, 14.7 m in the 2nd second, 24.5 m in the 3rd second, and so on.
    Here, the ball covers increasingly larger distances in successive seconds as it falls down.
    Thus, the motion of a freely falling body is non-uniform.

    Examples of non-uniform motion are: 

    (i) A stone dropped from the top of a building.

    (ii) A ball thrown vertically upwards.

    (iii) The motion of a train as it leaves the station.

    (iv) The motion of a bus as it approaches a bus-stop.

    (v) The motion of a ball rolling down an inclined plane.

    Question 149
    CBSEENSC9005686

    Define the term speed. What are its various units? Is it a scalar or vector quantity?

    Solution

    Speed is defined as the distance travelled by a body per unit time.
    Thus,
    Speed space equals space fraction numerator Distance space travelled over denominator Time space taken end fraction space space
space space space space space space space space space box enclose straight v space equals space straight s over straight t end enclose

    The SI unit of speed is ms-1 whenever the distance covered by body is expressed in metre and time in second.
    A smaller unit of speed is cm s-1.
    A larger unit of speed is km h-1
    Speed is a scalar quantity because it has only magnitude and no direction.
    The speed of a body tells us how fast or how slow the body is moving. But it does not give the direction of motion.

    Question 150
    CBSEENSC9005687

    Define uniform and non-uniform speeds.

    Solution

    Uniform speed: 

    When the speed of an object remains constant with time, that is it does not change in regular intervals of time, the speed is said to be uniform. 
    That, is if an object  covers equal distances in equal intervals of time, the body is moving with a unifrom speed. 

    E.g., A cyclist is covering 5 km uniformly every 1 hour. 

    Non-uniform speed:

    A body is said to be moving with non-uniform speed when an object covers unequal distances in equal inervals of time. 

    E.g., A soccer player moving in the football ground.

    Question 151
    CBSEENSC9005688

    Define the term average speed.

    Solution

    Average speed is the total distance travelled by a body, divided by the total time taken to cover that distance. Thus,

    Average speed = Total distance travelledTotal time taken

    Question 152
    CBSEENSC9005689

    Define the term velocity. What is its SI unit ? Is it a scalar or vector quantity?

    Solution

    Velocity is a physical quantity that gives both the speed and direction of motion of the body.

    Defintion: Velocity of a body is defined as the displacement produced per unit time. Velocity is also defined as the speed of a body in a given direction.

    If s is the distance travelled by a body in a given direction and t is the time taken to travel that distance, then the velocity v is given by,

    Velocity = displacement / time = s / t

    Velocity is a vector quantity because it requires both magnitude and direction of a body. 

    Question 153
    CBSEENSC9005690

    When is a body said to have

    (i) uniform velocity and 
    (ii) variable velocity ?

    Solution

    (i) Uniform velocity:

    If a body covers equal distances in equal intervals of time in a given direction, the body is moving with a ‘uniform velocity’. The velocity of the body does not change with time. So in uniform velocity, a body moves with uniform speed in a given direction.

    For example, light travels with a uniform velocity of 3×10 m/s through vacuum.

    (ii) Variable velocity:

    When a body travels unequal distances in equal intervals of time, such that its direction changes, or both change, velocity is said to be ‘variable velocity’.

    For example, a bus turning at a junction has variable velocity. 

    Question 154
    CBSEENSC9005691

    Define average velocity when the velocity of a body changes at a nonuniform rate and a uniform rate.

    Solution

    Average velocity:
    When the velocity of a body changes at a non-uniform rate, its average velocity is defined as the net displacement covered divided by the total time taken.
    Average space velocity space space equals space fraction numerator Net space displacement over denominator Total space time space taken end fraction

    When the velocity of a body changes at a uniform rate, then the average velocity is given by the arithmetic mean of initial velocity and final velocity for a given period of time.
    Average space velocity space space equals space fraction numerator Initial space velocity space space plus space Final space velocity over denominator 2 end fraction

    If u is the initial velocity and v is the final velocity, then the average velocity vav is given by,
                         v subscript av italic space italic equals italic space fraction numerator straight u plus v over denominator 2 end fraction

    Question 155
    CBSEENSC9005692

    Give some points of differences between speed and velocity.

    Solution

    Difference between speed and velocity: 

    Speed

    Velocity

    1. Speed is the distance traversed by a body per unit time in any direction.

    1. Velocity is the distance traversed by a body in a given direction. 

    2. It is a scalar quantity.

    2. It is a vector quantity.

    3. It is always positive or zero but never negative.

    3. It may be positive,  negative or zero.

    Question 156
    CBSEENSC9005693

    Under what condition is the magnitude of average velocity of an object equal to its average speed? 

    Solution

    When an object moves along a straight line in the same direction, its total path length is equal to the magnitude of displacement. Hence, its average speed is equal to the magnitude of average velocity. 

    Question 157
    CBSEENSC9005694

    What does the odometer of an automobile measure?

    Solution

    The odometer of an automobile measures the distance moved by it.

    Question 158
    CBSEENSC9005695

    What does the path of an object look like when it is in uniform motion?

    Solution

    When an object is in uniform motion, the body follows a straight line path.

    Question 159
    CBSEENSC9005696

    During an experiment, a signal from a spaceship reached the ground station in five minutes. What was the distance of the spaceship from the ground station ? The signal travels at the speed of light, that is, 3 × 108 ms-1.

    Solution

    Here we have,
    time, t = 5 minutes = 300 s,
    Speed with which the signal is travelling, v = 3 × 108 ms-1

    Now, using the formula, we get

    Distance of spaceship, s = vt
                                        = 3 × 108 × 300
                                        = 9 × 1010 m.

    Question 161
    CBSEENSC9005698

    The velocity of a car is 18 m/s. Express the velocity in km/h.

    Solution
    Given, velocity of car = 18 m/s

    Now, the velocity of the car in km/hr is given by, 

                              v = 181000×3600   = 64.8 km/hr

    Question 162
    CBSEENSC9005699

    Arrange the following speeds in increasing order:
    (i) Bicycle moving with a speed of 18 km/h.
    (ii) A fast runner running with a speed of 7 m/s.
    (iii) A car moving with a speed of 2000 m/min.

    Solution

    To compare the various speeds, we first express them in the same units:
    left parenthesis straight i right parenthesis space Speed space of space bicycle space space equals space 18 km over straight h space equals space fraction numerator 18 space cross times space 1000 over denominator 3600 space straight s end fraction space equals space 5 straight m divided by straight s

left parenthesis ii right parenthesis space Speed space of space fast space runner space equals space 7 space straight m divided by straight s

left parenthesis iii right parenthesis space Speed space of space car space equals space 2000 straight m over min space equals space fraction numerator 2000 space straight m over denominator 60 space straight s end fraction space equals space 33.3 space straight m divided by straight s
    Clearly, Speed of bicycle  < Speed of fast runner < Speed of car. 


    Question 166
    CBSEENSC9005703
    Question 167
    CBSEENSC9005704

    A 100 m long train crosses a 300 m long bridge at a speed of 90 km/hour. How much time will it take to cross the bridge completely?

    Solution

    Length of the train = 100 m
    Length of the bridge = 300 m
    Distance covered in crossing the bridge completely is, 
    Distance space covered space in space crossing space the space bridge space completely comma space

space straight s space equals space Length space of space train space plus space Length space of space bridge space equals space 100 space plus space 300 space equals space 400 space straight m

space straight v space equals space 90 space km divided by straight h space equals space fraction numerator 90 space cross times space 1000 over denominator 3600 end fraction space straight m divided by straight s space equals space 25 space straight m divided by straight s
therefore space space Time space required space to space cross space the space bridge space is space straight t space equals space straight s over v space equals space 400 over 25 space equals space 16 space straight s.

    Question 170
    CBSEENSC9005707
    Question 173
    CBSEENSC9005710

    The average time taken by a normal person to react to an emergency is one fifteenth of a second and is called the ‘reaction time’. If a bus is moving with a velocity of 60 km/h and its driver sees a child running across the road, how much distance would the bus had moved before he could press the brakes ? The reaction time of the people increases when they are intoxicated. How much distance had the bus moved if the reaction time of the driver were ½ s under the influence of alcohol?

    Solution

    Given, 
    Velocity of the moving bus = 60 km/hr
    Distance ravelled by the bus before brakes are applied = ?
    During the time of reaction, the velocity remains constant. 
    For space normal space person colon
space straight t space equals space 1 over 15 space straight s comma
space straight v space equals space 60 space km divided by straight h space equals space fraction numerator 60 space cross times space 1000 over denominator 3600 end fraction space equals space 50 over 3 straight m divided by straight s
Distance space moved space before space brakes space are space pressed comma space
space space space space space space space space straight s space equals space vt space equals space 50 over 3 cross times 1 over 15 equals 10 over 9 equals space 1.11 space straight m. space space space space space space space space space space space space space space space space space space space space space
space space space space space space space space space space space space space space space space space space


    When the driver is under the influence of alcohol, 
    T h a t space i s comma
space
F o r space intoxicated space driver colon space space
T i m e comma space straight t space equals space 1 half straight s comma space space
V e l o c i t y comma space straight v space equals space 60 space km divided by straight h space equals space 50 over 3 straight m divided by straight s 
    Distance moved before brakes are pressed, 
     straight s space equals space vt space equals space 50 over 3 cross times 1 half space equals 25 over 3 straight m space equals space 8.33 space straight m 

    That is the bus will be travel a lesser distance when the driver is under the influence of alcohol. 

    Question 176
    CBSEENSC9005713
    Question 177
    CBSEENSC9005714

    A train is moving with a speed of 72 km/h. What is its speed in m/s?

    Solution

    Given, 
    Velocity = 72 km/hr

    So, 
    Velocity in m/s will be given by, 

                  v = 72 × 10003600 = 20 m/s

    Question 178
    CBSEENSC9005715

    A train 40 m long passes over a bridge 200 m long at a speed of 36 km/h. How long will it take to pass completely over the bridge?

    Solution

    Length of the train = 40 m = 0.04 km
    Length of the bridge = 200 m = 0.2 km
    Speed of the train = 36 km/hr

    The total time taken by the train to completely pass over the bridge is given by time for the front of the train to travel 200 m at 36 km/hr + time for the back of train to travel 40 m at a speed of 36 km/hr. 

    Therefore, 

    Total time taken = s1v+s2v 

                             = 1vs1 + s2= 0.236+0.0436= 0.005 + 0.001= 0.006 hrs= 24 sec

    Question 179
    CBSEENSC9005716

    A car covers the first half of the distance between two places at a speed of 40 km/h and the second half at 60 km/h. What is the average speed of the car?

    Solution

    Let, the total distance covered the car be 'd'. 
    Given, 
    Speed in the first half = 40 km/hr
    Speed in the second half of distance = 60 km/hr

    Let the time taken by the car in the first half = t1
    Therefore, 
    Time, t1 = d/240= d80

    Time taken by car in the second half, t2d260=d120

    Now, average speed = Total distance travelledTotal time taken

                                  = dd80+d120= 80×12080+120= 9600200= 48 km/hr 


    Question 180
    CBSEENSC9005717

    A car travels 270 km in 4.5 hours. Calculate:

    (i) its speed;

    (ii) how far will it go in 7 hours at this speed; and

    (iii) how long will it take to travel 300 km at this speed.

    Solution

    Given, 
    Distance travelled by the car = 270 km 

    Total time for travel = 4.5 hours

    i) Speed of the car, v  = Distance travelledTime taken 

                                    =2704.5 = 60 km/hr


    ii) Speed = 60 km/hr

    Time taken = 7 hours

    Distance travelled = Speed × time 

                               =  420 km

    (iii) Speed = 60 km/hr

    Distance to be travelled = 300 km

    Time taken = DistanceSpeed

                      = 30060= 5 hours

    Question 181
    CBSEENSC9005718

    A car goes from a town A to another town B with a speed of 40 km/h and returns back to the town A with a speed of 60 km/h. Calculate the average speed during the whole journey.

    Solution

    Let the total distance travelled to and fro be d
    Speed while ging from town A to town B = 40 km/hr
    Speed while coming back from town B to town A = 60 km/hr
    Time taken to go to town B ,t1DistanceSpeed 

                                               td240 = d80

    Time taken to return to town A, t2 = d260=d120

    Average speed = Total distancetoatl time 

                          =dd80+d120 = 80×12080+120=9600200= 48 km/hr



    Question 182
    CBSEENSC9005719

    Define acceleration. Is it a scalar or a vector quantity? 

    Solution

    Acceleration is defined as the rate of change of velocity with time. 

    When the motion is non-uniform, the velocity of a body changes with time. It has different velocities at different instants of time and at different points of it's path.
    If the velocity of a body changes from u to v in time t, then
    Acceleration space equals space fraction numerator Change space in space velocity over denominator Time space taken 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 equals fraction numerator Final space velocity space minus space Initial space velocity over denominator Time space taken end fraction
rightwards double arrow space space space space space space space space space space space space space space space straight alpha space equals space fraction numerator v minus straight u over denominator straight t end fraction
    Acceleration is avector quantity and has both magnitude and direction. 

    Question 183
    CBSEENSC9005720

    What is the SI unit of acceleration?

    Solution

    The SI unit of acceleration is m/s2.

    Question 184
    CBSEENSC9005721

    When is the acceleration of a body positive?

    Solution

    When the acceleration of the motion is in the same direction as the velocity, acceleration is positive.

    E.g., Acceleration of a bus when it just leaves the bus-stop.

    Question 185
    CBSEENSC9005722

    What do you mean by the term retardation? Give two examples of such a motion.

    Solution

    When the velocity of a body decreases, its acceleration is negative. Negative acceleration is called ‘retardation’ or ‘deceleration’.

    For example, when a stone is thrown upwards, it is under retardation.
    Similarly, when a bus approaches a bus-stop, its motion gets retarded. 

    In both the cases mentioned above the acceleration decreases.

    Question 186
    CBSEENSC9005723

    When do you say a body is in
    (i) uniform acceleration, and
    (ii) variable acceleration?

    Solution

    (i) Uniform acceleration:

    When an object travels in a straight line and its velocity changes by equal amounts in equal intervals of time, the object is said to be in ‘uniform acceleration’.

    For example, a body falling freely to the ground has a uniform acceleration of 9.8 m/s2 as its velocity increases by 9.8 m/s after every one second.

    (ii) Variable acceleration:

    When the velocity of an object changes by unequal amounts in equal intervals of time, the object is said to have ‘variable acceleration’.

    For example, if the speed of a car travelling along a straight road increases by unequal amounts, then the car is moving with non-uniform acceleration.

    Question 187
    CBSEENSC9005724

    Give two examples of uniformly accelerated motions.

    Solution

    Examples of uniformly accelerated motions are:

    (i) An object moving with uniform speed along a circular path, has uniform acceleration. In this case the velocity of the object changes continuously due to change in its direction at each point.

    (ii) The motion of a ball rolling down an inclined plane is uniformly accelerated. 

    Question 188
    CBSEENSC9005725

    A bus decreases its speed from 80 km/h to 60 km/h in 5 seconds. Find the acceleration of the bus.

    Solution
    Initial space speed space o f space b u s comma space straight u space equals space 80 km over straight h space
space space space space space space space space space space space space space space space space space 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 80 space cross times space 1000 over denominator 3600 space straight s 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 equals space 800 over 36 ms to the power of negative 1 end exponent
Final space speed space o f space b u s comma space straight v space equals space 60 space km over straight h space
space space space space space space space space space space space space space space space space 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 60 space cross times space 1000 space straight m over denominator 3600 space straight s 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 equals space fraction numerator 600 over denominator 36 space end fraction ms to the power of negative 1 end exponent
T h e r e f o r e comma space
Acceleration comma space space straight alpha space equals space fraction numerator straight v minus straight u over denominator straight 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 equals space fraction numerator begin display style 600 over 36 end style space minus space begin display style 800 over 36 end style over denominator 5 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 equals space minus fraction numerator 200 over denominator 36 space cross times space 5 end fraction ms 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 space space space space space space space space space equals negative 10 over 9 ms 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 space space space space space space space space space space equals space minus 1.11 space ms to the power of negative 2 end exponent.
    So, the bus is decelerating at a rate of -1.11 m/s2
    Question 190
    CBSEENSC9005727

    What are the equations of motion for a body in linear motion?

    Solution

    When an object moves along a straight line with uniform acceleration, it is possible to relate its velocity, acceleration during motion and the distance covered by it in a certain time interval by a set of equations known as the equations of motion.

    Question 191
    CBSEENSC9005728

    A body starts initially with a velocity ‘u’ and is accelerated at constant rate ‘a’. Find an expression for final velocity after time ‘t’.

                                    Or

    Establish the relation: v = u + at.

    Solution

    First equation of motion:
    Let a body start with initial velocity 'u' and after time t, its velocity becomes v due to uniform acceleration a.
    From the definition of acceleration,
    Acceleration space equals space fraction numerator Change space in space velocity over denominator Time space taken 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 equals fraction numerator Final space velocity space minus space Initial space velocity over denominator Time space taken end fraction
rightwards double arrow space space space space space space space space space space space space space space space straight alpha space equals space fraction numerator straight v minus straight u over denominator straight t end fraction space space
rightwards double arrow space space space space space space space space space space space space space space at space equals space straight v minus straight u

rightwards double arrow space space space space space space space space space space space space space space space box enclose straight v space equals straight u space plus space at end enclose
 

    Question 192
    CBSEENSC9005729

    Deduce the expression for the distance travelled by a body moving with uniform acceleration, in a given time. 



    Solution

    Second equation of motion gives us the distance travelled by a body moving with uniform acceleration, in a given time.
    Suppose the body starts with initial velocity u and due to uniform acceleration a, its final velocity becomes v after time t.
    Then, 
    Average velocity is given by, 

    fraction numerator Initial space velocity space space plus space Final space velocity over denominator 2 end fraction space equals space fraction numerator straight u plus straight v over denominator 2 end fraction

So comma space the space distance space covered space by space the space body space in space time space straight t space is comma space

space straight s space equals space Average space velocity space cross times space Time

space space space space equals fraction numerator straight u plus straight v over denominator 2 end fraction cross times straight t space
space space space space equals space fraction numerator straight u plus left parenthesis straight u plus at right parenthesis over denominator 2 end fraction cross times space straight t space space space space space space space open square brackets because space straight v space equals space straight u space plus space at 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 space equals space fraction numerator 2 ut plus at squared over denominator 2 end fraction
rightwards double arrow space straight s space equals space u t plus 1 half a t squared
          


    Question 193
    CBSEENSC9005730

    Establish the relation v 2 – u2 = 2as, where ‘u’ is the initial velocity, ‘v’ the final velocity, ‘a’ the uniform acceleration and ‘s’ is the distance covered by the body.

    Solution

    Third equation of motion:
    Let a body start with initial velocity u. After covering distance 's' under uniform acceleration 'a', its velocity becomes 'v' in 't'  seconds.
    Then, 
    Average space velocity space space equals space fraction numerator straight u plus straight v over denominator 2 end fraction
    So space the space distance space covered space in space time space straight t space is space given space by comma

space straight s space equals space Average space velocity space cross times space Time space

space space space space equals space fraction numerator straight u plus straight v over denominator 2 end fraction cross times straight t
space space straight v plus straight u space equals space fraction numerator 2 straight s over denominator straight t end fraction space space space space space space space space... left parenthesis 1 right parenthesis
    Now, using the first equation of motion:
         v = u + at  
    rightwards double arrow  v - u = at      ... (2)
    Multiplying equations (i) and (ii), we get
     left parenthesis straight v plus straight u right parenthesis thin space left parenthesis straight v minus straight u right parenthesis space equals space fraction numerator 2 straight s over denominator straight t end fraction cross times space at space space space space space space

rightwards double arrow space space space box enclose straight v squared minus straight u squared space equals space 2 as end enclose


    Question 200
    CBSEENSC9005737

    A body starting from rest travels with uniform acceleration. If it travels 100 m in 5 s, what is the value of acceleration?

    Solution
    Here comma

I n i t i a l space v e l c o i t y comma space straight u space equals space 0

T i m e space o f space t r a v e l comma space straight t space equals space 5 space straight s

Distance space travelled comma space straight s space equals space 100 space straight m

Acceleration comma space straight a space equals space ?

therefore space space U sin g space t h e space s e c o n d space e q u a t i o n space o f space m o t i o n comma space
space space space space space space space space space space space space space space space space space space space space space straight s space equals space ut plus 1 half at squared
therefore space space space space 100 space equals space 0 plus 1 half straight a cross times 5 squared space space space space
rightwards double arrow space space space space 100 space equals space 25 over 2 straight a space space space space
rightwards double arrow space space space space space space space space space straight a space equals space fraction numerator 100 space cross times space 2 over denominator 25 end fraction space equals space space 8 space straight m divided by straight s squared
 
    Therefore, the body is accelerating at the rate of 8 m/s2
    Question 209
    CBSEENSC9005746

    What are the uses of graphical study of motion?

    Solution

    Graphical representation of motion is useful to study the nature of motion of bodies. In this type of representation, time is plotted along X-axis while distance or velocity is plotted along Y-axis.
    Graphs give more detailed information about the nature of motion than when motion is expressed in a tabular form.

    The uses of graphical representaion of motion are: 

    (i) From distance-time graph the position of the body at any instant of time can be determined.

    (ii) Distance covered by the body during particular interval of time can be seen from the graph. 

    (iii) The velocity of the body at any instant of time can be determined.

    (iv) By simply looking at the graph, one can tell whether motion is uniform or not.

    (v) Slope of velocity-time graph gives the acceleration of the body.

    (vi) Graphs are very useful for comparing the motions of two moving bodies. Distance-time graphs easily tell when and where one body crosses the other body.

    Question 210
    CBSEENSC9005747

    What is distance-time graph of a body? Draw distance-time graphs for, 

    (i) a stationary body,

    (ii) a body moving with uniform velocity, and

    (iii) a body moving with variable velocity.

    Solution

    Distance-time graph:
    It is a graph obtained by plotting distance travelled along Y-axis and time taken along X-axis.

    (i) Distance-time graph for stationary body:
    The position of a stationary body does not change with time. So distance-time graph for a stationary body is a straight line parallel to time axis, as shown in Fig. 8.7 (a). 

    (ii) Distance-time graph for uniform velocity:
    For a body moving with uniform velocity, the distance travelled is proportional to the time taken. So distance-time graph is a straight line inclined to the time axis, as shown in Fig. 8.7 (b).

    (iii) Distance-time graph for variable velocity:
    When a body moves with variable velocity, it covers unequal distances in equal intervals of time. So the distance-time graph is not a straight line but it is a curve, as shown in Fig. 8.7 (c). 

    Question 211
    CBSEENSC9005748

    Mention the uses of a distance-time graph.

    Solution

    Distance-time graph can be useful in the following ways: 

    (i) It tells the position of the body at any instant of time. 

    (ii) The distance covered by the body during a particular interval of time can be seen from the graph.

    (iii) The velocity of the body at any instant of time can be determined.

    Question 212
    CBSEENSC9005749

    Show that the slope of distance-time graph gives velocity of the body.

    Solution

    Consider the figure given below:


    This is a distance-time graph for a body moving with uniform velocity.

    Clearly, it covers distances S1 and s2 at times t1 and t2 respectively.
    Slope space of space line space AB space equals space tan space straight theta space equals space BC over AC
space space space space space space space space space space space space space fraction numerator straight s subscript 2 space minus space straight s subscript 1 over denominator space straight t subscript 2 space minus space straight t subscript 1 end fraction space equals space Displacement over Time space equals space Velocity.

    Hence, the slope of the distance-time graph gives velocity of the body.

    Question 213
    CBSEENSC9005750

    What conclusions do you draw about the motion of a body from the following displacement-time graphs: 


    Solution

    (a) From A to B, the body is at rest.
    (b) From O to A, the body is moving with a uniform velocity (positive).
    (c) From A to B, the body is moving with a uniform negative velocity.
    (d) From O to A, the body has positive variable velocity.
    (e) From O to A, the body has uniform positive velocity, from A to B the body has zero velocity and from B to C, the body has uniform negative velocity.
    (f) From O to A, the body has variable positive velocity and from A to B, the body has variable negative velocity.

    Question 214
    CBSEENSC9005751

    The following table shows the distance travelled by a car, moving with uniform acceleration, in time intervals of two seconds. Draw a distance-time graph for the motion of the car. Indicate whether the motion is uniform or not.



    Solution

    The distance-time graph for the motion of the car is shown in Fig.

    The motion of the car is non-uniform because it is a curved line. 

    Question 215
    CBSEENSC9005752

    What is velocity-time graph of a body? Draw velocity-time graphs for
    (i) a body moving with uniform velocity,
    (ii) a body moving with uniform acceleration, and
    (iii) a body moving with variable acceleration.

    Solution

    Velocity-time graph
    It is the graph obtained by plotting velocity of the body along Y-axis and time along X-axis.

    (i) Velocity-time graph for uniform velocity. For a body moving with a constant velocity, the velocity-time graph is a straight line parallel to the time-axis, as shown in Fig. (a).


    (ii) Velocity-time graph for uniform acceleration. For a body in uniform acceleration, the increase in velocity is proportional to the time taken. So velocity-time graph is a straight line inclined to the time-axis, as shown in Fig. (b).

    (iii) Velocity-time graph for variable acceleration. For a body having variable acceleration, the velocity increases by unequal amounts in equal intervals of time. So velocity-time graph is a curve, as shown in Fig.  (c).

    Question 216
    CBSEENSC9005753

    Mention the uses of a velocity-time graph.

    Solution

    The use of velocity-time graph are as follows:

    (i) To determine the speed of a body at any instant of time.

    (ii) To determine the acceleration of a body.

    (iii) To determine the total distance travelled by a body in a given time-interval.

    Question 217
    CBSEENSC9005754

    Show that the slope of velocity-time graph gives acceleration of the body.

    Solution
    Consider the figure given below.


    The fig. shows the velocity-time graph for a body in uniform acceleration.
    Graph is a straight line inclined to the time-axis.
    Body has velocities u and v at times t1 and t2respectively.
    Slope space of space line space AE space equals space tan space straight theta space equals space ED over AD
                          equals space fraction numerator straight v minus straight u over denominator straight t subscript 2 minus straight t subscript 1 end fraction space
equals space fraction numerator Change space in space velocity over denominator Time space taken end fraction
                           = Acceleration of the body
    Hence, the slope of the velocity-time graph gives the acceleration of the body.
    Question 218
    CBSEENSC9005755

    Draw velocity-time graph for a body moving with uniform velocity. Hence show that the area under the velocity-time graph gives the distance travelled by the body in a given time interval.

    Solution

    Consider the figure as below. 

    Distance covered as area under the velocity-time graph. 
    In Fig., line PQ is the velocity-time graph of a body moving with a uniform velocity, OP = v.
    Area of rectangle ABCD
    = AD × AB = OP × AB
    = v × (t2 – t1)
    = Velocity x time
    = Distance travelled in time interval (t2 – t1)
    Hence, the area under the velocity-time graph gives the distance travelled by the body in the given time interval.



    Question 219
    CBSEENSC9005756

    Draw a velocity-time graph for a body in uniform acceleration. Hence, show that the area under the velocity-time graph gives the distance travelled by the body in the given time interval.

    Solution

    Distance covered as area under the velocity-time graph.

    Fig. 8.14: Velocity-time graph for uniform acceleration.
    The figure above shows the velocity-time graph for a body in uniform acceleration.
    The graph is a straight line inclined to the time-axis.
    The body has velocities u and v at times tand t2 respectively.

    Area space under space the space velocity minus time space graph space AE

space space space space space equals space Area space of space trapezium space ABCE

space space space space space equals space Area space of space rectangle space ABCD space plus space Area space of space triangle space ADE

space space space space space equals space AB space cross times space space BC space plus space 1 half AD space cross times space DE
space space space space space equals space AB space cross times space BC space plus space 1 half BC space cross times space DE

space space space space space equals space open parentheses AB space plus space 1 half DE close parentheses BC

space space space space space equals space open parentheses straight u space plus space fraction numerator straight v minus straight u over denominator 2 end fraction close parentheses space cross times space left parenthesis straight t subscript 2 space minus space straight t subscript 1 right parenthesis

space space space space space space equals open parentheses fraction numerator straight u plus straight v over denominator 2 end fraction close parentheses space left parenthesis straight t subscript 2 space minus space straight t subscript 1 right parenthesis
         = Average velocity x Time interval 
         = Distance travelled in the time interval left parenthesis straight t subscript 2 space minus space straight t subscript 1 right parenthesis

    Hence, the area under the velocity-time graph gives the distance travelled by the body in the given time interval.


    Question 220
    CBSEENSC9005757

    What types of motions are represented by the following velocity-time graphs?





    Solution

    (a) Since the velocity-time graph is a straight line, the body has uniform velocity.
    (b) The body has uniform acceleration and its initial velocity is zero because the lne starts from the origin.
    (c) The body has some initial velocity and is under uniform retardation.
    (d) The body has some initial velocity and uniform acceleration.
    (e) The body has zero initial velocity and it has variable acceleration.
    (f) The body is at rest from O to A, it has uniform acceleration from A to B, it has uniform velocity from B to C and from C to D, the body is under uniform retardation. 

    Question 221
    CBSEENSC9005758

    What types of motions are represented by the following velocity-time graphs? 





    Solution

    Fig. (a) represents the velocity-time graph for a body whose velocity is decreasing uniformly with time. 
    Fig. (b) represents the velocity-time graph of the oscillating motion of body with time.

    Question 222
    CBSEENSC9005759

    What is the nature of the distance-time graphs for uniform and non-uniform motion of an object?

    Solution

    The distance-time graph for uniform motion is a straight line inclined with the time axis. 
    The distance-time graph is a curved line for non-uniform motion. 

    Question 223
    CBSEENSC9005760

    What can you say about the motion of an object whose distance-time graph is a straight line parallel to the time-axis?

    Solution

    The object is at rest, if the line is parallel to the time-axis. 

    Question 225
    CBSEENSC9005762

    What is the quantity which is measured by the area occupied below velocity-time graph?

    Solution

    The area occupied below velocity - time graph is the distance covered by the body in the given time interval. 

    Question 226
    CBSEENSC9005763

    Derive the equations of motion for uniformly accelerated motion from velocity-time graph.

    Solution

    Equations of motion by graphical method:
    Consider an object moving along a straight line with initial velocity 'u' and uniform acceleration a.
    Suppose, it travels distance s in time t.
    As shown in Fig. 8.19, its velocity-time graph is straight line.


    Here,
    OA = ED = u
    OC = EB = v and
    OE = t = AD.

    1. Equation for velocity-time relation.
    Acceleration is given by slope of velocity-time graph AB
    space space straight a space equals space DB over AD space equals space DB over OE space equals space fraction numerator EB minus ED over denominator OE end fraction space equals space fraction numerator straight v minus straight u over denominator straight t end fraction

i. e. comma space space space space straight v minus straight u space equals space at space

rightwards double arrow space space space space space space box enclose straight v space equals space straight u space plus space at end enclose

    Hence, the first equation of motion is proved. 

    2. Equation for position-time relation:
    From the first part, we have
    Acceleration comma space straight a space equals space DB over AD space equals DB over straight t
rightwards double arrow space space space space space space space space space space space space space space space space space DB space equals space at

Distance space travelled space by space the space object space in space time space straight t space is space g i v e n space b y comma space

space straight s space equals space Area space of space the space trapezium space OABE

space space space space equals space Area space of space rectangle space OADE space plus space Area space of space triangle space ADB
      space space space space space space space space space space equals space OA space cross times space OE space plus space 1 half DB cross times AD
space space space space space space space space space equals space ut space plus space 1 half at cross times straight t
rightwards double arrow space straight s space equals space u t space plus space 1 half a t squared 
    This proves the second equation of motion. 

    3. Equation for position-velocity relation:
    The distance travelled by object in time t is,
    s = Area of trapezium OABE 
      equals 1 half open parentheses E B plus O A close parentheses cross times O E

equals space 1 half open parentheses E B plus E D close parentheses cross times O E space
    Acceleration, a = slope of velocity-time graph AB.
    space a equals DB over AD equals fraction numerator E B minus E D over denominator O E end fraction

rightwards double arrow space O E space equals space fraction numerator E B space minus space E D over denominator a end fraction

therefore space D i s tan c e space t r a v e l l e d comma space s space equals space 1 half open parentheses E B plus E D close parentheses cross times fraction numerator left parenthesis E B minus E D right parenthesis over denominator a 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 equals fraction numerator 1 over denominator 2 a end fraction open parentheses E B squared space minus space E D squared close parentheses
space space space space space space space space space space 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 1 over denominator 2 a end fraction open parentheses v squared minus u squared close parentheses space
i. e. comma space space space space space space space space space space space v squared space minus space u squared space equals space 2 a s
    Hence, the third equation of motion is also proved. 

    Question 229
    CBSEENSC9005766

    A racing car has a uniform acceleration of 4 ms-2. What distance will it cover in 10 seconds after start?

    Solution
    Here comma

Initial space velocity space of space racing space car comma space straight u space equals space 0

R a t e space o f space u n i f o r m space a c c e l e r a t i o n comma space straight a space equals space 4 space ms to the power of negative 2 end exponent

Time space of space travel comma space straight t space equals space 10 space straight s

Using space the space second space equation space of space motion comma space we space have
space space space space space space space space space space space space space straight s space equals space ut plus 1 half at squared
space space space space space space space space space space space space space space space space equals space 0 space cross times space 10 space plus space 1 half cross times space 4 space cross times space left parenthesis 10 right parenthesis squared space
space space space space space space space space space space space space space space space space equals space 200 space straight m.
    So, distance travelled by the car in 10 seconds = 200 m
    Question 231
    CBSEENSC9005768

    Plot the distance-time graph of a body at rest (that is, not moving), at a distance of 100 m from the origin.

    Solution

    For a body at rest, distance-time graph is a straight line parallel to the time-axis.


    the graph obtained is shown in the fig.

    Question 232
    CBSEENSC9005769

    Discuss the graphs A, B and C shown in Fig. 8.21. Compare the total distance travelled and the displacements. Which graph represents a motion in which total displacement is zero?

    Solution

    (i) In graph A,
    Total distance travelled = 15 + 7.5 + 7.5 + 12.5
    = 42.5 m
    Graph B, 
    Total distance travelled = 15 + 15 = 30 m
    Graph C,
    Total distance travelled in graph C = 15 + 7.5 = 22.5 m.

    (ii) Displacement is the shortest distance between the initial and final positions.
    ∴  Displacement in graph A = 12.5 – 15 = – 2.5 m.
    Displacement in graph B = 0 – 0 = 0m. [ Because the object is returning back to the initial point]
    Displacement in graph C = – 7.5 - 15 = – 22.5 m.

    (iii) In graph B,
    Before t = 4 s, the velocity is positive and after t = 4 sec, velocity is negative. 
    So the acceleration is negative at t = 4 s in graph B. 

    Question 233
    CBSEENSC9005770

    Plot distance-time graphs of bodies moving with uniform speeds of 4 m/s and 7 m/s. Compare the graphs.

    Solution

    For a body moving with a uniform speed of 4 m/s

    For a body moving with a uniform speed of 7 m/s

    The graphical representation for the objects moving with uniform speed is a straight line. 

    For the body moving with a uniform speed of 4 m/s, the distance-time graph is the straight line AB .
    For the body moving with a uniform speed of 7 m/s, the distance-time graph is the straight line A'B'.
    Clearly, slope of graph A'B' is more than that of graph AB. 
    Question 234
    CBSEENSC9005771

    In this figure shows the distance-time graphs of two objects A and B moving with uniform speeds. Which of the two objects is moving faster?






    Solution

    Given, distance-time graph of two objects A and B moving with uniform speed.
    Draw space perpendiculars space PS comma space QS comma space RS space on space the space co minus ordinate space axes space as space shown.

Speed space of space body space straight A space equals space Slope space of space PQ space

space space space space space space space 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 QS over PS space
space space space space space space space 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 50 minus 10 over denominator 10 space colon 30 space minus space 9 space colon 30 end fraction space equals space 40 space km divided by straight h

Speed space of space body space straight B space equals space Slope space of space PR space

space space space space space space 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 RS over PS space

space space space space space space 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 30 minus 10 over denominator 10 colon 30 space minus space 9 colon 30 end fraction space equals space 20 space km divided by straight h
    Thus we can see that distance-time graph for body A has greater slope.
    Hence, object A is moving faster than body B. 

    Question 236
    CBSEENSC9005773

    The positions of a ball rolling down an inclined plane at the end of each second are given in the following table:

    Plot a displacement-time graph for the motion of the ball. Indicate whether the motion is uniform or non-uniform.

    Solution

    The displacement-time graph for the motion of the ball is as shown in Fig. below.


    Fig. 8.25. Displacement-time graph for non-uniform motion.
    The displacement-time graph is a curve, so motion of the ball is non-uniform. 
     

    Question 237
    CBSEENSC9005774
    Question 238
    CBSEENSC9005775

    The following is the distance-time table of a moving car.

    (i) Use a graph-paper and plot the distance travelled by the car versus time.

    (ii) When was the car travelling with the greatest speed ?

    (iii) What is the average speed of the car?

    (iv) What is the speed between 11.25 a.m. and 11.40 a.m.?

    (v) During a part of the journey, the car was forced to slow down to 12 km/h. At what distance did this happen?

    Solution

    (i) The distance-time graph for the motion of the car is as shown in the figure below.

    (ii) The car is travelling with maximum speed between 10.40 a.m. and 10.50 a.m.
    During this time the distance-time graph has maximum slope.
    Here

    Speed of the car =  fraction numerator 22 minus 12 over denominator 10 end fraction space equals space fraction numerator 10 over denominator bevelled 10 over 60 end fraction space equals space 60 space k m divided by h r
    (iii) Average speed of car between 10.05 a.m. and 11.40 a.m.

    equals space fraction numerator Total space distance over denominator Tota space time end fraction space equals space fraction numerator left parenthesis 42 minus 0 right parenthesis space km over denominator 1 space straight h space 35 space min end fraction space equals space fraction numerator 42 space km over denominator begin display style 95 over 60 end style straight h end fraction
equals space fraction numerator 42 space cross times space 60 over denominator 95 end fraction space equals space 26.52 space km divided by straight h. space

    (iv) Speed between 11.25 a.m. and 11.40 a.m.
           equals space fraction numerator 4 space km over denominator 15 space min end fraction space equals space fraction numerator 4 space km over denominator begin display style 15 over 60 end style straight h end fraction space equals space 16 space km divided by straight h.

    (v) Between 11.00 a.m. to 11.10 a.m., the car was forced to slow down to 12 km/h.
    So, this happened at a distance of 26 km. 

    Question 239
    CBSEENSC9005776

    Plot speed-time graph of bodies moving with uniform speeds of 4 m/s and 7 m/s. Compare the graphs.

    Solution

    When a body is moving with a uniform speed, its speed-time graph is a straight line parallel to time-axis.
    In Fig. below, the line AB represents the speed-time graph of a body moving with a uniform speed of 4 m/s, while line A'B' represents the speed-time graph of a body moving with uniform speed of 7 m/s.
    The graph A 'B' is at larger distance from time-axis than the graph AB.


    Speed-time graphs for uniform speeds of 4 m/s and 7 m/s.

    Question 240
    CBSEENSC9005777

    Both figures are shows the speed-time graphs of two cars (a) and (b) respectively.

    Using these graphs answer the following questions:

    (i) What is the acceleration of the car (a) and car (b) in the first two hours, in the next two hours and in the last two hours?

    (ii) What is the total distance travelled by the two cars?

    (iii) What is the average speed of the two cars? 



    Fig. 8.29

    Solution

    i) Acceleration of car (a) in first two hours is given by, 
    fraction numerator Change space in space speed over denominator time space taken end fraction space equals space fraction numerator 30 minus 0 over denominator 2 minus 0 end fraction equals 30 over 2 space equals space 15 space km divided by straight h squared
    Acceleration of car (A) in the next two hours is given by, 
    fraction numerator 60 minus 30 over denominator 4 minus 2 end fraction space equals space fraction numerator 30 space k m divided by h r over denominator 2 space h end fraction space equals space 15 space km divided by straight h squared
    Acceleration of car (a) in the last two hours, 
    fraction numerator 0 minus 30 over denominator 8 minus 6 end fraction space equals space fraction numerator negative 30 space k m divided by h r over denominator 2 space h end fraction equals negative space 15 space k m divided by h squared
    The negative accelerations means that the car is deccelerating. 
    ii) Total distance travelled by the car (a) in 8 hours is, 
    Area of increment ABC =  1 half cross times b a s e cross times h e i g h t
    1 half cross times left parenthesis 8 minus 0 right parenthesis space cross times space left parenthesis 60 minus 0 right parenthesis space equals space 240 space k m 
    iii) Average speed of the car (a) is,  
    equals space fraction numerator Total space distance space travelled over denominator Time space taken end fraction
equals 240 over 8
equals space 30 space km divided by hr

    Question 241
    CBSEENSC9005778

    A body moves from rest from a point A. It moves to another point B in 10 s, the motion being accelerated one and achieves a velocity 60 m/s at B. It then moves on to C with the uniform velocity for 20 s when it is immediately brought to rest. Draw a graph showing the complete motion and calculate the total distance travelled.

    Solution

    Given, a body from rest.


    The above figure illustrates the motion of body at different points.
    AB represents the accelerated part of motion.
    BC represents the uniform motion.
    At C, velocity of the body is abruptly made zero (point Q).

    Distance travelled during the first 10 s is, 
    equals space space space Area space of space increment ABP
equals space 1 half cross times space AP space cross times space BP space
equals space 1 half cross times 10 cross times 60
equals space 300 space straight m
    Distance space travelled space during space the space next space 20 space straight s space is space given space by comma space

space Area space of space rectangle space PBCQ space equals space 20 space cross times space 60 space equals space 1200 space straight m

therefore space space space Total space distance space travelled space equals space 300 space plus 1200
space space space space space space space space space space space space space space space space space space space space space space space space space 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 1500 space straight m

    Question 242
    CBSEENSC9005779

    Plot the velocity-time graph of a body whose initial velocity is 5 m/s and is moving with a retardation of 1 m/s2. Calculate the distance covered by it.

    Solution

    AB is the velocity-time graph of a body.
    Initial velocity of the body = 5 m/s
    The body is moving with a retardation of 1 m/s2 as shown in Fig.
          

    Distance space travelled space by space the space body space equals Area space of space increment AOB
space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space 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 cross times space OA space cross times space OB
space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space 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 cross times space 5 space cross times space 5
space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space 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 12.5 space straight m.
 


    Question 244
    CBSEENSC9005781

    A body moves with a velocity of 2 m/s for 5 s, then its velocity uniformly increases to 10 m/s in next 5 s. Thereafter its velocity begins to decrease at a uniform rate until it comes to rest after 10 s.

    (a) Plot a velocity-time and distance-time graph for the motion of the body.

    (b) Mark the portions of the graph to show when the motion of the body is uniform and when it is non-uniform.

    (c) From the graph find the total distance moved by the body after 2 s and 12 s and in the last 10 s.

    Solution

    (a) Fig. 8.34 shows the velocity-time graph and Fig. 8.35 shows the distance-time graph for the motion of the body.

    Fig. 8.34. Velocity-time graph.

    Fig. 8.35. Distance-time graph. 

    (b) From velocity-time, graph as shown in the fig., we note that the motion is uniform in portion AB and non-uniform in portions BC and CD.

    (c) From distance-time graph, we observe the following:

    Distance moved after 2 s = 4 m.

    Distance moved after 12 s = 50 m.

    Distance moved in last 10 s = 90 - 40
     
                                               = 50 m.
    Question 245
    CBSEENSC9005782

    What is uniform circular motion? Is this motion accelerated or not? What is the direction of velocity at any point? Give an examples of uniform circular motion.

    Solution

    Uniform circular motion:
    when a body moves with a uniform speed along a circular path, it is said to be in uniform circular motion.
    The direction of velocity changes at every point but the magnitude of the body remains constant.
    Therefore, unifrom circular motion is an example of accelerated motion.
    The direction of velocity at any point of the circular path is along the tangent at that point as shown in Fig. below.


    Examples of uniform circular motion:
    (i) A cyclist moving along a circular track at constant speed.
    (ii) Motion of the moon around the earth.
    (iii) Motion of a satellite in circular orbit around the earth.

    (iv) Motion of a stone being rotated at the end of a string. 

    Question 246
    CBSEENSC9005783

    Express the velocity of a body in uniform circular motion in terms of its time period T.

    Solution

    Suppose a body of mass m rotates in a circle of radius r with velocity v.
    The rotaing body completes one revolution in time T.
    Then,
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    Question 247
    CBSEENSC9005784

    Define the terms angular displacement, angular velocity, time period and frequency in respect of uniform circular motion?

    Solution

    Angular displacement (θ) is the angle swept by the radius in a given time interval.
    Mathematically, it is given by
    straight theta left parenthesis rad right parenthesis space equals space Arc over Radius space equals straight s over straight r.

    Angular velocity (ω)  is the angle swept out by the radius per unit time is called angular velocity. 
    straight omega space equals space fraction numerator Angular space displacement over denominator Time end fraction space equals space straight theta over straight t rad divided by straight s
    Time period  is the time taken for one complete revolution is called .
    Frequency is the number of revolutions completed per unit time is called frequency (v).
    straight omega space equals space fraction numerator 2 straight pi over denominator straight T end fraction space equals space 2 πv

    Question 248
    CBSEENSC9005785

    Derive a relation between linear velocity and angular velocity?

    Solution

    Consider a particle moving with uniform angular velocity co along a circular path of radius r.
                     
    Suppose it moves from A to B to cover distance s along arc AB.
    therefore space space space space space Linear space velocity comma space space space straight v space equals space Distance over Time space equals space straight s over straight t
    Let space straight theta space be space the space angle space subtended space by space the space arc space AB space at space centre space straight O. space Then
Angular space velocity comma space straight omega space equals space straight theta over straight t
But comma
space space space space space space space space straight theta left parenthesis rad right parenthesis space equals space Arc over Radius space equals space straight s over straight r

therefore space space space space space space space space space space space space straight omega space equals space straight s over rt space equals space straight v over straight r space space space space space space space space space space space space space space open square brackets because space space straight s over straight t space equals straight v close square brackets

rightwards double arrow space space space space space space space space space space space space straight v space equals space straight r space straight omega

T h a t space i s comma space l inear space velocity space equals space Radius space cross times space space Angular space velocity



    Question 250
    CBSEENSC9005787

    If the tyre of a car has a diameter of 0.8 m and the wheel has an angular velocity of 20 rad/s, what is the speed of the car in m/s?

    Solution

    Given that,
    Diameter = 0.8 m.
    Therefore, radius (r) = 0.4 m.
    Angular velocity, ω = 20 rad/s
      therefore space S p e e d space o f space t h e space c a r comma space straight v space equals space rω space equals space 0.4 space cross times space 20 space equals space 8 straight m divided by straight s. space

    Question 253
    CBSEENSC9005790

    Are rest and motion absolute or relative terms?

    Solution
    Rest and motion are relative terms.
    Question 254
    CBSEENSC9005791

    Can an object be at rest as well as in motion at the same time?

    Solution

    Rest and motion are relative terms. So, an object may be at rest relative to one object and at the same time it may be in motion relative to another object.
    Hence, an object can be at rest as well as in motion relatively at the same time.

    Question 255
    CBSEENSC9005792

    Is the motion of the earth visible to us?

    Solution

    The motion of the earth is not visible to us because it is a relative motion.

    Although the phenomenon like formation of day and night indicates the motion of the earth about its own axis.

    Question 256
    CBSEENSC9005793

    Give two examples of scalar quantities.

    Solution

    Two examples of scalar quantities are mass and volume.

    Question 257
    CBSEENSC9005794

    Give two examples of vector quantities.

    Solution

    Examples of vector quantities are velocity and acceleration.

    Question 258
    CBSEENSC9005795

    Define the term displacement.

    Solution

    Displacement is the shortest distance travelled by a body in the direction from initial to final position.

    Question 259
    CBSEENSC9005796

    Is displacement a scalar quantity?

    Solution

    No, displacement is a vector quantity as it has both magnitude and direction.

    Question 260
    CBSEENSC9005797

    If the displacement of a body is zero, is the distance covered by it necessarily zero?

    Solution

    No, it is not necessary that the distance covered by a body is zero when its displacement is zero. Displacement can be zero even when distance is zero. 

    Question 261
    CBSEENSC9005798

    Can the displacement be greater than the distance travelled by an object?

    Solution

    No, the displacement of an object can be either equal to or less than the distance travelled by the object. 

    Question 262
    CBSEENSC9005799

    Under what condition will the distance and displacement of a moving object have the same magnitude? 

    Solution

    When the object moves along the same straight line in the same fixed direction, distance and displacement are equal. 

    Question 263
    CBSEENSC9005800
    Question 264
    CBSEENSC9005801

    Ravi told his friend that his house is 1 km towards south from the Main Post Office. Express displacement and the distance moved by the friend from the Post Office when he arrives at Ravi’s house. Mention the reference point chosen.

    Solution

    We have considered Post Office as the reference point or origin.

    (i) Distance moved by the friend in going from Post Office to Ravi’s house = 1 km.

    (ii) Displacement moved by the friend in going from Post Office to Ravi’s house = 1 km, due south.

    Question 266
    CBSEENSC9005803
    Question 267
    CBSEENSC9005804

    What is the SI unit of displacement?

    Solution

    SI unit of displacement is metre (m).

    Question 268
    CBSEENSC9005805

    Define the term speed.

    Solution

    Speed is defined as the distance travelled by a body per unit time. 

    Question 269
    CBSEENSC9005806

    Is speed a vector quantity?

    Solution

    No, speed is a scalar quantity as it possesses only magnitude and not direction.

    Question 273
    CBSEENSC9005810

    Is electric potential a scalar or a vector quantity?

    Solution

    Electric potential is a scalar quantity.

    Question 274
    CBSEENSC9005811

    A body is moving with a velocity of 10 m/s. If the motion is uniform, what will be the velocity after 10 s?

    Solution

    Given, that the motion is uniform. Therefore, the velocity will remain the same after 10 seconds. That is 10 m/s. 

    Question 275
    CBSEENSC9005812

    Can the average speed of a moving body ever be zero?

    Solution

    Speed is a scalar quantity and is always positive. So, it's average speed can never be zero. 

    Question 276
    CBSEENSC9005813

    What is the angle between two vectors when their sum is maximum?

    Solution

    The angle between two vectors when the sum is maximum is 0°. 

    Question 277
    CBSEENSC9005814

    What is the angle between two vectors when their sum is minimum?

    Solution

    Angle between two vectors is 180° when their sum is minimum. 

    Question 278
    CBSEENSC9005815

    Give the name of the physical quantity that corresponds to the rate of change of displacement.

    Solution

    Rate of change of displacement is known as velocity. 

    Question 279
    CBSEENSC9005816

    Define velocity. What is its SI unit?

    Solution

    The distance travelled by a body per unit time in given direction is called its velocity. The SI unit of velocity is m/s.

    Question 280
    CBSEENSC9005817
    Question 281
    CBSEENSC9005818

    What does the reading of the odometer of a vehicle indicate?

    Solution

    Odometer measures the distance moved by the vehicle.

    Question 282
    CBSEENSC9005819

    Can the speed of a body vary if its velocity is constant?

    Solution

    When the velocity of a body is constant, speed is also constant. 


    Question 283
    CBSEENSC9005820

    Can a body have constant speed but variable velocity?

    Solution

    Yes,  a body can have constant speed but variable velocity.

    For example, a body in uniform circular motion has constant speed but its velocity changes at every point during the course of motion.

    Whenever, the direction changes, velcoity changes. 

    Question 284
    CBSEENSC9005821

    Define the term acceleration.

    Solution

    Rate of change of velocity with time is called as acceleration. 

    Question 285
    CBSEENSC9005822

    What is the SI unit of acceleration?

    Solution

    SI unit of acceleartion is m/s2.

    Question 286
    CBSEENSC9005823

    What is the SI unit of retardation?

    Solution

    SI unit of retardation is m/s2

    Question 287
    CBSEENSC9005824

    Out of energy and acceleration which one is a vector?

    Solution

    Acceleration is a vector quantity.

    Question 288
    CBSEENSC9005825

    What is acceleration of a body moving with uniform velocity?

    Solution

    For a body moving with uniform velocity, acceleration is zero. 

    Question 289
    CBSEENSC9005826

    When is a body said to have uniform velocity?

    Solution

    When a body covers equal distances in equal intervals of time, in a given direction, the body is said to be moving with a uniform velocity. 

    Question 290
    CBSEENSC9005827

    A body is moving with a velocity of 10 m/s. If the motion is uniform, what will be the velocity after 10 s?

    Solution

    We have a body moving with velocity 10 m/s and the motion is a uniform motion. 

    So, after 10 seconds, the body will move with the same velocity, i.e., 10 m/s. 

    Question 291
    CBSEENSC9005828

    Give one example of uniformly accelerated motion.

    Solution

    A body falling freely to the ground has a uniform acceleration of 9.8 m/sis an example of uniformly accelerated motion.

    Question 292
    CBSEENSC9005829

    What is called negative acceleration?

    Solution

    Negative acceleration is known as Retardation.

    Question 293
    CBSEENSC9005830

    Define uniform circular motion.

    Solution

    When a body moves along a circular path with uniform speed, the motion is called uniform circular motion.

    Question 294
    CBSEENSC9005831

    What can you say about the motion of a body if its distance-time graph is a straight line?

    Solution

    If the distance-time graph is a straight line, the body possesses uniform speed. 

    Question 295
    CBSEENSC9005832

    What does the slope of a distance-time graph indicate?

    Solution

    The slope of the distance-time graph gives us the speed of the body. 

    Question 296
    CBSEENSC9005833

    If the distance-time graph is parallel to the time-axis, what can you say about the motion of the body?

    Solution

    If the distance-time graph is parallel to the time axis, the body is at rest.

    Question 297
    CBSEENSC9005834

    What does the slope of a speed-time graph indicate?

    Solution

    Slope of a speed-time graph gives us acceleration.

    Acceleartion = dv/dt

    Question 298
    CBSEENSC9005835

    What do we get from the area between the speed-time graph of a body and the time-axis?

    Solution

    The area between the speed-time graph f a body and time axis measures the distance travelled by the body. 

    Question 299
    CBSEENSC9005836
    Question 300
    CBSEENSC9005837

    A body goes round the sun with constant speed in a circular orbit. Is the motion uniform or accelerated?

    Solution

    The motion of body around the sun in a circular orbit is accelerated. 

    Question 301
    CBSEENSC9005838

    A satellite goes round the earth in a circular orbit with constant speed. Is the motion accelerated?

    Solution

    Yes, the motion of satellite around the earth in a circular orbit is accelerated.

    Question 302
    CBSEENSC9005839

    Why is the motion in a circle at a constant speed called accelerated motion?

    Solution

    The velocity of the rotating body changes continuously due to the change in its direction at every point of its motion. Hence, a circular motion is called an accelerated motion. 

    Question 303
    CBSEENSC9005840

    A graph drawn between speed and time for a moving body is found to be a straight line parallel to the time-axis. What conclusion can you draw from the graph about the speed of the body?

    Solution

    The body is moving with a constant speed if the speed-time graph is a straight line parallel to the time axis. 

    Question 304
    CBSEENSC9005841

    What is meant by uniform motion ? Can you think of an example of a body in uniform motion?

    Solution

    If a body covers equal distances in equal intervals of time it is said to be in uniform motion. 

    For example, if a body in uniform motion covers 100 m in 50 s, then it must cover 10 m in every 5 s, 1.0 m in every 0.5 s, 0.1 m in every 0.05 s and so on. 

    Question 305
    CBSEENSC9005842

    Can a body have:

    (i) a uniform speed, but still have acceleration, and

    (ii) a zero velocity, but still have acceleration? Give an example of a each kind?

    Solution

    (i) A body can have acceleration even when it has uniform speed. A body in uniform circular motion has uniform speed but at every point its direction of velocity changes, so its motion is accelerated.

    (ii) Yes, a body can have acceleration even when its velocity is zero. When a body is thrown up, at highest point its velocity is zero but it has acceleration equal to acceleration due to gravity.

    Question 306
    CBSEENSC9005843

    What can you say about the motion of a body, if its velocity-time graph is a straight line?

    Solution

    Here two possibilities exist:

    (i) The body has a uniform velocity, if the velocity-time graph is a straight line parallel to the time-axis.

    (ii) If the velocity-time graph is inclined to the time-axis at certain angle, the body has uniform acceleration.

    Question 307
    CBSEENSC9005844

    What can you conclude from the following graphs about the state of motion of an object?

    (i) Displacement-time graph is parallel to the time-axis

    (ii) Velocity-time graph is a straight line making an angle with the time-axis. 

    Solution

    The conclusions are: 

    (i) The body is at rest when the displacement-time graph is parallel to the time axis.

    (ii) The body has uniform acceleration when the velocity-time gaph is slightly inclined with the time axis. 

    Question 309
    CBSEENSC9005846

    A boy goes to post office with uniform velocity, there he posts a letter and comes back home with the same velocity. Draw a graph of distance and time for his velocity.

    Solution

    Distance-time graph for the motion of the boy is shown in the fig. below. 

    Since, the boy is moving with a uniform velocity, distance-time graph is a straight line. 

    Question 310
    CBSEENSC9005847

    In this fig. is the distance-time graph of an object. Do you think it represents a real situation? Why?



    Solution

    This distance-time graph does not represent real situation because after 10 hours, the object occupies two different positions simultaneously.
     
    For example, at 10 hours object is at distance 0 km as well as at 75 km which is not physically possible.

    Question 312
    CBSEENSC9005849

    A few speed-time graphs for non-uniform motion of objects moving along a straight line are shown in Figs. 8.48(a), (b), (c) and (d).

    Tell how the speed is changing with time in each of these graphs? 

    Solution

    (a) Speed is increasing with time.
    (b) Speed is decreasing with time.
    (c) Speed is uniform.
    (d) Speed is oscillating. That is, it alternately increases and decreases with the passage of time.

    Question 313
    CBSEENSC9005850

    The state of motion of an object is represented by the following graphs. What conclusion do you derive about the state of motion of the object in each graph? 





    Solution

    (a) Here, distance-time graph is a straight line. Therefore, the body is at rest.
    (b) The body has uniform velocity because distance-time graph is a straight line. 
    (c) Here, the velocity-time graph has a negative slope. Hence, the body has some initial velocity and is under uniform retardation.

    Question 314
    CBSEENSC9006282

    Look at the activities listed below. Reason out whether or not work is done in the light of your understanding of the term ‘work’:

    (i) Suma is swimming in a pond.

    (ii) A donkey is carrying a load on its back.

    (iii) A wind mill is lifting water from a well.

    (iv) A green plant is carrying out photosynthesis.

    (v) An engine is pulling a train.

    (vi) Food grains are getting dried in the sun.

    (vii) A sailboat is moving due to wind energy.

    Solution

    (i) Yes, Suma is doing work by pushing water in the backward direction. 

    (ii) No, because the force exerted by donkey in the upward direction is perpendicular to the horizontal displacement of the load. Hence, no work is done.

    (iii) Yes, work is done in lifting water against the force of gravity. 

    (iv) No, because the leaves of plants remain at rest during photosynthesis.

    (v) Yes, engine is doing work in pulling the train. Both the applied force and displacement are in same direction.

    (vi) No, because food grains remain at rest.

    (vii) Yes, work is done by the wind in moving the sailboat.

    Question 315
    CBSEENSC9006283

    An objects thrown at a certain angle to the ground moves in a curved path and falls back to the ground. The initial and the final points of the path of the object lie on the same horizontal line. What is the work done by the force of gravity on the object.

    Solution

    The amount of work done by the force of gravity on the object is zero. This is because the net displacement of the object is in the horizontal direction while the force of gravity acts in the vertical downward direction. 

    Question 316
    CBSEENSC9006284

    A battery lights a bulb. Describe the energy changes involved in the process.

    Solution

    When a battery lights a bulb, first the battery converts chemical energy into electrical energy. This electrical energy is converted into heat and light. 

    Question 317
    CBSEENSC9006285

    Certain force acting on a 20 kg mass changes its velocity from 5 ms-1 to 2 ms-1. Calculate the work done by the force.

    Solution

    Given,

    Mass, m = 20 kg
    Initial velocity, u= 5 ms-1 
    Final velocity, v= 2 ms-1

    Work done = change in K. E.

                 W = 12 mv2 - 12 mu2

                     = 12 m ( v- u2)

                     = 12 x 20 x (22 - 52)

                     = 10 (4-25)

                     = -10 x 21

                     = -210 J

    Here, the negative sign implies the retarding nature of the applied force. 

    Question 318
    CBSEENSC9006286

    A mass of 10 kg is at a point A on a table. It is moved to a point B. If the line joining A and B is horizontal, what is the work done on the object by the gravitational force? Explain your answer.

    Solution

    We have,
     
    Mass m kept on table, m = 10 kg

    Displacement of mass from A to B is horizontal.

    Force of gravitation acts vertically downwards. That is, force acts perpendicular to displacement.

    ∴ W = Fs cos 90° = 0. 

    Work done on the object by the gravitaional force is 0.

    Question 319
    CBSEENSC9006287

    The potential energy of a freely falling object decreases progressively. Does this violate the law of conservation of energy? Why?

    Solution

    No, the law of conservation of energy is not violated. The loss in potential energy appears as an equal gain in kinetic energy of the object. Hence, the law of conservation of energy remains conserved.

    Question 320
    CBSEENSC9006288

    What are the various energy transformations that occur when you are riding a bicycle?

    Solution

    For pulling the bicycle, muscular energy is used. Muscular enrgy is further converted into kinetic energy. A part of the muscular energy is used in doing work against friction on the road. This part of the muscular energy changes into heat. 

    Question 321
    CBSEENSC9006289

    Does the transfer of energy take place when you push a huge rock with all your might and fail to move it ? Where is the energy you spend going?

    Solution

    While pushing the rock, transfer of muscular energy takes place. This energy is entirely spent in doing work against friction between the rock and the ground. 

    Since, the rock does not move, no work is done by the body.

    Question 322
    CBSEENSC9006290

    A certain household has consumed 250 units of energy during a month. How much energy is this in joules?

    Solution

    Energy consumed by the household = 250 units

    1 unit of energy = 1 kWh = 3.6 x 106 J

    ∴ 250 units of energy = 250 x 3.6 x 106

                                      = 9 x 108 J.

    Question 324
    CBSEENSC9006292

    What is the work done by the force of gravity on a satellite moving round the earth? Justify your answer.

    Solution

    When the satellite moves around the earth, the force of gravity acts on it along the radius of its orbit. The direction of motion is along the tangent to the orbit at any point. Thus, force acts perpendicular to displacement. Therefore, the work done on the satellite is zero. 

    Question 325
    CBSEENSC9006293

    Can there be displacement of an object in the absence of any force acting on it? Think. Discuss this question with your friends and teacher.

    Solution

    Yes, there can be displacement of an object in the absence of any force acting on it.

    We know that F = m x a, then m x a = 0 for F = 0

    Since mass 'm' cannot be zero, therefore, when force F = 0, then acceleration 'a' = 0. In such a situation, either the object is at rest or it is in uniform motion in straight line. For example, rain drops falling on the earth while the net force on them is zero. There is displacement of the water droplets.

    Question 326
    CBSEENSC9006294

    A person holds a bundle of hay over his head for 30 minutes and gets tired. Has he done some work or not? Justify your answer.

    Solution

    No work has been done because the bundle of hay remains stationary. That is, the displacement is zero. So, the work done is also zero. 

    Question 328
    CBSEENSC9006296

    Illustrate the law of conservation of energy by discussing the energy changes, which occur when we draw a pendulum bob to one side and allow it to oscillate. Why does the bob eventually comes to rest? What happens to its energy eventually? Is it a violation of the law of conservation of energy?

    Solution

    Conservation of energy during the oscillations of a simple pendulum. 

    Fig. Conservation of energy in a simple pendulum. 
    i) As shown in the fig. below, a simple pendulum consists of a spherical metal bob suspended by a thread from a fixed support.
    ii) As the bob is displaced to end A, it gains potential energy. As it is released from rest, its kinetic energy begins to increase.
    iii) At an intermediate position like C, the energy is partly kinetic and partly potential.
    iv) At mean position O, the energy is totally kinetic. At the end B, again the energy becomes totally potential. So, total mechanical energy remains constant (= mgh) at all points.
    The energy of the body is gradually used in doing work against the force of friction at the point of suspension and also against friction of air. The energy spent is converted into heat. There is no violation of the law of conservation of energy.
    Due to decrease in the energy of the bob, its amplitude of oscillation decreases with time and eventually the bob comes to rest.

    Question 330
    CBSEENSC9006298

    An object of mass m is moving with a constant velocity v. How much work should be done on the object in order to bring the object to rest?

    Solution

    Work done on the object is equal to the change in kinetic energy of the object.
    straight K. straight E equals space 1 half m v squared space minus space 1 half straight m left parenthesis 0 right parenthesis squared space
space space space space space space equals space 1 half m v squared.

    Question 331
    CBSEENSC9006299

    Calculate the work required to be done to stop a car of 1500 kg moving at a velocity of 60 km/h?

    Solution
    G i v e n comma space

M a s s space o f space t h e space b o d y comma space straight m space equals space 1500 space k g

V e l o c i t y comma space space straight v space equals space 60 space k m divided by straight h space equals space fraction numerator 60 space cross times space 1000 space straight m over denominator 3600 space straight s end fraction space equals space 50 over 3 straight m divided by straight s
    Work required to be done to stop the car, 
    W = Change in K.E. of the car
        equals space 1 half m v squared space minus space 1 half straight m left parenthesis 0 right parenthesis squared space
equals space 1 half m v squared
equals space 1 half cross times 1500 space cross times space open parentheses 50 over 3 close parentheses squared
space equals space 208333.3 space straight J. 
    Question 332
    CBSEENSC9006300

    In each of the following a force F is acting on an object of mass m. The direction of displacements is from west to east shown by the longer arrow. Observe the diagrams carefully and state whether the work done by the force F is negative, positive or zero.

     

     

    Solution

    Given, 
    Force is acting on an object of mass m.
    (a) We have,

    Force is perpendicular to displacement,
                straight W space equals space Fs space cos space 90 degree space equals space 0 space
    (b) Here, Force and displacement are in the same direction.
                       
               straight W space equals space Fs space Cos space 0 degree space equals space Fs space equals space straight a space positive space work.
    (c) In this case, Force and s are in opposite direction. 
    Therefore, 
    straight W space equals space Fs space cos space 180 degree space

space space space space space equals space Fs left parenthesis negative 1 right parenthesis space

space space space space space equals space minus Fs space comma space straight a space negative space work.

    Question 333
    CBSEENSC9006301

    Soni says that the acceleration in an object could be zero even when several forces are acting on it. Do you agree with her? Why?

    Solution

    Yes, the acceleration of the object would be zero when several forces acting on the object add up to give a zero resultant force.
    That is,
                     straight a space equals space straight F over straight m space equals space 0 over straight m space equals space 0

    Question 335
    CBSEENSC9006303

    A freely falling object eventually stops on reaching the ground. What happens to its kinetic energy on reaching the ground?

    Solution

    When a freely falling object stops on reaching the ground, kinetic energy of the object changes into heat and sound. 

    Question 336
    CBSEENSC9006304

    Define the term work. State the conditions that need to be satisfied for work to be done. Give some situations in which work is done.

    Solution
    Work is the amount of force exerted on a body such that it moves in the direction of the force. 

    Two conditions need to be satisfied for work to be done are as follows: 

    (i) A force should act on the body. 

    (ii) The body must be displaced in the direction of the applied force. 

    Work is not done, if either of the above two conditions is not satisfied.

    Let consider some examples from the daily life. 

    (i) Push a pebble lying on a surface. The pebble moves through some distance. Here we apply a force on the pebble and the pebble gets displaced. So, we have done work on the pebble.

    (ii) We apply a force to lift a book through a height. The book rises up. We have done work in moving up the book.

    (iii) A bullock is pulling a cart. The cart moves. There is a force on the cart and the cart has moved. The bullock has done work on the cart.

    Question 337
    CBSEENSC9006305

    How is work done measured when a body moves in the direction of the applied force?

    Solution

    Work done by a constant force when a body moves in the direction of applied force.
    As shown in Fig., suppose a force F acts on a body.

    Fig. Work done, when force and displacement are in same direction.
    The body moves through a distance s in the direction of the force.
    Work done = Force   cross times space d i s p l a c e m e n t
    i.e.,    W = F.s
    So, work done by a force acting on a body is equal to the magnitude of the force multiplied by the distance moved in the direction of the force. 

    Question 338
    CBSEENSC9006306

    When is the work done by a force said to be
    (i) positive and (ii) negative?

    Solution
    (i) Positive work: when the force acting on the object is in the direction of displacement, then work done by a force is positive. 

    Positive work done is given by, 

                       W = Fs cos 0° = Fs
    Examples:

    (i) When a body falls freely under gravity (θ = 0°), the work done by the force of gravity is positive.

    (ii) When a horse pulls a cart, the applied force and displacement are in the same direction, the work done by the horse is positive.

    (ii) Negative work: when the force acts in a direction opposite to the direction of displacement, work done is negative. 

    Negative work done is given by, 

                W = Fs cos 180°
                     = Fs (-1)
                     = - Fs

    Examples

    (i) When a body is lifted up, the work done by the gravitational force is negative. Gravitational force is acting downwards and weight is lifted up. 

    (ii) When brakes are applied to a moving vehicle, the work done by the braking force is negative. The braking force acts in the backward direction while the displacement of the vehicle is in the forward direction.
    Question 339
    CBSEENSC9006307

    Name and define the SI unit of work.

    Solution

    The SI unit of work is joule (J).

    1 Joule is the amount of work done on an object when a force of 1 N displaces it by 1 m along the line of action of the force.

    Thus, 1 joule = 1 newton x 1 metre
     
                    1 J = 1 Nm.

    Question 340
    CBSEENSC9006308

    Is work done a scalar or vector quantity?

    Solution

    Work done is a scalar quantity because it has only magnitude and no direction. 

    Question 341
    CBSEENSC9006309

    A force of 7 N acts on an object. The displacement is say 8 m in the direction of the force. Let us take it that the force acts on the object through the displacement. What is the work done in this case?

     

    Solution

    Force acting on the body, F = 7 N
    Displacement, s = 8 m
    Therefore, 
    Work done = Force x Displacement
                     = 7Nx8m
                     = 56 Nm
                     = 56 J.

    Question 342
    CBSEENSC9006310

    What is the work done when the direction of displacement and the direction of force F acting on a body are perpendicular to each other?

    Solution

    When direction of displacement and direction of force F acting on a body are both perpendicular to each other, work done is zero. The component of the force acting on a body in the direction of its displacement is zero in this case.

    Mathematically,

                       W = Fs cos 90° = 0.

    Question 343
    CBSEENSC9006311

    Is it possible that a force is acting on a body but still the work done is zero? Explain giving one example.
    OR
    What is the work done by a coolie walking on a level road with a load on his head? 

    Solution

    Yes, it is possible that a force is acting on a body but the work done is zero.

    E.g., Consider a coolie carrying a load on his head across a level road. He exerts an upward force equal to the weight of the load. But the displacement of the load is in the horizontal direction. Thus the angle between force F and displacement s is 0°.

    Hence, 
                       W = Fs cos 90° = 0. 

    Question 344
    CBSEENSC9006312

    Is any work done on a body in uniform circular motion? Justify your answer.

    Or

    A body is moving along a circular path. How much work is done by the centripetal force?

                      

     

     

     

    Solution

    As shown in the fig. below,  
                         
    Consider a body being rotated at the end of a string in a circle.
    The centripetal force F acts along the radius towards the centre O.
    The displacement s acts along the tangent.
    Thus, the angle between force F and displacement s is 90°.
    Hence,

    W = Fs cos 90° = 0.

    Thus, no work is done by the centripetal force on a body in uniform circular motion.

    Question 345
    CBSEENSC9006313

     When do we say that work is done?

    Solution

    Whenever a force acts on a body and the body is displaced in the direction of the force, work is said to be done.

    Question 346
    CBSEENSC9006314

    Write an expression for the work done when a force is acting on an object in the direction of its displacement.

    Solution

    Work done when a force is acting on an object in the direction of its displacement is given by, 

                Work done = Force x Displacement

    Question 347
    CBSEENSC9006315

    Define 1 Joule (J) of work.

    Solution

    One joule of work is done when a force of one newton displaces a body through a distance of 1 metre in its own direction.

    Question 348
    CBSEENSC9006316

    A pair of bullocks exerts a force of 140 N on a plough. The field being ploughed is 15 m long. How much work is done in ploughing the length of the field?

    Solution

    We have,

    Force exerted, F = 140 N
    Displacement, s = 15 m

    Therefore,

    Work done, W = Fs

                          = 140 N x 15 m

                          = 2100 J.

    Question 349
    CBSEENSC9006317

    A force of 5 N is acting on an object. The object is displaced through 2 m in the direction of the force, as shown in Fig. 11.5. If the force acts on the object all through the displacement, then what is the work done?

     (Textbook Example 11.1)

     

    Solution

    Consider the object on which the force is acting upon as shown in the fig. below. 

    We have,
    Force, F = 5 N
    Displacement, s = 2 m
    Therefore, 
    Work done, W = Fs = 5N cross times space 2 straight m = 10 Nm = 10 J

    Question 350
    CBSEENSC9006318

    Determine the work done in pushing a cart through a distance of 50 m against the force of friction equal to 150 N.

    Solution

    We have,

    Force, F = 150 N
    Displacement of the cart, s = 50 m

    Work done, W = Fs

                          = 150 N x 50 m

                          = 7500 J. 

    Question 351
    CBSEENSC9006319

    A body of mass 10 kg is displaced through a distance of 2 m under an acceleration of 5 m/s2. Calculate the work done.

    Solution

    Given,

    Mass of the body, m - 10 kg

    Acceleration, a = 5 m/s2

    Displacement, s = 2 m

    Therefore,

    Work done, W = Fs 

    Force on the body, F = ma 

                                     10×5 = 50 N
    Therefore,

    Work done is given by, W = 50 x 2 J

                                          = 100 J.

    Question 353
    CBSEENSC9006321

    A porter lifts a luggage of 15 kg from the ground and puts it on her head 2 m above the ground. Calculate the work done by her on the luggage? Take g - 10 ms-2.

    Solution

    Force applied is equal to the weight of object = mg = 15 x 10 = 150 N

    Displacement, s = 1.5 m

    Work done, W = Fs = 150 N x 1.5 m

                          = 225 Nm
                          = 225 J.
    Question 354
    CBSEENSC9006322

    An object of mass 5 kg is lifted vertically to a height of 2.0 m above the ground. If the acceleration due to gravity at the place is 10 m/s2, what is the amount of work done?

    Solution

    Mass of the object, m = 5 kg
    The object is lifted vertically to a height, h = 2 m
    Force applied is equal to the weight of object = mg = 5 x 10 = 50 N
    Work space done comma space straight W space equals space Fs
space
space space space space 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 50 space cross times space 2.0 space

space space space space 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 100 space straight J. space

    Question 360
    CBSEENSC9006328

    Define the term energy. Give some examples of bodies possessing some energy. 

    Solution

    The energy of a body is measured by it's capacity to do certain amount of work. A body does work, implies more energy has been expended.  

    Some examples of bodies possesing energy are:

    (i) When a fast moving cricket ball hits a stationary wicket, it throws away the wicket. The moving ball possesses energy and hence does work on the wicket.

    (ii) A hammer raised through a certain height possesses energy. When it falls on a nail placed on a piece of wood, it drives the nail into the wood.

    (iii) When a child winds a toy car and place it on the floor, the toy car begins to move. On winding the toy car gains energy and does work in moving against the force of friction.

    In the examples given above, different means of doing work is acquired by the object. Hence, these objects are said to posses energy. 

    Question 361
    CBSEENSC9006329

    How does the energy of an object change when it performs work or work is done on it? 

    Solution

    An object loses energy, when work is done by it. When work is done on the object, object gains energy. 

    Question 362
    CBSEENSC9006330

    How does an object with energy do work? 

    Solution

    When an object exerts force on another object, energy is transferred from the former to the latter. The object onto which the enrgy is transferred may start moving as it receives energy and starts moving. Therefore, the first object has the capacity to do some work. This implies, any object that possesses energy can do work. 

    Question 363
    CBSEENSC9006331

    Is energy a scalar or vector quantity?

    Solution

    Energy is a scalar quantity as it has only magnitude and no direction.

    Question 364
    CBSEENSC9006332

    Why are the units of energy the same as those of work ? What is the SI unit of energy?

    Solution

    Energy is measured by the amount of work that a body can do, so units of energy are the same as those of work.

    The SI unit of energy is joule (J). 

    Question 365
    CBSEENSC9006333

    Define 1 J of energy.

    Solution

    One joule of energy is the amount of energy required to do 1 joule of work.  

    Question 366
    CBSEENSC9006334

    Name a larger unit of energy. How is it related to joule?

    Solution

    A larger unit of energy is kilojoule (kJ).

    1 kilojoule = 1000 joule

         1 kJ = 1000 J

    Question 367
    CBSEENSC9006335

    Name the biggest natural source of energy. Name some other sources of energy also. 

    Solution

    The sun is the biggest natural source of energy. We can also get energy from the nuclei of atoms, the interior of the earth and the tides. 

    Question 368
    CBSEENSC9006336

    Name the different forms of energy.

    Solution

    Different forms of Energy are:

    Mechanical energy, sound energy, heat energy, light energy, chemical energy, nuclear energy, electric energy, magnetic energy, solar energy, etc. 

    Question 369
    CBSEENSC9006337

    What is mechanical energy? What are its two forms?

    Solution

    The energy assosciated with motion and positon of the object is known as mechanical enrgy.

    It has two forms :

    (i) Kinetic energy and (ii) Potential energy.

    Question 370
    CBSEENSC9006338

    What do you understand by kinetic energy of a body? Give some examples.

    Solution

    Kinetic energy:
    Kinetic energy is the energy possessed by a body by virtue of its motion. Work is done by a moving object.
    The amount of work which a moving object can do before coming to rest is equal to its kinetic energy.
     
    Fig. Conversion of energy into kinetic energy by a hammer. 

    Examples:
    (i) A moving hammer drives a nail into the wood. Being in motion, it has kinetic energy or ability to do work, as shown in Fig. above.

    (ii) A fast moving stone can break a window pane. The stone has kinetic energy due to its motion and so it can do work.

    (iii) The kinetic energy of air is used to run windmills.

    (iv) The kinetic energy of running water is used to run water mills.

    (v) A bullet fired from a gun can pierce a target due to its kinetic energy.

    Question 371
    CBSEENSC9006339

    Show that the kinetic energy of a body of mass m moving with velocity v is given bystraight E subscript straight k space equals space 1 half mv squared.

     

    Solution

    Expression for kinetic energy:
    The kinetic energy of a body can be determined by calculating the amount of work required to set the body into motion with the velocity v from its state of rest.

    U s i n g space t h e space 3 to the power of r d end exponent space e q u a t i o n space o f space m o t i o n comma space

space space space space space space space straight v squared space minus space straight u squared space equals space 2 a s

therefore space space space straight v squared space minus space 0 squared space equals space 2 a s space

rightwards double arrow space straight a space equals space fraction numerator straight v squared over denominator 2 straight s end fraction
    Let m be the mass of the body,
    u is the initial velocity of the body,
    F = force applied on the body,
    a = acceleration produced in the body in the direction of force,
    v = final velocity of the body, and
    s = distance covered by the body
    Now, work done is given by, considering that force and displacement are in the same direction.

     

    straight W space equals space Fs space

From space Newton apostrophe straight s space first space law space of space motion comma space straight F space equals space ma

That space is comma space

straight W space equals space ma. space space straight s space
space space space space space equals space straight m. space space fraction numerator straight v squared over denominator 2 straight s end fraction cross times space straight s space
space space space space space equals space 1 half mv squared
    This work done appears as the kinetic energy of the body, given by
    therefore space space space straight E subscript straight k space equals space 1 half mv squared., is the kinetic energy of the body. 

     

    Question 372
    CBSEENSC9006340

    State the factors on which the kinetic energy of a body depends.

    Solution

    Kinetic energy of a body depends upon:

    (i) K.E. of a body is directly proportional to its mass.

                         K.E. ∝ m.

    (ii) K.E. of a body is directly proportional to the square of its velocity.

                         K.E. ∝ v2.

    Question 373
    CBSEENSC9006341

    What is the kinetic energy of an object?

    Solution

    Kinetic energy is the energy possessed by an object by virtue of its motion. 

    Question 374
    CBSEENSC9006342

    Write an expression for the kinetic energy of an object.

    Solution

    The kinetic energy possessed by an object of mass 'm' moving with a uniform velocity 'v' is given by, 

                             Ek = 12mv2


    where, 
    m is the mass of the object
    v is the velocity with which the object is moving. 

    Question 376
    CBSEENSC9006344

    What happens to the kinetic energy of an object when its velocity is doubled?

    Solution

    The kinetic energy becomes four times its initial value when the velocity of the object is doubled .

                               Ek ∝ v2

    Question 385
    CBSEENSC9006353

    A mass of 10 kg is dropped from a height of 50 cm.

    Just as it reaches the ground, does the velocity depend on the mass of the particle? Explain.

    Solution

    The velocity does not depends on the mass of the particle because the acceleration due to gravity under which the particle falls does not depend on mass.

    Question 387
    CBSEENSC9006355

    A car is moving with uniform velocities: 18 km/h, 36 km/h, 54 km/h and 72 km/h at some intervals. Calculate the kinetic energy of the boy of 40 kg sitting in the car at these velocities. Plot a graph between the kinetic energy and the velocity. What is the nature of the curve?

    Solution

    Given space that comma space

Velocity comma space straight v subscript 1 space equals space 18 space km divided by hr space equals space fraction numerator 18 cross times 1000 over denominator 3600 end fraction space equals space 5 space straight m divided by straight s
Kinetic space energy comma space straight K. straight E space equals space 1 half mv 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 space space space space space equals space 1 half cross times 40 cross times left parenthesis 5 right parenthesis squared space equals space 500 space straight J
Velocity comma space straight v subscript 2 space equals space 36 space km divided by hr space equals space 10 space straight m divided by straight s

straight K. straight E space equals space 1 half mv subscript 2 squared

space space space space space space space space equals 1 half cross times 40 cross times left parenthesis 10 right parenthesis squared space equals space 2000 space straight J

Velocity comma space straight v subscript 3 space equals space 54 space km divided by hr space equals space 15 space straight m divided by straight s
straight K. straight E space equals space 1 half straight m subscript 3 straight v subscript 3 squared space

space space space space space space space space equals space space 1 half cross times 40 cross times left parenthesis 15 right parenthesis squared space equals space 4500 space straight J

Velocity comma space straight v subscript 4 space equals space 20 space straight m divided by straight s

straight K. straight E space equals space 1 half mv subscript 4 squared space
space space space space space space space space equals space space 1 half cross times 40 cross times left parenthesis 20 right parenthesis squared space equals space 8000 space straight J space

    Fig. 11.9. Graph between K.E. and velocity.
    The graph between K.E. and velocity is a parabolic curve as shown in the fig. below. 

    This is because kinetic energy has a direct dependence on square of velocity. 

     

     

    Question 388
    CBSEENSC9006356

    What is meant by potential energy of a body? Give some examples.

    Solution

    Potential energy is the energy possessed by a body by virtue of its position or shape.

    Examples of P.E. due to position:

    (i) Water stored in dam has potential energy,

    (ii) A stone lying on the roof of the building has potential energy.

    Examples of P.E. due to shape:

    (i) In a toy car, the wound spring possesses potential energy. As the spring is released, its potential energy changes into kinetic energy which moves the toy car.

    Fig. 11.10. P.E. stored in a wound spring.

    Fig. 11.11. P.E. stored in a stretched bow.

    (ii) A stretched bow possesses potential energy. As soon as it is released, it shoots the arrow in the forward direction with a large velocity. The potential energy of the stretched bow gets converted into the kinetic energy of the arrow.

    Question 389
    CBSEENSC9006357

    What is gravitational potential energy.

    Deduce an expression for the gravitational P.E. of a body of mass m placed at a height h above the ground.

    Solution

    Gravitational potential energy:
    The energy of an object increases when it is raised through a certain height above the ground. This is because work is done on it against gravity. The energy present in such an object is called gravitational potential energy.
    The amount of work done in raising the object from the ground to that point against gravity gives us the gravitational potential energy.

    Fig. 11.12. Measurement of P.E.

    Expression for potential energy
    Consider a body of mass m lying at point A on the earth's surface, where its potential energy is taken as zero.
    Weight of object = mg, acting vertically downwards.
    In order to lift the body to another position B at a height h, we have to apply a minimum force equal to mg in the upward direction.
    So work is done on the body against the force of gravity, is given by
    Work done  = Force x Displacement

    rightwards double arrow     W  =   Fs
    But,

    F = mg   and s = h,
    Therefore,
            W = mg x h = mgh

    The work done (mgh) on the body is equal to the gain in energy of the body.
    This is the potential energy (Ep)) of the body.
    Thus,
                             Ep = mgh

     

    Question 390
    CBSEENSC9006358

    Does the gravitational potential energy of a body depend on the path along which the body is moved ? Illustrate your answer with the help of a suitable example.

    Solution

    Consider a body of mass m being raised to a height h vertically upwards from position A to B, as shown in the fig. below.


    In the above fig.
    (a) The work done against gravity = mgh.

    Suppose the body is taken from position A to B along the path ACDEB, as shown in Fig. (b).
    During horizontal motions along paths CD and EB, the force of gravity is perpendicular to displacement, so work done is zero.
    Work is done only along vertical paths AC and DE.
    Therefore, the net work done = mgh, because AC + DE = h. 

    Hence, we can see that the work is done against gravity and is independent of the path along which the body is moved.
    The work done depends only on the initial and final positions of the body.
    That is why, gravitational force is called a conservative force.

    Question 391
    CBSEENSC9006359

    Find the energy possessed by an object of mass 10 kg when it is at a height of 6 m above the ground.
    (The value of g = 9.8 ms-2).

    Solution

    Given,

    Mass of the object, m = 10 kg

    Height of the object, h = 6 m

    Acceleration due to gravity, g = 9.8 ms-2

    Potential energy of the object, Ep = mgh

                                                       = 10 x 9.8 x 6 J

                                                       = 588 J.

    Question 393
    CBSEENSC9006361

    A boy weighing 50 kg climbs up a vertical height of 100 m. Calculate the amount of work done by him. How much potential energy does he gain ? Take g = 9.8 m/s2.

    Solution

    Mass of the boy, m = 50 kg

    Acceleration due to gravity, g = 9.8 m/s2

    Vertical height that the boy climbed up, h = 100 m

    Work done by the boy = mgh

                                      = 50 x 9.8 x 100

                                      = 49000 J

                                      = 4.9 x 104 J


    Gain in P.E. = Work done = 4.9 x 104 J.
    Question 396
    CBSEENSC9006364

    A body of mass 25 kg is raised to the top of a building 10 m high and then dropped freely under gravity (g = 10 m/s2):

    (a) Calculate the work done in raising the body to the top of the building.

    (b) What is the value of the gravitational potential energy at the top of the building?

    (c) By what factor will the gravitational potential energy of the same body increase if it is raised to the top of a multistorey building 30 m high?

    (d) When will the kinetic energy of the body be maximum?

    Solution

    Given that,
    Mass of the body, m = 25 kg
    Height to which the object is rasied up, h = 10 m
    Acceleration due to gravity, g = 10 m/s2

    (a) Work done  = mgh
                           = 25 x 10 x 10 J
                           = 2500 J. 
    (b) Gravitational P.E. = work done = 2500 J 
    (c) Gravitational P.E. at a height of 30 m is,
         P.E= 25 x 30 x 10 J = 7500 J
    therefore space space space space fraction numerator straight P. straight E. space at space 30 space straight m space height over denominator straight P. straight E. space at space 10 space straight m space height end fraction space equals space fraction numerator 7500 space straight J over denominator 2500 space straight J end fraction space equals space 3

space Thus comma space space straight P. straight E. space increases space by space 3 space times. space
    (d) K.E. of the body will be maximum just before the body strikes the ground. 

    Question 397
    CBSEENSC9006365

    Two bodies A and B of equal masses are kept at heights of h and 2h, respectively. What will be the ratio of their potential energies?

    Solution

    Let the mass of each body be m.
    P.E. of body A = mgh
    P.E. of body B = mg. 2h
    therefore space space Ratio space of space the space potential space energies space space equals space fraction numerator straight P. straight E. space of space body space straight A over denominator straight P. straight E. space of space body space straight B 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 space space space space space space space space space space space space space space space space equals space fraction numerator mgh over denominator mg. space 2 straight h 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 space space space space space space space space space space space space space space space space equals 1 half space equals space 1 colon 2

    Question 401
    CBSEENSC9006369

    Are various forms of energy inter-convertible?

    Solution

    Yes, energy can be changed from one form to another. The total energy before and after remains the same after conversion of it's forms. 

    Question 402
    CBSEENSC9006370

    Mention the various energy transformations that take place when we throw a ball.

    Solution

    When we throw a ball, the muscular energy of our body gets converted into the kinetic energy of the ball, thus providing it's motion.



    Question 403
    CBSEENSC9006371

    A body is thrown up vertically upwards. Its velocity goes on decreasing. What happens to its kinetic energy as its velocity becomes zero?

    Solution

    When a body is thrown upwards, its kinetic energy gradually changes into potential energy. At the highest point where its velocity becomes zero, whole of the kinetic energy gets converted into potential energy.

    Question 404
    CBSEENSC9006372

    What kind of energy transformation takes place when a body is dropped from a certain height. Explain.

    Solution

    When a body is dropped from a certain height, its potential energy gradually changes into kinetic energy. On reaching the ground, the whole of its potential energy gets converted into kinetic energy. But when the ball strikes the ground, the kinetic energy is converted into sound, light and heat. 

    Question 405
    CBSEENSC9006373

    What kind of energy transformation takes place in the following devices: 

    (i) Electric heater,
    (ii) electric motor,
    (iii) dynamo,
    (iv) microphone,
    (v) headphone,
    (vi) hydro-electric power station,
    (vii) steam engine,
    (viii) solar battery,
    (ix) photoelectric cell,
    (x) electric cell. 

    Solution

    (i) Electric heater: Electrical energy into heat. 

    (ii) Electric motor: Electrical energy into mechanical energy.

    (iii) Dynamo: Mechanical energy into electrical energy.

    (iv) Microphone: Sound into electrical energy.

    (v) Headphone: Electrical energy into sound.

    (vi) Hydroelectric power station: Mechanical energy into electrical energy.

    (vii) Steam engine: Heat into mechanical energy.

    (viii) Solar battery: Solar energy into electrical energy.

    (ix) Photoelectric cell: Light into electrical energy.

    (x) Electric cell: Chemical energy into electrical energy.

    Question 406
    CBSEENSC9006374

    State the law of conservation of energy. Hence describe the relationship between work and energy.

    Solution

    The law of conservation of energy states that energy can neither be created nor destroyed.
    It can only be converted from one form to another.

    Work and energy are correlated. Whenever work is done, the object that does work loses energy and the object on which work is done, gains some energy.

    For example, when we kick a ball, we lose some energy and the ball gains an equal amount of energy. The energy gained by the ball makes it to move. 

    Question 407
    CBSEENSC9006375

    Show that when a body is dropped from a certain height, the sum of its kinetic energy and potential energy at any instant during its fall is constant.

    Solution

    Conservation of energy during the free fall of a body.
    Let a body of mass m lying at position A be moved to a position B through a height h as shown in the fig. below.

    Fig. The free fall of the body is hown below. 
    The work done on the body is mgh.
    The body is at rest at position B.
    Therefore,
    straight K. straight E space at space straight B space equals space 0 space

straight P. straight E space at space straight B space equals space mgh

Total space energy space at space straight B space equals space mgh space plus space 0 space equals space mgh

If space the space body space falls space from space straight B space to space straight C space through space straight a space distance
straight x comma space the space body space possess space both space straight P. straight E space and space straight K. straight E. space

Now comma space straight P. straight E space at space straight C space equals space mg space left parenthesis straight h minus straight x right parenthesis

Let space straight v space be space the space velocity space acquired space by space the space body space when
it space reaches space straight C. space

Now comma space using space the space third space equation space of space motion comma space

space space space space space space straight v squared space minus space straight c squared space equals space 2 gx space

rightwards double arrow space straight v squared space equals space 2 gx

So comma space

straight K. straight E space at space straight C space equals space 1 half mv squared space equals space 1 half straight m cross times 2 gx space equals space mgx
So comma space total space energy space at space straight C space equals space mg left parenthesis straight h minus straight x right parenthesis space plus space mgx space equals space mgh

When space the space body space reaches space the space initial space position space straight A comma space

straight P. straight E space at space straight A space equals space 0 space

Let comma space straight v space be space the space veocity space of space the space body space at space straight A.

Velocity space is space given space by comma space

straight v squared space minus space 0 space equals space 2 space gh space

straight i. straight e. comma space straight v squared space equals space 2 gh space

So comma space straight K. straight E space at space straight A space equals space 1 half mV 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 equals 1 half straight m cross times 2 gh

space space 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 mgh space space

Therefore comma space total space energy space at space straight A space equals space mgh space space

Thus comma space throughout space the space motion comma space total space energy space
of space the space body space is space constant. space

This space verifies space the space law space of space conservation space
of space energy. space space

    Question 408
    CBSEENSC9006376

    Explain qualitatively how the total energy of a freely falling body is conserved.

    Solution

    Let us consider an object possesing mass 'm' falling freely from a height 'h'. Initially, 

    P.E of the body = mgh 

    K.E of the body = 0 [ body is at rest] 

    Therefore, 

    Total energy of the body = mgh+0 = mgh

    As the process tails, its potential energy will change into kinetic energy.

    At a given instant, v is the velocity of the moving body. 

    So, K.E of the body = 12mv2 

    During the fall, the potential energy would decrease while the kinetic energy would increase. When the object is about to reach the ground, h = 0 and velocity will be maximum. Therefore, the kinetic energy would be maximum and potential energy minimum. However, the total energy of the system, that is, the sum of potential and kinetic energy remains constant. 

    That is, 

                    mgh + 12mv2 = constant

    Thus, the law of conservation of energy is satisfied. 

    Question 409
    CBSEENSC9006377

    Explain the term power. Is it a scalar or vector quantity?

    Solution

    The rate of doing work or rate of transfer of enrgy is known as power. If a body does work W in time t, then it's power P is given by, 

    Power = Worktime 

    Power is a scalar quantity because it is a ratio of two scalarquantities. 

    At different intervals of time, an agent may perform work. In these cases, average power is the ratio of total energy consumed to the total time taken.

    Average power   =  Total energy consumedtotal time 

     

    Question 410
    CBSEENSC9006378

    Name and define the SI unit of power. Give the bigger power units of power.

    Solution

    The SI unit of power is watt (W). 

    The power of a body is one watt if it does work at the rate of 1 joule per second.

    Also, power is 1 watt when the rate of consumption of energy is 1 Js-1.

    1 watt   = 1 Joule1 second

    Bigger units of power are kilowatt, megawatt and horse power.

    1 kilowatt = 1000 watt;
           1 kW = 1000 W

    1 megawatt = 106 watt;
             1 MW = 106 W

    1 horse power = 746 watt;
                 1 H.P. = 746 W.

    Question 411
    CBSEENSC9006379

    What is power?

    Solution

    The rate of doing work or the rate of transfer of energy is termed as power. 

    Question 412
    CBSEENSC9006380

    Define 1 watt of power.

    Solution

    When a body works at the rate of 1 joule per second, the power done is 1 watt. 

    Question 413
    CBSEENSC9006381

    A lamp consumes 1000 J of electrical energy in 10 seconds. What is its power?

    Solution
    Electrical energy consumed by lamp, W = 1000 J
    Time taken, t = 10 seconds
     
    Power, P = Worktime= Wt = 100010 = 100 Js-1  = 100 W
    Question 414
    CBSEENSC9006382

    Define average power.

    Solution

    Average power of a body is defined as the total energy consumed by it divided by the total time taken.

    Average power   =  Total energy consumed

                                            Total time taken

    Question 415
    CBSEENSC9006383

    Name and define the commercial unit of energy. Express it in joules. Where is it used?

    Or

    Define one kilowatt hour. Express it in joules.

    Solution

    The commercial unit of electric energy is kilowatt-hour (kWh).

    One kilowatt hour is defined as the electric energy consumed by an appliance of power 1000 watt in one hour.

     1 kWh = 1 kW x 1 h
                = 1000 W x 1 h
                = 1000 Js-1 x 3600 s
     1 kWh = 3600000 J
                = 3.6 x 106 J

    The electrical energy used in households, industries and commercial establishments is expressed in kilowatt hour.

    Electrical energy used is expressed in terms of ‘units’ in our monthly bills.
    Here 1 unit means 1 kilowatt hour.

    Question 416
    CBSEENSC9006384

    A man does 50 J of work in 5 seconds. What is his power?

    Solution

    Work done, W  = 50  J
    Time, t   =  5 s

    Therefore,

    Power, P = Wt = 505 = 10 W

    Question 417
    CBSEENSC9006385

    The work done by the heart for each beat is 1 joule. Calculate the power of the heart if it beats 72 times in a minute.

    Solution

    Work done by the heart per beat = 1  J

    Work done by the heart in 72 beats, W = 1 x 72 

                                                               =  72 J

    Time Taken, t= 1 min = 60 s

    Power, P  = Wt = 7260 = 1.2 W 

    Question 418
    CBSEENSC9006386

    The heart does 1.5 J of work in each heart beat. How many times per minute does it beat if its power is 2 watt?

    Solution

    Work done during each heartbeat = 1.5 J

    Time taken, t = 1 min = 60 s 

    Power of heartbeat, P = P x t

                                     = 2 Js-1 x  60  s

                                     = 120 J

    Number of times heart beats per minute = Total work doneworked done in each beat = 1201.5 = 80 times

    Therefore, the heart beats 80 times per minute. 

    Question 419
    CBSEENSC9006387

    Calculate the power of a pump which can lift 100 kg of water to store it in a water tank at a height of 19 m in 25 sec. (Take the value of g = 10 m/s2).

    Solution
    Mass of water which can be lifted by pump, m 100  kg

    Height of the water tank, h = 19  m

    Time taken to fill the tank, t = 25 s

    Acceleration due to gravity, g  =  10 m/s2

    Therefore, using the formula,  

    Power is given by, 

    P = Work donetime taken = mght=100×10×1925 = 760 Watt 
    Question 420
    CBSEENSC9006388

    Water is falling on the blades of a turbine at the rate of 6 x 103 kg per minute. The height of the fall is 10 m. Calculate the power given to the turbine. Take g = 10 m/s2.

    Solution

    Here,  

    Mass of water falling on the turbine, m = 6 x 103 kg
    Rate at which the water is flowing, t = 1 min =  60 s
    Height of the fall, h = 10 m, g = 10 m/s2

    So, power given to the turbine, P = Work donetime taken 

    mght = 6×103×10×1060= 104 W =10 kW

    Question 421
    CBSEENSC9006389

    A woman pulls a bucket of water of total mass 5 kg from a well which is 10 m deep in 10 s. Calculate the power used by her.

    Solution

    Mass of the bucket of water, m = 5 kg

    Depth of the well, h = 10 m

    Time taken to draw water from the well, t = 10 s

    Acceleration due to gravitu, g = 10m/s2

    Power is given by, P = Work doneTime taken 

                                  = mght = 5×10×1010 = 50 W

    Question 422
    CBSEENSC9006390

    Two girls each of weight 400 N climb up a rope through a height of 8 m. We name one of the girls as A and the other as B. Girl A takes 20 seconds while B takes 50 seconds to accomplish this task. What is the power expended by each girl?

    Solution
    (i)  Case 1: Girl A

    Given,

    Weight of the girl, mg  = 400  N

    Height which girl A climbs up, h = 8 m

    Time taken, t= 20 s

    So, power expended by girl A, P is given by, 

    Work donetime taken = mght = 400×820 = 160 W

    (ii) Case 2: Girl B 

    Given, 

    Weight of the girl, mg = 400 N

    Height which girl B climbs up, h =  8  m

    Time taken, t=  50 s

    Power expended by girl B is given by, 

    P = Work donetime taken = mght= 400×850 = 64 W 
    Question 423
    CBSEENSC9006391

    A boy of mass 50 kg runs up a staircase of 45 steps in 9 seconds. If the height of each step is 15 cm, find his power. Take g = 10 ms-2.

    Solution

    Here we have, 

    Mass of the body, m = 50 kg 

    Total number of steps on the staircase = 45 
    Height of each step = 15 cm 

    So, height of the staircase, h = 45×15100 = 45 × 0.15 = 6.75 m 


    Time taken to climb each step = 9 sec

    Now, weight of the boy, mg  = 50 x 10  = 500  N

    So, power of the boy, P = Work doneTotal time taken

                                       = mght= 500×6.759 = 375 W

    Question 424
    CBSEENSC9006392

    What is the power of an engine which supplies 18 kJ of energy per minute?

    Solution

    Energy of the engine, E = 18 kJ

                                        = 18,000 J

    Time taken to supply energy, t = 1  min = 60  s

    Power, P = Energytime taken = 1800060 = 300 W, is the required power of the engine.

    Question 425
    CBSEENSC9006393

    A tube-well pumps out 2400 kg of water per minute. If water is coming out with a velocity of 3 ms-1, what is power of the pump ? How much work is done, if the pump runs for 10 hours?

    Solution
    Here  we have,

    Mass of the water being pumped, m = 2400 kg

    Time taken to pump out the water, t = 1 min
                                                            = 60  s

    Velocity with which the water is coming out, v = 3 m/s

    Pump runs for 10 hours.

    Power of the pump, P = Work donetime taken = K.Et = 12mv2t

                                     = 2400×3×32×60 = 180 W 
    Question 426
    CBSEENSC9006394

    A horse exerts a pull on a cart of 300 N so that the horse-cart system moves with a uniform speed 18 km/h on a level road. Calculate the power developed by the horse in watt and also find its equivalent in horsepower.

    Solution
    Force exerted on the cart, F = 300 N

    Speed with which the horse-cart system is moving = 18 km/hr = 18×10003600 = 5 m/s 

    Power developed by the horse, P = Work donetime 

                                               Fst= Fv        = 300 ×5       = 1500 W

    Power in terms of horsepower = 1500746=2.01 H.P 
    Question 427
    CBSEENSC9006395

    An electric bulb of 60 W is used for 6 h per day. Calculate the units of energy consumed in one day by the bulb.

    Solution
    Given, 

    Power of the bulb, P = 60 W = 601000  kW = 0.06 kW

    Time taken, t = 6 hrs

    Therefore, 

    Energy consumed per day  = P × t 

                                             = 0.6 x 6 = 0.36 kWh
                                             = 0.36 units 
    Question 428
    CBSEENSC9006396

    Calculate the energy consumed by a 100 W refrigerator in 30 days, if it works for 16 hours a day.

    Solution

    The refrigerator works for 16 hours on 30 days.

    Power of the refrigerator, P = 100 W = 0.1 kW

    Total time, t = 30 x 16 hours 

    Now, usig the formula for energy, we get

    Energy consumed = P x t

                                = 0.1 kW x 30 x 16 h

                                = 48 kWh.
    Question 429
    CBSEENSC9006397

    A household has a 100 W lamp lighted for 2 hours, two 60 W lamps lighted for 4 hours and an electric fan of 50 W working for 8 hours a day. Calculate the electric units consumed each day.

    Solution
    Given, 

    A 100 W lamp is lighted for two hours.
    Two 60 W lamps lighted for 4 hours. 
    Electric fan of 50 W working for 8 hours a day.

    Now, using the formula,

                   Energy = Power × time

    Therefore,

    Electric energy consumed by 100 W lamp is, 

               E= 100 W  x 2 h

                 = 200 Wh

                 =  2001000=  0.20 kWh

    Electric energy consumed by two 60 W lamps is, 

             E= 2 x 60 W x 4 h

               = 480 Wh

               =4801000 

               =  0.48  kWh 

    Electric energy consumed by electric fan is, 

            E= 50 W x 8 h

              = 400 Wh

              = 4001000

              = 0.40 kWh

    Thus, total electric energy consumed each day is

    T.E = 0.20 + 0.48 + 0.40 

         = 1.08 kWh

         = 1.08 units
    Question 430
    CBSEENSC9006398

    A 100 W electric bulb is lighted for 2 hours everyday and five 40 W tubes are lighted for 4 hours everyday. Calculate,

    (i) the energy consumed for 60 days and

    (ii) the cost of electricity consumed at the rate of Rs. 3 per kWh. 

    Solution

    Given, 

    100 Watt bulb lighted for 2 hours.
    Five 40 Watt tubes are lighted for 4 hours everyday.

    (i) Energy consumed by a 100 W bulb each day is,

              E= 100 W x 2 h

                = 200  Wh

                = 2001000 

                =  0.2  kWh


    Energy consumed by five 40 W tubes each day, 

                  E = 5 x 40 W x 4 h

                     = 800  Wh

                     = 8001000 

                     = 0.8 kWh

    Therefore,

    Total energy consumed each day = 0.2 + 0.8  =  1.0 kWh 

    Total energy consumed in 60 days = 1.0 x 60 = 60 kWh

    (ii) Cost fo 1 kWh of electricty = Rs. 3

    So, cost of 60 kWh = 3 x 60 = Rs. 180.

    Question 431
    CBSEENSC9006399

    State the two essential conditions which should be satisfied for work to have been done on an object.

    Solution

    The two conditions to be followed for work done is given below: 

    (i) a force acts on the body, 

    (ii) the force has to be applied in the direction of the force. 

    Question 432
    CBSEENSC9006400

    What is the SI unit of work?

    Solution

    Joule

    Question 433
    CBSEENSC9006401

    Define joule.

    Solution

    One joule of work is said to be done when a force of one newton displaces a body through a distance of 1 metre in its own direction. 

    Question 434
    CBSEENSC9006402
    Question 435
    CBSEENSC9006403

    Moment of force and work done by a force have the same units. Then, what is the difference between them? State at least one point of difference.

    Solution

    Moment of force produces rotatory motion in a body, whereas when a force does work on a body, it produces translatory motion in it. 

    Question 436
    CBSEENSC9006404

    A coolie is walking on a railway platform with 20 kg load on his head. What is the amount of work done by him? Justify your answer. 

    Solution

    Work done is zero because the load does not get displaced from the coolie's head. Hence, the work done is zero. 

    Therefore, using the formula,

                       W = Fs cos 90° = 0.

    Question 437
    CBSEENSC9006405

    What should be the angle between force and displacement for doing maximum and minimum work?

    Solution

    For maximum work, the angle between force and displacement should be θ= 0°.

    For minimum work, angle between force and displacement θ = 90°. 

    Question 438
    CBSEENSC9006406

    A man is holding a suitcase in the state of rest. What is the work done by him?

    Solution

    Work done by the man = 0

    Man is in a state of rest and there is no displacement. 

    Using the formula,

     W = Fs cos θ
         = F x 0 x cos θ = 0.

    Question 439
    CBSEENSC9006407

    What is the amount of work done by a man in pressing a rigid wall with a force of 500 N?

    Solution

    Zero. Because there is no displacement. 

    Question 440
    CBSEENSC9006408

    What is the work done when a body is moved horizontally along a frictionless surface?

    Solution

    Work done when a body is moved horizontally along a frictionless surface is zero. 

    Question 441
    CBSEENSC9006409

    In a tug of war, one team is slowly giving way to the other. What work is being done and by whom?

    Solution

    The work is done by the winning team and is equal to the product of resultant force applied by the two teams and the displacement suffered by the losing team. 

    Question 442
    CBSEENSC9006410

    What is the amount of work done when an artificial satellite revolves around the earth?

    Solution

    An artificial satellite revolves around the earth because of the force of gravity which acts perpendicular to the direction of motion of the satellite.
    Therefore, work done is zero when a satellite revolves around the earth. 

    Question 443
    CBSEENSC9006411

    A boy is running along a circular path with a uniform speed. How much work is done?

    Solution

    Displacement is zero for a circular path because the initial and final point is same. Therefore, the work done is zero. 

    Question 444
    CBSEENSC9006412

    Define energy.

    Solution
    Energy possesed by an object is measured as it's capacity to do work. 
    Question 445
    CBSEENSC9006413

    What is the unit of energy?

    Solution

    The SI unit of energy is Joule.

    Question 446
    CBSEENSC9006414

    Does an object in motion have ability to do work?

    Solution

    Yes, an object in motion is capable of doing work because it possesses kinetic energy. Kinetic energy is converted in the form of work done. 

    Question 447
    CBSEENSC9006415

    Out of energy and acceleration, which one is vector?

    Solution
    A body is accelarated or deccelarated in a particular direction. Hence, acceleration is a vector quantity. 
    Question 448
    CBSEENSC9006416

    Write a formula for the measurement of kinetic energy.1

    Solution

    Kinetic energy of a body, of mass 'm' moving with velocity 'v' is given by,

                         K.E. = 1/2 mv2.

    Question 449
    CBSEENSC9006417

    Write down the expression for gravitational potential energy of a body of mass m placed at a height h above the surface of the earth.

    Solution

    Gravitational P.E. = mgh

    where, 
    m is the mass of the body, 
    g is the accelaration due to gravity, and
    h is the height of the object.

    Question 450
    CBSEENSC9006418

    The speed of a car increases four times. What is the increase in its kinetic energy?

    Solution

    Kinetic energy of a moving body is given by,

                                  K.E. ∝ v2.

    So, when the speed of  car increases 4 times, kinetic energy becomes 16 times of it's initial value.

    Question 451
    CBSEENSC9006419

    A body is thrown vertically upward. Its velocity goes on decreasing. What happens to its kinetic energy as its velocity becomes zero?

    Solution

    As the velocity of the body becomes zero when it is thrown vertically upwards, whole kinetic energy gets converted into potential energy. That is, its kinetic energy becomes zero.

    Question 452
    CBSEENSC9006420

    Name the form of energy which a wound up watch spring possesses.

    Solution

    Since the body is in a state of rest, it posses potential energy.

    Question 453
    CBSEENSC9006421

    Give one example of potential energy due to position.

    Solution

    Water stored in the reservoir of a dam possesses potential energy.

    Question 454
    CBSEENSC9006422

    Give an example where a body possesses both kinetic energy and potential energy.

    Solution

    An aeroplane flying at a height has both potential energy and kinetic energy. 

    Question 455
    CBSEENSC9006423

    Give an example where a body has potential energy due to change of shape.

    Solution

    When a bow is stretched, due to it's change of shape there is a potential energy. 

    Question 456
    CBSEENSC9006424

    Does a man standing at rest on a moving car possess kinetic energy?

    Solution

    The man sitting inside the car shares the motion of the car. Hence, he posses kinetic energy. 

    Question 457
    CBSEENSC9006425

    Can a body have momentum without possessing energy?

    Solution

    Any body which has a momentum is in motion. And a moving body has kinetic energy.  

    Question 458
    CBSEENSC9006426

    Can the kinetic energy of an object be negative?

    Solution

    A kinetic energy of a moving body cannot be negative because both m and v2 are always positive.

    Question 459
    CBSEENSC9006427

    Can the potential energy of an object be negative?

    Solution

    When force involved on the bodies are attractive, then potential energy is negative. 


    Question 460
    CBSEENSC9006428

    Does the potential energy of a spring decrease or increase when it is compressed? 

    Solution

    When a spring is compressed, work is done on it and hence the potential energy increases. 

     

    Question 461
    CBSEENSC9006429

    An electric cell converts one form of energy into another. Name two forms.

    Solution

    An electric cell converts chemical energy into electrical energy. 

    So the two forms are:

    1. Chemical energy
    2. Electrical energy

    Question 462
    CBSEENSC9006430

    Name the instrument which transforms electrical energy into mechanical energy.

    Solution

    Electric motor transforms electrical energy into mechanical energy.

    Question 463
    CBSEENSC9006431

    What energy transformation takes place when an electric bulb is switched on?

    Solution

    When an electric bulb is switched on, electric energy changes into light and heat. 

    Question 464
    CBSEENSC9006432

    State the energy transformation in an electric fan.

    Solution

    In an electrical fan, electric energy is converted into mechanical energy.

    Question 465
    CBSEENSC9006433

    A bullet strikes a target. The bullet stops but a certain portion of the target is burnt. What is the main energy transformation that has taken place?

    Solution

    When the bullet hits a target, kinetic energy of the bullet gets converted into heat. 

    Question 466
    CBSEENSC9006434

    What kind of energy transformation takes place at thermal power station?

    Solution

    In a thermal power station, heat energy is converted into electrical energy. 

    Question 467
    CBSEENSC9006435

    What kind of energy transformation takes place at hydro-electric power plant?

    Solution

    In a hydro-electric power plant, the potential energy of falling water changes into kinetic energy which then changes into electrical energy.

    Question 468
    CBSEENSC9006436

    What is the SI unit of power.

    Solution

    SI unit of power is Watt (W).

    Question 469
    CBSEENSC9006437
    Question 470
    CBSEENSC9006438

    State the relation between horse power and watt.

    Solution

    1 horse power = 746 watt.

    Question 471
    CBSEENSC9006439

    What is the relationship between watt and kilowatt?

    Solution

    1 kilowatt = 1000 watt.

    Question 472
    CBSEENSC9006440

    Name the commercial unit of energy.

    Solution

    Kilowatt hour (kWh) is the commercial unit of energy. 

    Question 473
    CBSEENSC9006441

    Write relation between Si unit of energy and 1 kWh.

    Solution

    Relation between SI unit of energy and 1 kWh is given by, 
                            1 kWh = 3.6 x 106 J.

    Joule is the SI unit of energy. 

    Question 474
    CBSEENSC9006442

    What does B.O.T. unit stand for ?

    Solution

    B.O.T. unit stands for Board of Trade Unit. 

    Question 475
    CBSEENSC9006443

    How much work is done when we push an immovable stone? Justify. 

    Solution

    Work done is zero because the distance moved by the immovable stone in the direction of force is zero. 


    Question 476
    CBSEENSC9006444

    Discuss whether or not work is done in the following cases:

    (i) When we twist a wire.

    (ii) When we press a football.

    (iii) When we push a table.

    (iv) When a person carries a hand bag and walks on a level road.

    (v) When a person holds a book in his hand and keeps it stationary.

    Solution

    (i) Twisting a wire changes it's shape. So, work is done against internal forces of cohesion between the molecules of the wire. 

    (ii) The volume of a football changes when we press it. So, work is done against the pressure of the gas inside the bladder of the football. 

    (iii) Work is done against the force of friction between the legs of the table and the floor, when we push a table between the legs of the table and the floor.

    (iv) No work is done in this case. The force the person is applying against the weight of the bag is acting vertically upwards and his motion is in the horizontal direction.Thus, no motion in the direction of the force is applied.

    (v) In this case, though the force is constantly being applied, there is no displacement and hence work done is zero.

    Question 477
    CBSEENSC9006445

    Explain by an example that a body may possess energy even when it is not in motion.

    Solution

    Consider a stone which is lying on the roof. The stone which is at rest, has a potential energy. The stored potential energy in it is used to lift the stone against the force of gravity.
    The stone has a capacity to do an equal amount of work, when it is allowed to fall. 

    Question 478
    CBSEENSC9006446
    Question 479
    CBSEENSC9006447

    How is energy stored in a watch?

    Solution

    When a watch is winded, work is done in compressing the spring which is stored in it as potential energy. As the spring expands, it does work and moves the hands of the watch. 

    Question 480
    CBSEENSC9006448

    Why does a driver speed up his vehicle when he moves up a hill? 

    Solution

    When a vehicle is moving up the hill, its kinetic energy gradually changes into potential energy. So, to move it up further, it needs more of kinetic energy which is obtained by increasing the speed of the vehicle. 

    Question 481
    CBSEENSC9006449

    If a body moves along a frictionless path, its energy remains always the same. Why? 

    Solution

    A body does not experience any resisting force, while moving ina frictionless surface. Therefore, no work is done by the body. Hence, it's energy remains constant. 

    Question 482
    CBSEENSC9006450

    Which has greater effect on the kinetic energy of an object : doubling the mass or doubling the velocity?

    Solution

    Kinetic energy is directly proportional to the square of the velocity. 
                                  K.E. ∝ vand 

                                  K.E. ∝ m

    Therefore, doubling the velocity has greater effect on the kinetic energy of the object than doubling its mass.

    Question 483
    CBSEENSC9006451

    Define kinetic energy and potential energy. Write an expression for the kinetic energy of a body of mass m moving with a velocity v. Find the kinetic energy of a body of mass 40 kg moving with a velocity 15 m/s.

    Solution

    Kinetic energy: Energy possessed by a body by virtue of its motion is called its kinetic energy.

    Mathematically, kinetic energy is expressed as, 

                           K.E  = 12mv2 

    where, 

    m is the mass of the moving body,
    v is the velocity of the body.

    Potential energy: Energy possessed by a body by virtue of its position or shape is called its potential energy.

    For a body of mass m lying at a height h above the ground level, gravitaional potential energy is given by, 

                             P.E = mgh 

    Given,

    Mass of the body, m = 40 kg
    Velocity of the body, v = 15 m/s

    So,  K. E. = 1 mv2  = 1 x 40 x (15)2 = 4500 J.
                      2             2

    Question 484
    CBSEENSC9006452

    When an arrow is shot, where from the arrow will acquire its kinetic energy?

    Solution

    A stretched bow possesses potential energy due to the change in its shape. The bow is released to shoot an arrow. The potential energy of the bow is converted into the kinetic energy of the arrow. 

    Question 485
    CBSEENSC9006453

    What change should be affected in the velocity of a body to maintain the same kinetic energy if its mass is increased four times?

    Solution

    Kinetic energy of a body is given by,

                             K.E. = 12 mv2.            ... (1)

    From equation (1) we can say that if mass is increased four times, the velocity must be reduced to half to maintain the same kinetic energy.                                                                                  

    Question 486
    CBSEENSC9006454

    A boy and girl do the same work in 5 minutes and 10 minutes respectively. Which of these two has more power and why?

    Solution

    Power is given by,

                     Power = Work doneTotal time taken

    That is, power and time are inversely proportional to each other. 

    Here, in the given case, as the boy takes less time to do the same work as compared to the girl, so power of the boy is more than that of the girl.

    Question 487
    CBSEENSC9006455

    Why do we say that a lift delivers more power in taking a man up than that delivered by the man climbing stairs through the same height? 

    Solution

    Power is given by, 

                    Power = Work donetime taken

    As the lift takes less time in taking the man up, so it delivers more power. 

    Question 488
    CBSEENSC9006456

    Water at the bottom of water-fall is warmer than at the top. Why?

    Solution

    When water is falling from a height, the mechanical energy (K.E. + P.E.) of the falling water is converted into heat energy. Therefore, because of the heat energy the temperature of water at the bottom of water-fall increases.

    Question 489
    CBSEENSC9006457

    Our hands become warm when we rub them against each other. Why?

    Solution

    Work is done against the force of friction between the hands, when we rub our hands against each other. The work done, which is the mechanical energy is converted into heat energy. So, our hands get warm. 

    Question 490
    CBSEENSC9006458

    The head of nail becomes warm when it is hammered into a plank of wood. Explain the series of energy transformations that have taken place.

    Solution

    When the moving hammer strikes the head of the nail, the kinetic energy of the moving hammer is converted into heat energy. As a resukt of the heat energy, the temperature of head of the nail increases. 

    Question 491
    CBSEENSC9006459

    In a thermal power station, coal is used for the generation of electricity. Mention how energy changes from one form to another before it is transformed into electrical energy?

    Solution

    Energy transformations in a thermal power station is as follows:

    The heat energy produced due to combustion of coal converts water into steam. The heat energy of steam is converted into mechanical energy when it turns blades of a turbine. The mechanical energy so obtained is converted into electrical energy by the generators. 

    Question 492
    CBSEENSC9006460

    A girl sits and stands repeatedly for 5 minutes. Draw a graph to show the variation of potential energy of her body with time.

    Solution

    Consider the sitting position of the girl as the position of zero potential energy.
    Let,
    h = position of centre of gravity while standing above the sitting position.
    m = mass of the girl.
    Potential energy while standing = + mgh
    P.E while sitting = zero.
    There is no acceleration or deceleration while standing and sitting.
    Given that, sitting and standing is repeated after every minute.
    The graph for the repeated motion is shown below: 


     

    Question 493
    CBSEENSC9006461

    Distinguish between work, energy, and power. State the SI units for each of these quantities.

    Solution

    When a force acts on a body and the body moves in the direction of the force, work is done. 

    SI unit of work is joule.

    The ability of a body to do work is known as energy.

    SI unit of energy is Joule. 

    The rate of doing work is called power.

    SI unit of power is watt.

    Question 494
    CBSEENSC9006462
    Question 501
    CBSEENSC9006469

    Joule is the unit of

    • power

    • work

       

    • momentum

    • force.<

    Solution

    B.

    work

     

    Question 505
    CBSEENSC9006473

     A body at rest can have

    • speed

    • velocity

       

    • momentum

    • energy

    Solution

    D.

    energy

    Question 519
    CBSEENSC9006852

    Give reason for the following observation:

    The smell of hot sizzling food reaches you several meters away, but to get the smell from cold food you can have to go close.

    Solution

    The rate of diffusion is very high in case of gases. The particle of hot sizzling food mixes with the particles of air and reaches us several meters away. In case of cold food the food particles do not mix with the particles of air easily. The temperature increases the rate of diffusion. Hence we have to go close to it get the smell.

    Question 520
    CBSEENSC9006853

    A diver is able to cut through water in a swimming pool. Which property of matter does this observation show?

    Solution

    This shows that liquids are compressible. The intermolecular space between the particles of liquid is greater as compared to the solids. Hence when the diver into water the particles of water to get compressed.

    Question 521
    CBSEENSC9006854

    What are the characteristics of the particles of matter?

    Solution

    (i) Particles of matter are continuously moving, that is they possess kinetic energy.
    (ii) Particles of matter intermix on their own with other. They do so by getting into the spaces between the particles.

    Question 522
    CBSEENSC9006855

    The mass per unit volume of a substance is called (density = mass/volume).
    Arrange the following in order of increasing density- air, exhaust from chimney, honey, water, chalk, cotton, and iron.

    Solution

    The given substance in the increasing order of their densities can be represented as:
    Air< Exhaust form chimney< Cotton< Water< Honey< Chalk<Iron.

    Question 523
    CBSEENSC9006856

    Tabulate the differences in the characteristics of states of matter.

    Solution

    S.No

    Solid state

    Liquid state

    Gaseous state

    1

    Definite shape and volume

    No definite shape. Liquids attain the shape of the vessel in which they kept.

    Gases have neither a definite shape nor a definite volume.

    2.

    There is little space between the particles of solid.

    These particles have a greater space between them.

    The space between gas particles is the greatest.

    3.

    Incompressible

    Compressible

    Highly compressible

    4.

    These particles attract each other very strongly

    The force of attraction between liquid particles is less than solid particles

    The force of attraction is least between gaseous particles.

     

    5.

    Rigid

    These particles move freely

    Gaseous particles are in a continuous, random motion.

     

     

     

     

    Question 524
    CBSEENSC9006857

    Comment upon the following: rigidity, compressibility, fluidity, filling a gas container, shape, kinetic energy, and density,

    Solution

    Rigidity: Rigidity is property shown by solids. It relates to tendency of solids to resist any change in their shape when an external force is applied. If enough force is applied solids may breaks but do not change their shapes, liquids are not at all rigid as they change their shape.
    Compressibility: compressibility means reduction in the volume of a substance on applying pressure. Solids and liquid are incompressible. This is because there is very empty space between the particles of solids and liquids. For example, you cannot fill more petrol in your car petrol tank even if you apply pressure. On the other hand, gases are easily compressible on applying pressure. It is because there are large empty space between the particles of gases.
    Fluidity: fluidity is a property shown by liquids and gases. It means tendency to flow. Liquid and gases can change their shapes when a force is applied externally. This is the reverse of rigidity.
    Filling a gas container; a gas container can be filled with a large amount of gas when an external pressure is applied on a gas while filling. In the gas container, the gas particles are compressed into a small volume.
    Shape: solids have definite shapes and definite boundaries. Liquid and gases do not have fixed shapes. They acquire the shape of the container in which they are kept.
    Kinetic energy: kinetic energy is also called energy of motion since particles of matter always remaining motion, they have kinetic energy.
    In solids, the particles are bound to each other with strong inter particle forces of attraction, they have least kinetic energy. In liquid, the particles are less tightly bound to each other as the interparticles force of attraction is lesser then in  case of solid particle. The K.E of solid particles and hence the particles of liquid of liquid can move more freely. In gases, the particle are bound by very weak forces of attraction. Gas particles can, therefore move almost freely and have a much higher value of kinetic energy. Also temperature increase, kinetic energy of particles of all kinds of matter increases.
    Density: density is mass per unit volume. Density means how compact a substance is. Solid, because their particles are tightly packed, occur smaller volume and have greater densities. Gases have particles which can move around freely and therefore they can occupy more volume therefore, gases have least density.

    Question 525
    CBSEENSC9006858

    Give reasons:

    A gas fills completely the vessel in which it is kept.

    Solution

    This is because the gases have large inter-particle spaces and the inter-particle attraction force is very weak. The gaseous particle can move freely into the empty spaces of another gas as particles of gas have very high kinetic energy. This property of gases is called diffusion. A gas can mix with any other gas easily and move rapidly from one part of vessel to another part of vessel. Therefore, gas particles can spread easily and fill the whole vessel.

    Question 526
    CBSEENSC9006859

    Give reasons:

    A gas exerts pressure on the walls of the container. 

    Solution

    Particles of gas move randomly in all directions at high speed. As a result, the particles hit each other and also hit the walls of the container with a force. Therefore, gas exerts pressure on the walls of the container.

    Question 527
    CBSEENSC9006860

    Give reasons:
    A wooden table should be called a solid.

    Solution

    A wooden table has a definite shape and volume. It is very rigid and cannot be compressed i.e., it has the characteristics of a solid. Hence, a wooden table should be called a solid. 

    Question 528
    CBSEENSC9006861

    Give reasons:
    We can easily move our hand in air but to do the same through a solid block of wood we need a karate expert.

    Solution

    Particles of air have large spaces between them. On the other hand, wood has little space between its particles. Also, it is rigid. For this reason, we can easily move our hands in air, but to do the same through a solid block of wood, we need a karate expert.

    Question 529
    CBSEENSC9006862

    Convert the following temperature to celsius scale:
    (a) 300 K.        (b) 573 K.

    Solution

    The conversation factor for temperature in celsius scale to kelvin scale is:
    Temperature in kelvin  = 273 + Temperature in celsius
    Temperature of celsius = Temperature in kelvin - 273
    (a) 300 K = 300 - 273 =  27 degree straight C
    (b) 573 K = 573 - 273 =  300 degree straight C

    Question 530
    CBSEENSC9006863

    What is the physical state of water at:
    left parenthesis straight a right parenthesis space 250 degree straight C space space space space space space space space space space left parenthesis straight b right parenthesis space 100 degree straight C

    Solution

    (a) The physical state of water above  100 degree straight C is gaseous. It exists in the form of steam at  250 degree straight C.
    (b) Water at exactly  100 degree straight C starts boiling at 1 atmospheric pressure. At this temperature, water exists in two states liquid as gaseous.

    Question 531
    CBSEENSC9006864

    For any substance, why does the temperature remain constant during the change of state?

    Solution

    During change of state, heat energy is absorbed and used for conversion of one physical state to another. Heat absorbed increase the kinetic energy of particles to overcome the interparticle forces of attraction. 

    Question 532
    CBSEENSC9006865

    Suggest a method to liquefy atmospheric gases.

    Solution

    In liquid, the particles are closer to each other and have lower kinetic energy than gases. When pressure is applied to gas particles they move closer and loose K.E. and thus change to their liquid state. It is possible to liquefy any gas. i.e., convert gas into liquid by increasing the pressure and decreasing the temperature.

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