Physics Part I Chapter 3 Motion In Straight Line
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    NCERT Solution For Class 11 Physics Physics Part I

    Motion In Straight Line Here is the CBSE Physics Chapter 3 for Class 11 students. Summary and detailed explanation of the lesson, including the definitions of difficult words. All of the exercises and questions and answers from the lesson's back end have been completed. NCERT Solutions for Class 11 Physics Motion In Straight Line Chapter 3 NCERT Solutions for Class 11 Physics Motion In Straight Line Chapter 3 The following is a summary in Hindi and English for the academic year 2021-2022. You can save these solutions to your computer or use the Class 11 Physics.

    Question 2
    CBSEENPH110024930

    Define v=u+at from velocity - time graph

    Solution

    The variation in velocity with time for an object moving in a straight line can be represented by Velocity Time Graph. In this graph, time is represented along the x axis and the velocity is represented along the y axis.

    If the object moves at uniform velocity, the height of its velocity time graph will not change with respect to time

    Question 3
    CBSEENPH11016866

    What is rest?

    Solution
    When an object does not change its position with time, with respect to it's surroundings. In such a case the object is said to be at rest. 
    Question 4
    CBSEENPH11016867

    What is motion?

    Solution
    When an object changes it's position with time, with respect to it's surroundings, the object is said to be in motion. 
    Question 5
    CBSEENPH11016868

    Is absolute rest or absolute motion possible?

    Solution
    Absolute rest and absolute motion is not possible because there is no point in the universe which is at rest and can be taken as point of reference. The planet Earth is in continuous motion. 
    Question 6
    CBSEENPH11016869

    What is the most essential thing to study the motion of a body?

    Solution
    The most essential thing to study the motion is reference frame. Rest and motion are relative terms. It means that an object in one situation can be in rest but in another situation, the same object will be in motion. 
    Question 7
    CBSEENPH11016870

    Can a body be always at rest in some frame?

    Solution
    Yes, a body is always at rest in a frame which is fixed with the body itself. 
    Question 8
    CBSEENPH11016872

    Can a body be in motion in the frame associated with the body itself?

    Solution
    A body is always at rest in the frame fixed with the body itself. A body cannot be in motion with respect to a particular frame. 


    Question 9
    CBSEENPH11016874

    When an object in motion can be considered as particle or point object?

    Solution
    An object can be considered as a point object if during motion in a given time, it covers distances much greater than it's own size. 
    Question 10
    CBSEENPH11016875

    Can you think of physical phenomena involving the moon in which moon cannot be treated as particle?

    Solution
    During a solar eclipse, the moon partially or fully blocks the sun. Hence, the moon cannot be treated as particle. 
    Question 11
    CBSEENPH11016876

    A train is crossing the bridge. Can it be taken as a particle?

    Solution
    A train cannot be treated as a particle when it is crossing a bridge. 
    Question 12
    CBSEENPH11016878

    Can a train be taken a particle? Give an example in support of your answer.

    Solution
    Yes, a train can be taken as particle when it runs between say Delhi and Jalandhar. The train has covered distances more than it's own size. 
    Question 13
    CBSEENPH11016880

     A particle is moving along curved line in space. Is the motion one dimensional?

    Solution
    The motion of a particle along curved line in a space is a three dimensional motion.
    Question 15
    CBSEENPH11016884

    Define distance.

    Solution
    The length of the actual path traversed by an object during motion in a given interval of time is called as distance travelled by that object. 
    Question 16
    CBSEENPH11016885

    Define displacement.

    Solution
    The displacement of an object in a given interval of time, is defined as the shortest distance between the two positions of the object in a particular direaction during a given time. 
    Question 17
    CBSEENPH11016887

    Can magnitude of displacement be greater than distance travelled by particle?

    Solution
    No, magnitude of the displacement is always less or equal to the distance travelled by a particle. 

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    Question 18
    CBSEENPH11016888

    A particle moves from A to B along the path APB, When the distance APB will equal to magnitude of displacement AB ? 

    Solution

    The magnitude of displacement AB will be equal to the length of path APB, if P lies in between AB on the line joining the points A and B. 

    Question 19
    CBSEENPH11016889

    A body is thrown with a velocity of 9.8 m/s in the vertical upward direction. What is the velocity after one second?

    Solution

    We have, 
    Initial velocity, u = 9.8 m/s 
    Acceleration, a = - 9.8 m/s2
    Time taken, t = 1 sec
    Final velocity, v = ? 
    Using the formula, 
    Velocity after 1 sec, 
    v = u + at 
       = 9.8 + (-9.8) x 1
       = 0 

    Question 20
    CBSEENPH11016890

    How does the displacement of a particle change, when reference point for the position is shifted?

    Solution
    The displacement of the particle will not change because, displacement of a particle is independent of choice of reference point. 
    Question 21
    CBSEENPH11016892

    A particle moves from A to B along three different paths. Is the displacement along three paths different?

    Solution
    No, the displacement of the particle is same along different paths. 
    Question 22
    CBSEENPH11016893

    Can the distance travelled by a particle be negative?

    Solution
    No, the distance travelled by a particle cannot be negative or zero. It is always positive. 
    Question 23
    CBSEENPH11016894

    Can a body undergo some displacement without travelling any distance? 

    Solution
    No, it is not possible for a body undergo some displacement without travelling any distance. 
    Question 24
    CBSEENPH11016895

    Can a body travel certain distance without undergoing any displacement?

    Solution
    Yes, the body can travel certain distance without undergoing any displacement.
    Question 26
    CBSEENPH11016897

    Define speed.

    Solution
    Speed of an object is defined as the time rate of change of position of the object in any direction. 

    Speed is measured by the distance travelled by the object in unit time. 
    Question 27
    CBSEENPH11016898

    Define velocity.

    Solution
    Velocity is defined as the time rate of change of displacement of the object. It is also defined as the speed of an object in a given direction. 
    Question 28
    CBSEENPH11016899
    Question 29
    CBSEENPH11016900

    Define uniform velocity.

    Solution
    A particle is said to be moving with a uniform velocity, if it undergoes equal displacements in equal intervals of time. 
    Question 30
    CBSEENPH11016901

    Define variable velocity.

    Solution
    A particle is said to be moving with variable velocity, if it undergoes unequal displacements in equal intervals of time. 
    Question 31
    CBSEENPH11016902

    Define average velocity.

    Solution
    The average velocity of an object is equal to the ratio of the displacement, to the time interval for which the motion takes place. 

    Mathematically it is given by, 

              Average velocity = Displacement Time taken 
    Question 32
    CBSEENPH11016904

    Define instantaneous velocity.

    Solution
    Instantaneous velocity is the velocity of the particle at any instant. It is measured as the limiting case of average velocity as the limit ∆t → 0.

                            ν=limtOΔxΔt=dxdt
    Question 33
    CBSEENPH11016905

    Is instantaneous speed always equal to the magnitude of instantaneous velocity?

    Solution
    The magnitude of instantaneous velocity is equal to the instantaneous speed at the given instant. 
    Question 34
    CBSEENPH11016906

    A body is moving without any efforts. What is the nature of motion of body?

    Solution
    The nature of motion of the body is uniform motion.
    Question 35
    CBSEENPH11016908

    Write the equation of kinematics for uniform motion.

    Solution
    The equation for kinematics of uniform motion is given by,

                             x
    0 = x + ut. 
    where, 

    u is the velocity with which the particle is moving, 
    t is the time of travel, 
    x is the distance covered by the object in time t, and

    xo is the position of the object at time t = 0. 
    Question 36
    CBSEENPH11016909

    Which is greater - average speed or magnitude of average velocity.

    Solution
    Average speed is greater than magnitude of average velocity.
    Question 37
    CBSEENPH11016910

    Name two quantities which may be zero during motion over some interval of time.

    Solution
    Displacement and average velocitymay be zero during motion over some interval of time. 
    Question 38
    CBSEENPH11016911

    A car travels along a straight line. For first half of time it moves with speed vand for the second half of time with speed vWhat is the mean speed v of the car?

    Solution
    Speed of the car during it's first half = v1
    Speed of the car during it's second half journey = v2

    Therefore, 

    Mean speed of the car is given by, 

                                   v=v1+v22
    Question 39
    CBSEENPH11016913

    A car travels along a straight line with speed v1 from A to B and returns back from B to A with speed v2. What is the average speed of the car?

    Solution
    Here, the body covers equal distances with different speeds. 

    Avergae speed = Total distance travelled total time taken

    Total distance = S1 + S2 = 2S ( because S1 = S2)

    Total time = S1v1+S2v2 = S 1v1+1v2

    Then, 

    Average speed, vav2SS1v1+1v2 

    That is,             vav=2v1v2v1+v2

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    Question 40
    CBSEENPH11016915

    Define acceleration.

    Solution
    Acceleration is defined is the time rate of change of velocity of the object. 

    Acceleration, a = dvdt 
    Question 41
    CBSEENPH11016916

    When the motion of a particle is said to be uniform accelerated motion?

    Solution
    When a particle undergoes equal change in velocity in equal interval of time, an object is said to be undergoing uniform acceleration. 
    Question 42
    CBSEENPH11016917

    A body describes unequal distances in equal intervals of time. Is it necessary that the motion be uniformly accelerated?

    Solution
    No, the motion of the object may or may not be uniformly accelerated. 
    Question 43
    CBSEENPH11016918

    When the average velocity of a body is mean of initial and final velocities?

    Solution
    When body moves with a constant acceleration then the average velocity of the body is mean of initial and final velocities. 
    Question 44
    CBSEENPH11016919

    How many times, the time occurs in unit of velocity and acceleration?

    Solution
    Velocity is given by dsdt and, 

    Acceleration is given by ds2dt2

    So, time occurs once in velocity and twice in acceleration.
    Question 45
    CBSEENPH11016920

    Is there any significance of defining the rate of change of acceleration?

    Solution
    No, because basic laws of motion require only the concept of acceleration. So, there is no significance of the rate of change of acceleration. 
    Question 46
    CBSEENPH11016921

    Is it possible for a body to be accelerating, though its speed is constant?

    Solution
    Yes, a particle can be accelerating when the speed is constant. 

    For e.g., A particle revolves with constant angular velocity in circular orbit. Speed is constant for the particle but the body is accelearting. 
    Question 47
    CBSEENPH11016922

    Write two basic equations for the kinematics of uniform accelerated motion.

    Solution

    Two basic equations for the kinematics of uniformly accelerated motion are, 
    
space space space space space space space space straight v equals straight u plus at
space space space space space space space space straight x equals straight x subscript 0 plus ut plus 1 half at squared

    Question 48
    CBSEENPH11016923

    A particle starts with velocity u and accelerates uniformly at the rate a. Write an expression for the displacement undergone by particle in nth second.

    Solution

    Initial velocity of the particle = u 
    Rate of acceleration = a
    So, dispalcement undergone by the particle is given by,
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    Question 49
    CBSEENPH11016925

    What is the velocity-time graph for uniform motion?

    Solution
    The velocity-time graph for uniform motion is a straight line parallel to time axis. 
    Question 50
    CBSEENPH11016926

    What is position-time graph for uniform motion?

    Solution
    The position-time graph of a uniform motion is an oblique straight line.

    The slope of the graph gives us the velocity of the object. 
    Question 51
    CBSEENPH11016927

    What is the velocity-time graph for uniformly accelerated motion?

    Solution
    Velocity-time graph for uniformly-accelerated motion is a straight line. 
    Question 52
    CBSEENPH11016928

    What is the velocity-time graph for uniform accelerated motion?

    Solution
    The velocity-time graph for uniform accelerated motion is an oblique straight line. 
    Question 53
    CBSEENPH11016929

    What is position-time graph for uniform accelerated motion?

    Solution
    Position-time graph for uniformly accelerated motion is a parabola. 

    Position-time graph for positive and negative acceleration have both parabolic shapes. 
    Question 54
    CBSEENPH11016930

    Can we find the distance travelled by a particle from velocity-time graph?

    Solution
    Yes. Distance travelled by the body in a given interval of time is equal to the total area of velocity-time graph. 

    Even if the body is moving with a negative velocity, area is taken as positive for the measurement of distance travelled by a body. 
    Question 55
    CBSEENPH11016931

    How can you determine the instantaneous velocity from displacement time graph?

    Solution
    Instantaneous velocity of particle at any instant is equal to the slope of displacement-time graph at that instant. 
    Question 56
    CBSEENPH11016932

    What does the area under velocity-time graph represent?

    Solution
    The are under velocity-time graph gives the  displacement of the particle. 
    Question 57
    CBSEENPH11016933

    What does the area under velocity-time graph represent?

    Solution
    The area under velocity-time graph gives the acceleration of the particle.
    Question 58
    CBSEENPH11016934

    Can we find the displacement from speed-time graph?

    Solution
    No, speed time graph will give us the distance covered and not the displacement of the particle. 
    Question 59
    CBSEENPH11016935

    Can we find the distance travelled by particle from speed-time graph?

    Solution

    Yes, the area of the speed-time curve gives us the distance.  

    Question 60
    CBSEENPH11016936

    Can we find displacement from the velocity-time graph?

    Solution
    The total area of velocity-time graph for a given interval of time gives us the displacement of the body. 
    Question 61
    CBSEENPH11016937

    Under what condition does the relationship ∆ x = v ∆t hold exactly?

    Solution
    When the particle is moving with a constant velocity, relationship ∆ x = v.∆t holds good. 
    Question 62
    CBSEENPH11016940

    Under what condition does the relationship ∆v = a ∆t hold exactly?

    Solution
    Relationship ∆ v=a.∆ t holds good if particle is moving with constant acceleration. 
    Question 63
    CBSEENPH11016941

    When average velocity of body over interval is equal to half of the sum of initial and final velocity? 

    Solution
    When a motion is uniformly accelerated motion, then average velocity is half the sum of initial and final velocities. 
    Question 64
    CBSEENPH11016942

    A particle is moving with constant velocity of 5m/s. What will be the slope of (a) position time graph (b) velocity time graph?

    Solution

    Given, a particle is moving with a constant velocity 5 m/s. 

    (a) Slope of position-time graph will give us the speed of the particle. Speed will be equal to 5m/s.

    (b) Slope of velocity-time graph will be zero because velocity is constant. 

    Question 66
    CBSEENPH11016945

    The odometer of car reads 40km at the beginning of trip and 76km after 40min. What is the average velocity of car?

    Solution

    Initial distance covered = 40 km 
    Distance covered after 40 mins of the trip = 76 km
    Time taken = 40 min
    Therefore, average speed of car = fraction numerator Total space distance space travelled over denominator Total space time space taken end fraction   
                                         

space space space space space space space equals fraction numerator 76 minus 40 over denominator 40 divided by 60 end fraction equals 54 space km divided by hr

    Question 67
    CBSEENPH11016947

    If the slap is defined as the rate of change of acceleration, then what is the unit of slap in SI system of units?

    Solution
    Slap is defined as rate of change of acceleration.

    SI unit of acceleration = m/s2

    Therefore SI unit of slap = m/s2.s = m/s
    3
    Question 68
    CBSEENPH11016948

    Can an object have eastward velocity while experiencing a westward acceleration?

    Solution
    Yes, acceleration and velocity can be in different direction.

    There is no definite relationship between direction of acceleration and velocity.
    Question 69
    CBSEENPH11016949

    What does the change in velocity of a body depends upon? 

    Solution
    Change in velocity of a body is determined by change in acceleration of the body. 
    Question 70
    CBSEENPH11016950

    Can a body have zero velocity but still accelerating?

    Solution
    When body is thrown up with some velocity, velocity at the highest point is zero. But, the acceleration is non-zero for the object when it is moving. 
    Question 71
    CBSEENPH11016951

    Can a body be at rest, at any instant of motion?

    Solution
    Yes, when a body is thrown in upward direction, at the highest point, the body is momentarily at rest.The velocity at the highest point is zero for the particle. 
    Question 72
    CBSEENPH11016952

    When a body is in equilibrium?

    Solution
    A body is in equilibrium when it is at rest or is moving with a constant velocity. 
    Question 73
    CBSEENPH11016953

    A particle moves along a straight line path. After some time it comes to rest. What is the direction of acceleration w.r.t. direction of displacement?

    Solution
    For a particle which is moving in a straight line, the acceleration is negative throughout the motion with respect to the displacement.
    Question 74
    CBSEENPH11016954

    What do you mean by kinematics and dynamics?

    Solution

    Kinematics is the branch of classical mechanics which describes the motion of points, bodies and system of bodies without considering the masses of these objects not the forces that has caused the motion. 

    Dynamics is the study of motion of particle by considering the forces and their torque and their effect on motion. 

    Question 75
    CBSEENPH11016955

    Can a particle be simultaneously at rest as well as in motion? Give one example.

    Solution
    Rest and motion are relative terms. A  particle can be simultaneously at rest as well as in motion. 

    For example, a person standing on the earth is at rest with respect to the earth and same person is in motion with respect to the moon.
    Question 76
    CBSEENPH11016956

    Explain by giving an example that rest and motion are the relative terms i.e. a body can be simultaneously at rest as well as in motion.

    Solution
    Rest and motion are relative terms. A body can be simultaneously at rest as well as in motion.

    Let two persons A and B be sitting in a moving train and a third person be at the platform. Here the position of A does not change with respect to B. Therefore, A is at rest as per B. But, position of A changes with respect to C with time. Hence A is in motion with respect to C.
    Question 77
    CBSEENPH11016957

    List down the similarities and difference between  displacement and distance?

    Solution

    Displacement and distance can be compared as follows: 

    (i) Both have units and dimensions same as that of length.

    (ii) The change in position of particle in particular direction is known as displacement while distance is the total length of path travelled by particle.

    (iii) Displacement is a vector quantity while distance is a scalar quantity. 

    (iv) Displacement can be negative, positive or zero while distance covered by particle is always positive.

    (v) Displacement of a particle between two points is a unique path, while distance is not a unique path .

    Question 78
    CBSEENPH11016958

    If displacement of body is zero, then is the distance covered being necessarily zero? Explain.

    Solution
    Displacement is the total change in position and distance is total length of path travelled by a particle.  

    For example, consider a student who start from his home. After attending the college he again comes back to home.

    In this motion, displacement is zero but distance travelled is not zero and has a finite number. 
    Question 79
    CBSEENPH11016960

    Can the speed of a body be negative?

    Solution
    Speed is defined as the ratio of total length of path travelled to the time taken to traverse the path.

    The distance travelled by a paticle is always positive. Therefore, speed is positive quantity. 

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    Question 80
    CBSEENPH11016961

    Distinguish between speed and velocity.

    Solution
    Speed Velocity
    Speed changes only when the magnitude of the object changes. Velocity changes whenever there is a change in direction and speed of the particle. 
    Speed is a scalar quantity. Velocity is a vector quantity. 
    Speed is either zero or positive. It cannot be negative.  Velocity can be positive, negative or zero
    Question 81
    CBSEENPH11016962

    When velocity is constant, does the average velocity over any interval differ from instantaneous velocity at any instant?

    Solution
    Velocity at all the instants is same because the velocity is constant. Therefore, instantaneous velocity of a particle at any instant is equal to the average velocity. 
    Question 82
    CBSEENPH11016963

    What is the difference between average and instantaneous velocity? 

    Solution
    Average velocity Instantaneous velocity

    1. The velocity with which the body covers the same distance in the same time as the body actually covered with variable velocity is known as average velocity. 

    It is measured as the ratio of total displacement to total time taken.

    1. Instantaneous velocity of a body is defined as the velocity of the body at a particular instant and it is the limiting case of average velocity as the time interval approaches zero.

    That is instantaneous velocity is, 

    v =limt0ΔSdt

    2. Average velocity between two times is equal to the slope of the line joining the two points on the position vs. time graph.  2. Instantaneous velocity of an object at a particular time is equal to the slope of it's position vs. time graph at that time. 
    Question 83
    CBSEENPH11016965

    What is the ratio of displacement to distance if the particle is moving in a straight line with constant velocity? 

    Solution
    When a particle moves in a straight line with constant velocity, then magnitude of displacement is equal to the distance. Direction of displacement is same as that of velocity.

    So, the ratio of displacement to distance  gives us the direction of velocity. 
    Question 84
    CBSEENPH11016966

    Plot x-t graph for uniform motion and show that the slope of x-t graph is equal to the velocity of the particle.


    Solution

    Kinematics of uniform motion is given by

                    x = x0+ut                          ...(1)

    Therefore x - t graph will be a straight line. Slope of graph is given by, 
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    Hence the result. 

    Question 85
    CBSEENPH11016967

    Plot velocity-time graph for uniform motion and show that the area under the graph is equal to displacement of the particle.

    Solution

    When a particle moves with a constant velocity, displacement is given by, 

                          S = x - x= ut ...(1)

    For uniform motion, the velocity time graph is straight line parallel to time axis.



    Now area under v-t graph from t = t1 to  t = t2 is given by = Area of rectangle ABCD 

                  = ut

                  = x - x0, which is the displacement of the particle. 

    Question 86
    CBSEENPH11016971

    The position time graph of a particle is moving along a straight line is shown here. Find the velocity of particle from graph.
    Also find the displacement of particle from t = 4s to t = 10s.


    Solution
    Consider two points A and B corresponding to time t =4s and t =10s on the graph given below.



    From the graph,

    Position of particle at t = 4s is 10m

    Position of particle at t = 10s is 17.5m.

    Therefore, the velocity of parties is given by the slope of the graph. 

    Slope = v=BCAC                   =17.5-101.-4                   =1.5m/s

    Therefore, sisplacment of particle from t = 4s to t = 10s is given by,

                      S = x10-X4

                         =17.5-10 

                         = 7.5m
    Question 87
    CBSEENPH11016972

    A particle completes one revolution in 6 seconds along the circumference of radius 20cm. What is the total length of the path travelled by the particle in 15 seconds and displacement in 15 seconds?

    Solution

    Given,  

    Time taken by the particle to complete 1 revolution = 6 s 

    Radius of the circle = 20 cm 

    No. of revolutions made in 15 seconds = 2.5 revolution

    After 2.5 revolutions, the particle will be at the point diagonally opposite to the starting point. 

    So, Displacement of particle = 2r = 40 cm

    Distance travelled by the particle = Total length of the particle. 

    Distance = total length of the path = 2.5x2πr = 100 π cm 

    Question 88
    CBSEENPH11016973

    In the figure shown here, find the displacement and distance travelled by particle in going from P to Q.




    Solution

    Displacement of partcle is, 
    straight S with rightwards harpoon with barb upwards on top equals 10 space straight m comma space along space PQ with rightwards harpoon with barb upwards on top space direction.
    Distance traveled by particle is, 
    D i s t a n c e italic equals italic space P L italic space italic plus italic space l e n g h t italic space o f italic space a r e italic space stack L K Q with italic overbrace on top
                 equals 10 plus fraction numerator 3 straight pi over denominator 2 end fraction straight x 10
                  = 57.1m

    Question 89
    CBSEENPH11016975

    A particle covers half of circle of radius r in time t. Show that ratio of distance travelled to magnitude of displacement is numerically equal to right angle in radian.

    Solution

    Given, 
    Radius of the circle = r 
    Distance covered = half of the circle =  πr

    Magnitude of displacement is 2r. 

      DistanceDisplacement=πr2r=π2, which is also numerically eqal to right angle in radian. 

    Question 90
    CBSEENPH11016985

    Particle P completes half revolution in circular orbit of radius 10m in 5s and another particle Q is moving along a straight line with constant speed 4 m/s. Which of the two particles has greater average speed?

    Solution
    Given, particle P completes half revolution in a circular orbit of radius 10m in 5s.
    Therefore,
    Particle P travels the distance πr =10π m in 5s.
    Therefore,
    Average speed of particle P = 2π m/s.
    Particle Q is moving with constant speed 4m/s.
    Therefore its average speed is also 4m/s. 

    Hence, particle P has a greater average speed. 
    Question 91
    CBSEENPH11016987

    A body starts from x0 = 5m with velocity u = 10 m/s and B body starts from x0' = 10 m and with velocity u' = 5 m/s. Can we find when and where A will overtake B from graph? Explain.



    Solution
    For body A,

    Initial distance, xo = 5 m
    Initial velocity, u = 10 m/s 

    For body B, 

    Distance,xo' = 10 m 
    velocity, u' = 5 m/s

    From the graph, point where the position time graphs of two bodies meet, corresponds to position and time where the A will overtake B. 
    Question 92
    CBSEENPH11016989

    A particle travels half of distance with velocity v1 and next half with velocity v2. Find the average velocity of motion.

    Solution

    Let the particle travel a total distance of 2S.

    Distance covered in first half is S with a velocity v1 and next half with velocity v2

    Time taken to travel first half distance is, t1=Sν1

    Time taken to travel second half distance is, t2=Sv2

    Total time taken to travel 25 distance is, 

    t =t1+t2=Sv1+Sv2=S(v1+v2)v1v2

     Average velocity =Total distance travelledTotal time taken 

                                 = 2SSv1+v2v1v2 = 2v1v2v1 + v2
          

    Question 93
    CBSEENPH11016992

    A body travels for half of time with velocity v1 and for next half of time with velocity v2. What is the average velocity of motion?

    Solution

    Let the body travel for total time to 2t.
    The distance travelled by body in first t time is,
                            s subscript 1 equals straight v subscript 1 straight t
    The distance travelled by body in next t time is, 
                            s subscript 2 equals straight v subscript straight v straight t
    Total distance travelled by the body is, 
    straight S equals s subscript 1 plus s subscript 2 equals straight v subscript 1 straight t plus straight v subscript 2 straight t

therefore space A v e r a g e space v e l o c i t y space equals space fraction numerator T o t a l space d i s t a n c e space t r a v e l l e d over denominator T o t a l space t i m e space t a k e n end fraction

space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space equals space fraction numerator straight v subscript 1 straight t plus straight v subscript 2 straight t over denominator 2 t end fraction equals fraction numerator straight v subscript 1 plus straight v subscript 2 over denominator 2 end fraction

    Question 94
    CBSEENPH11016995

    The light travels 100 m in vacuum and then 100 m in media in which velocity of light is 2.4 x 108 m/s. What is the average velocity of light?

    Solution

    Distance travelled by light in vacum is, S= 100m
    Distance travelled by light in media is, S= 100m
    Velocity of light in vacum is c = 3 x 108 m/s
    Velocity of media is v = 2.4 x 10m/s
    Time taken by light to cover 100m in vacum is,
    straight t subscript 1 fraction numerator 100 over denominator 3 cross times 10 to the power of 8 end fraction equals 1 third straight x 10 to the power of negative 6 end exponent straight s
    Time taken by light to cover 100m in media is,
    straight t subscript 2 equals fraction numerator 100 over denominator 2.4 cross times 10 to the power of 8 end fraction equals 5 over 12 cross times 10 to the power of negative 6 end exponent straight s
    Total time is given by, 
    straight t equals straight t subscript 1 plus straight t subscript 2
space equals open parentheses 1 third plus 5 over 12 close parentheses cross times 10 to the power of negative 6 end exponent
space space equals 3 over 4 cross times 10 to the power of negative 6 end exponent straight s
     
    Now average velocity is given by, 

    top enclose straight v space equals fraction numerator 200 over denominator begin display style 3 over 4 cross times 10 to the power of negative 6 end exponent end style end fraction

space space space equals 8 over 3 cross times 10 to the power of 8 straight m divided by straight s
           

    Question 96
    CBSEENPH11016999

    A train travels from one station to another at a speed of 40 km/hr and returns to first station at the speed of 60 km/hr. Calculate the average speed and average velocity of the train.  

    Solution

    Let the distance between two stations be 'd' km.

    After the journey train returns to the first station. 

    Total distance travelled = 2d 
    Net displacement of the train = 0 

    Time taken by train to travel from one station to another, t1=d40hrs

    Time taken by train to come back,  t2=d60hrs

    Total time taken by train to complete the round trip is given by, 

    t=t1+t2=d40+d60=d24hrs

     Average speed =2dd/24=48 km/hr

    Average velocity = 0d/24=0 km/hrs 

    Question 97
    CBSEENPH11017000

    A drunkard walking in a narrow lane takes 7 steps forward and 4 steps backward followed by same strokes. Each step is 0.5m long and requires 1sec. Find the net displacement of drunkard in three such strokes and average velocity of the drunkard.

    Solution

    Given, the drunkard walks 7 steps forward and 4 steps backward.

    So, the net displacement undertaken by the drunkard = 3 steps in one stroke

    Also given, length of 1 step = 0.5 m 

    Time taken to take 1 step = 1 sec

    ∴ Displacement in 1 stroke = 3 x 0.5=1.5m

    Therefore,

    Displacement in 3 strokes = 3 x 1.5 =4.5m

    Time taken to walk one step is 1 sec.

    Since to complete one stroke, drunkard walks 7 steps forward and 4 steps backward i.e. total of 11 steps, therefore drunkard takes 11s to complete one stroke.

    So,

    Time taken to complete three strokes, t = 11 x 3 = 33s 

    Now, average velocity of drunkard is given by, 

                        v=4.533=15110m/s

    Question 98
    CBSEENPH11017003

    A drunkard walking in a narrow lane takes 5 steps forward and 3 steps backward, followed again by 5 steps forward and 3 steps backward, and so on. Each step is 1 m long and requires 1 s. Plot the x-t graph of his motion. Determine graphically and otherwise how long the drunkard takes to fall in a pit 13 m away from the start.

    Solution
    Length of 1 step taken by drunkard = 1 m

    Time to take 1 step = 1 sec 

    So, the speed of the drunkard = 1 m/s 

    When he walk 5 steps forward, its position is 5m at time 5s.
    After walking 3 steps backward, the position of drunkard at 8 sec becomes 2m. 

    Time, t (sec) Distance, x (m)
    5 5
    8 2
    13 7
    16 9
    24 6

    So, the position- time graph is as shown in figure. 



    Question 99
    CBSEENPH11017005

    A car is moving with a constant speed of 30km/hr along a straight line towards a vertical wall. A fly flying at a constant speed of 60km/hr starts from a wall towards the car when it is 90km away from the wall. The fly flies till it reaches the glass pane of car and then returns to wall at the same speed. After touching the wall again it flies towards car. Fly continues its motion till car just touches the wall. Find the total distance travelled by fly. How many trips the fly will make?

    Solution

    Speed of car, v = 30 km/hr
    Distance of wall, d= 90 km 
    Time taken by car to reach the wall is,
    straight t space equals space straight d over straight v equals 90 over 30 equals 3 hr
    Fly moves back and forth between car and the wall with constant speed 60km/hr.
    Therefore, distance travelled by fly is,

    S = ut
       = 60 x 3
       = 180 km
    To find: The distance travelled by the fly in it's one complete trip. 

    Consider, initially when the fly is flying from the wall towards the car, it is at a distance D from the wall. 
    Speed at which the car is approaching the fly = 30 km/hr
    Speed at which the fly is flying towards the car = 60 km/hr
    Let both car and fly meet after time t1.

    Thus distance travelled by car and fly respectively is,

     straight d subscript 1 straight c end subscript space equals space vt subscript 1 space equals space 30 space straight t subscript 1 space space space space space space space space space space space space space space space space space... space left parenthesis 1 right parenthesis space

straight d subscript 1 straight f end subscript space equals space vt subscript 1 space equals space 60 space straight t subscript 1 space space space space space space space space space space space space space space space space space... space left parenthesis 2 right parenthesis

Therefore comma space

straight d subscript 1 straight c end subscript space plus straight d subscript 1 straight f end subscript space equals space straight D

Now comma space solving space equations space left parenthesis 1 right parenthesis comma space left parenthesis 2 right parenthesis space and space left parenthesis 3 right parenthesis comma space
we space have

straight d subscript 1 straight c end subscript space equals space straight D over 3 space space space space space space space space and space straight d subscript 1 straight f end subscript space equals space fraction numerator 2 straight D over denominator 3 end fraction
    The fly will return to the wall. 
    Let t2 be the time taken by fly to come back to the wall. 
    straight t subscript 2 space equals space space straight d subscript 1 straight F end subscript over straight v space equals space fraction numerator 2 straight D over denominator 3 cross times 60 end fraction space equals space straight D over 90 space
    In time t2, the distance moved by the car towards the fly is
    straight d subscript 2 straight c end subscript space equals space Vt subscript 2 space equals space 30 straight D over 90 space equals space straight D over 3
    Distance travelled by the car after completing 1 trip is, 
    d = straight d subscript 1 straight c end subscript space plus space straight d subscript 2 straight c end subscript space equals space fraction numerator 2 straight D over denominator 3 end fraction
    When fly completes one trip, distance of car from the wall is, 
    straight D subscript 1 space equals space straight D space minus space fraction numerator 2 straight D over denominator 3 end fraction
    Thus, if at the beginning of motion, the car is at a distance D from the wall, then after completion of one trip of fly, it will be at a distance one-third of initial distance i.e. 1/3 D from the wall. 
    Thus, the distance of car from wall after n trips will be 
     space space space space 1 over 3 to the power of straight n space straight D space equals space 0

rightwards double arrow space straight n space equals space infinity
    Therefore, the fly will make infinitely many trips.
    Question 100
    CBSEENPH11017007

    Can the direction of velocity of a body change when its acceleration is constant?

    Solution
    Yes, the direction of velocity of  body changes when it's acceleration is constant.

    For example, when we throw a body in horizontal direction, the direction of velocity continuously changes but acceleration due to gravity is constant. 
    Question 101
    CBSEENPH11017015

    When the magnitude of acceleration is more - the accelerator of the car is pushed to floor or when the brake pedal is pushed hard? 

    Solution
    When the brake pedal is pushed hard, the magnitude of acceleration is more as compared to the accelerator of the car when pushed to floor.
    Question 102
    CBSEENPH11017016

    Show by giving an example that direction of motion is given by direction of velocity not by that of acceleration.

    Solution
    When a particle is thrown in upward direction, it's motion is in upward direction and so is it's velocity. But, the acceleration on the particle is acting downwards. 
    Question 103
    CBSEENPH11017027

    What does positive acceleration mean? Can a particle slow down if acceleration is positive?

    Solution
    When the velocity of particle is increasing with time, we have positive acceleration. Particle can speed up or slow down if acceleration is positive.

    e.g. If downward direction is taken positive, then acceleration due to gravity will be positive.

    When we throw the body in upward direction, acceleration slows down. And, acceleration speeds up when the body falls down. 
    Question 104
    CBSEENPH11017028

    Comment on the motion of particle revolving in circular path with constant angular velocity.

    Solution
    For a particle moving in a circular motion, speed remains constant but the particle undergoes an accelertion. The magnitude of acceleration is constant and direction changes continuously. Therefore, the motion is not uniformly accelerated. 
    Question 105
    CBSEENPH11017029

    Is it possible to have the rate of change of velocity constant while velocity itself does not change both in magnitude and direction?

    Solution
    This case is possible if the rate of change of velocity is zero or constant. Otherwise, if the rate of change of velocity is nonzero constant then velocity will change. 
    Question 106
    CBSEENPH11017030

    Read each statement carefully and state if it is true or false:

    A particle in one dimensional motion:

    (a) with zero speed at an instant may have non-zero acceleration at that instant;

    (b) with zero speed may have non-zero velocity;

    (c) with constant speed must have zero acceleration;

    (d) with positive value of acceleration must be speeding up.

    Solution

    (a) True; when the body is projected vertically upward, then the speed of the particle at the highest point is zero but has an acceleration equal to the acceleration due to gravity.

    (b) False; because

    |Displacement| ≤ Distance

    ∴|Velocity| ≥ Speed

    Thus, the speed of the body zero implies that the magnitude of velocity of the particle is zero. Hence the velocity of body is zero.

    (c) True; since motion is in one dimension, therefore there are only two possible directions of motion. Since speed is constant, therefore motion can be accelerated if  the direction of motion changes instantaneously.
    i.e. acceleration of particle is infinite, which is not possible in practice.

    (d) False; because if acceleration is positive, then it need not be necessary that particle will always speed up, it may slow down. The particle will speed up if velocity is also positive.

    Question 107
    CBSEENPH11017032

    Derive the relation v = u + at for uniform accelerated motion.


    Solution
    Initial velocity of the particle = u
    Rate of acceleration of the particle = a 
    Velocity of the particle after time 't' = v
    Now, using the formula, 
    space space space space space space space space space straight a space equals space dv over dt
therefore space space space space space dv space equals space straight a. dt space

Integrating space both space sides comma space we space get

space space space space space integral subscript straight u superscript straight v dv space equals space integral subscript 0 superscript straight t straight a space dt space equals straight a space integral subscript 0 superscript straight t dt space
rightwards double arrow space space space space space space right enclose straight v subscript straight u superscript straight v space equals space straight a right enclose straight t subscript 0 superscript straight t space

rightwards double arrow space space space space space space space space straight v minus straight u space equals space at space

rightwards double arrow space space space space space space straight v space equals space straight u space plus space at
    Question 108
    CBSEENPH11017039

    Two particles of different masses are dropped from the same point. Which will reach the ground earlier?

    Solution
    Both the particles will reach the ground at the same moment. Because, the fall is independent of the masses. 
    Question 109
    CBSEENPH11017040
    Question 110
    CBSEENPH11017041

    A stone thrown upwards with velocity u and rises to a height h. What will be its speed when it again reaches the point of throw?

    Solution
    The velocity of the stone will remain the same.

    Velocity = u m/s 
    Question 111
    CBSEENPH11017042

    A body is projected upwards with velocity 9.8 m/s. How long it will take to reach the highest point?

    Solution
    The velocity with which the body is projected = 9.8 m

    Acceleration due to gravity = 9.8 m/s2

    Time taken to reach to the highest point is given by, 

              v = u + at 

          v = at 
           t = v/a = 9.8/9.8 = 1 sec 
    Question 112
    CBSEENPH11017044

    A body is projected vertically upwards. How does the direction of acceleration and velocity differ during the upward motion of particle?

    Solution
    During an upward motion, direction of acceleration is opposite to the direction of velocity. Velocity is acting upwards and acceleartion is acting downwards. 
    Question 113
    CBSEENPH11017046

    A body is projected vertically upwards. How are the direction of acceleration and velocity related after the body reaches the highest point?

    Solution
    After reaching the highest point, the body starts falling in downward direction, therefore the direction of acceleration and velocity is same.
    Question 114
    CBSEENPH11017047

    A body is thrown vertically upwards in vacuum. Which of the two will have greater-time of ascent or descent?

    Solution
    When a body is thrown vertically upwards in vacuum, time of ascent is equal to the time of descent. 

    In vacuum there is no resistance of air. 
    Question 115
    CBSEENPH11017048

    A body is thrown vertically upwards in air. When air resistance is taken into account, then which of the two will be greater: time of ascent or descent?

    Solution
    When a body is thrown vertically upwards in air, then time of descent is greater than time of ascent.
    Question 116
    CBSEENPH11017049

    A body is thrown up with velocity 39.2m/sec. What is the direction of velocity and acceleration at t = 0, t = 3, t = 4, t = 6?

    Solution

    Given, a body is thrown up with velocity = 39.2 m/s.
    Acceleration due to gravity is acting vertically in a downward direction.  
    Acceleration, g = 9.8 m/s2

    The body thrown in upward direction with velocity 39.2 m/s will take 4s to reach the maximum height. 

    At t = 0 and t = 3, the velocity is in upward direction.
    At t = 4 sec velocity is zero and at t = 6 sec, velocity will be in downward direction. 

    Question 117
    CBSEENPH11017060

    From the foot of a tower 90m high, a stone is thrown up so as to just reach the top of the tower. Find the velocity with which the stone is thrown.

    Solution
    Height of the tower, h = 90 m 

    Velocity of the stone when it reaches the top = 0 

    Therefore, 

    velocity, v = 0

    acceleration, a = -9.8 m/s2

    distance travelled, s = 90 m

    Now, using the relation v2-u= 2as, we have 

    02-u2 = 2x(-9.8) x 90

             = -1764
     
        u= 1764

          u = 42 m/s, is the velocity with which the stone is thrown. 
    Question 118
    CBSEENPH11017063

    A stone falls from a cliff and travels 45m in last second before it reaches the ground. What is the height of cliff? (Given g = 10 m/s2)

    Solution

    Let, height of the cliff be 'h'. 

    Time taken by the stone to reach the ground = n 

    The distance travelled by particle in nth second is given by, 
                       Snth=u+a2(2n-1)

    Here we have, 

    Snth=45m,   u = 0,   a = 10 m/s2

       45=0+102(2n-1) 

         n = 5 s 

    i.e. Stone takes 5 s to reach the ground from top of cliff. 

    Now the height of cliff is,

    h =ut+12gt2   =0×5+12×10×(5)2   =125m

    Question 119
    CBSEENPH11017064

    A body of mass 5 kg is dropped from a height of 100m. With what velocity it will hit the ground? (Take g=10 m/s2)

    Solution

    Given, 

    Mass of the body, m = 5 kg 
    Height from where the mass is dropped = 100 m 
    Acceleartion due to gravity, g = 10 m/s2 

    Using the formula, 

                          v2-u= 2as
    Here, 
    u = 0, 

    a = g = 10 m/s2,

    s = 100m

     v2-(0)2=2×10×100=2000 

     v = 44.72 m/s, is the velocity with the body will hit the ground. 

    Sponsor Area

    Question 121
    CBSEENPH11017077

    A ball is thrown vertically upward and returns back to the hands of thrower in 6s. Find the velocity of throw and maximum height attained by the ball.

    Solution
    Let the ball be projected with velocity u from the ground.
    Let us take upward direction positive and ground as reference point.
    Thus,
    Displacement comma space straight y subscript 0 equals 0 comma space space straight y space equals space 0

straight g space equals space minus 9.8 space straight m divided by straight s squared

Time space of space travel comma space straight t equals 6 straight s 
    Now, using the formula,
        straight y space equals straight y subscript 0 plus u t plus 1 half a t squared 
    rightwards double arrow  0 equals 0 plus straight u cross times 6 plus 1 half left parenthesis negative 9.8 right parenthesis left parenthesis 6 right parenthesis squared
    rightwards double arrow     0 = 6u-176.4
    i.e.,      u = 29.4 m/s
    Let h be the maximum height attained by the ball.
    Using third equation of motion, we have
    therefore space space space space straight v squared minus straight u squared equals 2 a s 
    rightwards double arrow  (0)- (29.4)= 2 x(-9.8)h
    rightwards double arrow     h = 44.1 m, is the maximum height attained by the ball. 
    Question 122
    CBSEENPH11017081

    A ball thrown vertically a upwards with a speed 29.4 m/s from the top of tower reaches the ground in 8s. Find the height of the tower.

    Solution

    Given, a ball is thrown vertically upwards with a speed 29.4 m/s. 
    Let, ground be the reference point and vertically upward direction is taken as positive. 
    We have,
    Initial velocity, u = 29.4 m/c
    Acceleration due to gravity, a = g= -1m/s
    Time taken for the ball to reach the ground, t = 8s
    Initial distance, x = 0 
    Let, h be the height tower.
    Therefore,
    Displacement, xo = h 
    Using the equation of motion, we have 
    Now      straight x space equals space straight x subscript 0 plus ut plus 1 half at squared  
    therefore space 0 space equals space straight h plus 29.4 space straight x space 8 minus 1 half cross times 9.8 cross times left parenthesis 8 right parenthesis squared 
    rightwards double arrow 0 = h + 235.2 - 313.6
    rightwards double arrow h = 78.4m, is the required height of the tower. 

    Question 123
    CBSEENPH11017092

    A ball is thrown vertically upwards from the top of a tower 80m high and reach the ground in 6 seconds. Find the velocity with which the ball is thrown?

    Solution
    Consider, ground as reference point and vertically upward direction as positive. 
    Height of the tower, h = 80 m 
    Acceleration due to gravity, a = -10 m/s2 
    Time taken for the ball to reach the ground = 6 seconds
    Here, 
    straight x subscript 0 space equals space straight h space equals space 80 space straight m space space space space space space space straight a space equals space straight g space equals space minus 10 space straight m divided by straight s squared
straight t space equals space 6 space sec space space space space space space space space space space space space space space space space space space straight x space equals space 0 space
    Let  be the initial velocity of the ball. 
    So, usign equation of motion, we have
    space space space space space space space space straight x space equals space straight x subscript 0 space plus space ut space plus space 1 half at squared space
therefore space space space space space 0 space equals space 80 space plus space straight u cross times 6 space minus 1 half cross times 9.8 cross times left parenthesis 6 right parenthesis squared space
space space space space space space space space space 0 space equals space 80 space plus space 6 straight u space minus 1 half cross times 9.8 space cross times 36 space
space space space space space space space space space 0 space equals space 80 space plus space 6 straight u space minus 176.4 space space

space space space space space space space space space straight u space equals space 16.07 space straight m divided by straight s

    So, u = 16.07 m/s is the velocity with which the ball is thrown. 
     
     
    Question 124
    CBSEENPH11017119

    Plot position-time graph for uniform accelerated motion and show that area under v-t graph is equal to displacement of particle.


    Solution

    The velocity-time relation for uniform accelerated motion is given by,

                               v = u + at ...(1)

    The position-time graph for uniform accelerated motion is oblique straight line. 

    Now area under v-t graph from t = 0 to t = t is = Area under segment AB 
    = Area of rectangle ACDO + Area of ∆ABC.
    equals space OA cross times OD plus 1 half AC cross times BC
equals space ut plus 1 half straight t left parenthesis straight v minus straight u right parenthesis
equals ut plus 1 half left parenthesis at right parenthesis space space space space space space space space space space space space space space space space space space space space open square brackets because open parentheses straight v minus 4 close parentheses equals at close square brackets
equals ut plus 1 half at squared
equals S comma space is space the space distance space travelled.

    Question 125
    CBSEENPH11017120

    What is the sign of acceleration during interval OA, AB and BC and that of velocity during AB, BC and CD?


    Solution
      OA OB BC
    Acceleration positive Zero Zero
      AB BC CD
    Velocity Positive Negative Zero
    Question 126
    CBSEENPH11017122

    Find the average velocity of particle from displacement time graph?

    Solution
    Average velocity of a particle is defined as the ratio of total displacement to the total time taken. 

    From graph the particle undergoes displacement of (X
    2-X1) in time (t2 - t1).

     Average velocity = Total distance travelledTotal time taken                                    =x2-x1t2-t1                                   = ACBC =tanθ                                   = Slope of chord AB

    Hence, average velocity is the slope of chord joining the two points on displacement-time graph.
    Question 127
    CBSEENPH11017124

    Average speed is magnitude of average velocity vector and average speed is also defined as the ratio of total length of path travelled to the time elapsed. Does these two definitions have different meanings? If so, give an example. 


    Solution
    Yes, the two sentences imply different meaning except when direction of velocity is constant. 
    When the direction of velocity changes then according to the first definition,
    Average speed = AC over Δt any the second definition, 
    Average speed  = space space space space fraction numerator AC with overbrace on top over denominator Δt end fraction
    Since  top enclose AC greater than AC
    Therefore average speed by second definition is greater then the average speed by first definition.
    Question 128
    CBSEENPH11017128

    The velocity time graph of two objects A and B makes angle of 30° and 60° with time axis respectively. Compare their accelerations. 



    Solution

    Slope of velocity-time graph gives the acceleration of the object.
    Therefore, 

    Acceleration of object A, straight alphaA = tan 30°
    Acceleration of object B, straight alpha= tan 60°


    Now, straight alpha subscript straight A over straight alpha subscript straight S equals fraction numerator tan 30 over denominator tan 60 end fraction equals 1 third 

    Question 129
    CBSEENPH11017129

    Position time relation of a particle is given by x = (3t + 4)m. Find the velocity of particle from the graph.

    Solution

    Position of a particle at any instant, x = 3t+4



    The x -t graph is a straight line.
    The slope of the position-time graph gives us the velocity of the particle. 
    Here,  slope of the graph is,
        
space space space space space space Slope space equals space BC over AC equals 9 over 3 equals 3
space space space space space space space therefore space space straight v space equals space 3 space straight m divided by straight s


    Therefore, velocity of the particle is 3 m/s. 

    Question 130
    CBSEENPH11017130

    Two straight lines drawn on the same displacement time graph make angle 30° with each other as shown in the figure. For what value of straight theta velocity of A will be three times the velocity of B?


    Solution
    On analysing the given graph,
    Velocity of the particle A is given by, 
    straight v subscript straight A space equals space tan space left parenthesis straight theta plus 30 to the power of straight o right parenthesis space 
    Velocity of particle B is given by, 
    straight v subscript straight B space equals space tanθ 
    Here, v= 3 vB 
    tan space left parenthesis theta space plus space 30 to the power of o right parenthesis space equals space 3 space tan space theta
U sin g space t h e space f o r m u l a space o f space tan space left parenthesis A plus B right parenthesis comma space w e space h a v e

space space space space fraction numerator tan space theta space plus space space tan space 30 to the power of 0 over denominator 1 space minus tan theta space tan space 30 to the power of o end fraction space equals space 3 space tan space theta

rightwards double arrow space space tan theta space plus space tan 30 to the power of 0 space equals space 3 space tan theta space minus space 3 space tan squared theta space tan space 30 to the power of o space

rightwards double arrow space space 3 space tan space 30 to the power of o space tan squared theta space minus space 2 space tan theta space plus space tan space 30 to the power of o space equals space 0 space

rightwards double arrow space space space space space tan space theta space equals space fraction numerator 2 plus-or-minus square root of 4 space minus 12 space tan squared 30 to the power of o end root over denominator 6 space tan space 30 to the power of o end fraction

S u b s t i t u t i n g space tan space 30 to the power of o space equals space fraction numerator 1 over denominator square root of 3 end fraction comma space w e space g e t
space space space space space space space space tan space theta space equals space fraction numerator 1 over denominator square root of 3 end fraction space

rightwards double arrow space space space space space space space space space space space space theta space equals space 30 to the power of o
 
     
     
    Question 131
    CBSEENPH11017136

    State, with reasons, which of the following graphs cannot possibly represent one dimensional motion of particle: 



    Solution

    One-dimensional motion of the particle is represented by graphs (b) and (d).
    But, (a), (c), (e) and (f) cannot represent the motion of particle because, 

    (a) a particle cannot have two velocities at the same time,

    (c) a particle cannot have two positions simultaneously at the same time,

    (e) total length of path can never decrease with time and

    (f) a particle cannot have two different accelerations at the same time.

    Question 132
    CBSEENPH11017139

    The velocity-time graph of particle moving in straight line is given below. Three equal intervals t1 = t2 = tare shown. In which interval acceleration is constant, zero and variable? What is the acceleration at A and C? 

    Solution
    Acceleration is constant in interval t2, it is zero in the interval tand variable in the interval t1.
    At point A, the velocity is maximum. So, acceleration is zero. 
    The slope of v-t graph is zero at point C. Hence, acceleration is zero here. 
    Question 133
    CBSEENPH11017141

    Suggest the suitable physical situation for each of the following graphs:


    Solution

    (a) Graph (a) represents the path of a ball which is dropped from a certain height and on hitting the ground, it rebounds elastically. 

    (b) Graph (b) represents the path of a uniformly moving football turned back by hitting it for a very short interval of time. 

    Question 134
    CBSEENPH11017143

    The position-time (x-t) graph for two children A and B returning from their school O to their homes P and Q respectively is as shown in the figure.


    (a) Who lives closer to the school?

    (b) Who starts from the school earlier?

    (c) Who walks faster?

    (d) Who overtakes the other? 


    Solution

    (a) Since, OP < OQ, A lives closer to the school than B.

    (b) From the position-time graph, A starts at t = 0 and B at t = T. That is, A starts earlier than B. 

    (c) Velocity is given by the slope of x-t graph. Here, the slope of x-t graph for B is greater than A. Therefore B walks faster than A. 

    (d) B overtakes A.

    Question 135
    CBSEENPH11017147

    Look at the graphs (a) to (d) (Fig. 3.20) carefully and state, with reasons, which of these cannot possibly represent one-dimensional motion of a particle.

    Solution
    (a) From position-time graph we can find the position of the particle at any instant by drawing the perpendicular on time axis at that instant.
    There are two different positions of particle at a particular instant which is not possible because position is a single valued function. 

    (b) Since velocity is a single valued function, therefore at any instant, the particle cannot have two different velocities, hence the graph cannot represent motion. 

    (c) Since speed is magnitude of velocity which cannot be negative, therefore, graph is not possible.

    (d) Total length of path travelled by particle cannot decrease with the time. Therefore, graph is not possible.

    Question 136
    CBSEENPH11017149

    Figure 3.21 shows the x-t plot of one-dimensional motion of a particle. Is it correct to say from the graph that the particle moves in a straight line for t < 0 and on a parabolic path for t >0 ? If not, suggest a suitable physical context for this graph.

    Solution

    Position-time  graph represents the position of particle at any instant, not the trajectory of particle.
    For t < 0, the position-time graph is time axis which signifies that the particle is at rest.
    For t > 0, the position-time graph is a parabola which signifies that the body starts moving with constant acceleration.
    The given graph is a suitable context for a body is dropped from a certain height at t=0.

    Question 137
    CBSEENPH11017151

     Is the time variation of position shown in the figure observed in nature?


    Solution
    No, the slope dt over dx of horizontal portion of the graph is zero. This implies that the velocity of the particle is infinite, which can never be possible to obtain in practice. 
    Question 138
    CBSEENPH11017155

    Which of the following equations of kinematics are of:

    (i) uniform motion

    (ii) uniform accelerated motion

    (iii) general motion.

    (a) x = 4 - 7t2+4t3

    (b) v = 4 + 5t (c) x — 5 

    (d) a = 9t -5 (e) a = 4.

    Solution

    a) x = 4 - 7t1+4t3

    Differentiating 'x' twice w.r.t. time, we get

    a = 24 t- 14
    Since acceleration depends on time, therefore this is kinematics of general motion.

    (b) v = 4 + 5t

    ∴ a = 5

    Since acceleration of particle is constant, therefore the motion is uniform motion.

    (c) x = 5

    ∴ v = 0

    i. e. particle is at rest.

    (d) a = 9t-5

    Since acceleration depends on time, therefore this is kinematics of general motion.

    (e) a =4

    Because acceleration is constant, therefore motion is uniform accelerated motion.

    Question 139
    CBSEENPH11017167

    A car moving along a straight highway with speed of 126 km/hr is brought to a stop within a distance of 200 m. What is the retardation of the car (assumed uniform), and how long does it take for the car to stop?

    Solution

    Here,
    Initial velocity, u = 126 km/hr = 35m/s
    Final velocity, v = 0
    Distance travelled, s = 200m 
    Now using the equation of motion,
                          v2-u= 2as
    Substituting the value of v, u and s we get, 
                 (0)2-(35)= 2a x 200 
    rightwards double arrow                       a = -3.0625 m/s

    Time taken by the car to stop is given by, 
               straight t equals fraction numerator straight v minus straight u over denominator straight a end fraction equals fraction numerator 0 minus 35 over denominator negative 3.0625 end fraction equals 11.43 straight s

    Question 140
    CBSEENPH11017174

    A car starts from rest and acquires a velocity of 12m/s in 5 sec. Calculate the acceleration and distance moved.

    Solution

    Here,
    Initial velocity, u = 0
    Final velocity, v = 12m/s
    Time taken to acquire the velocity, t = 5s
    Using the equation, we have
                        v = u + at
    therefore space space space space space space a equals space fraction numerator straight v minus straight u over denominator straight t end fraction equals fraction numerator 12 minus 0 over denominator 5 end fraction equals 2.4 space straight m divided by straight s
and space space space straight s space equals space fraction numerator straight v squared minus straight u squared over denominator 2 straight a end fraction

space space space space space space space space space space space space space equals fraction numerator left parenthesis 12 right parenthesis squared minus left parenthesis 0 right parenthesis squared over denominator 2 cross times 2.4 end fraction space

space space space space space space space space space space space space space equals 30 straight m
    where, a is the acceleration of the moving car, and 
    s is the distance moved by the car. 

    Question 141
    CBSEENPH11017183

    The velocity of a particle at any instant is given by v = 5t + 3. From the v- t graph, find the displacement of the particle between t = 2 to t = 5.

    Solution

    Here,
    Velocity of a particle at any instant, v = 5t + 3
    So,
    Velocity|t=2 is v(2) = 13 m/s
    velocity|t=5 is  v(5) = 28 m/s 
    We know displacement of particle is equal to area under v-t graph.
    space space therefore Distance travelled = Area of trapezium ABCD
                                    space equals space 1 half left parenthesis AB plus BC right parenthesis cross times DC
space equals 1 half left parenthesis 13 plus 28 right parenthesis cross times 3 equals 61.5 straight m

    Question 142
    CBSEENPH11017184

    Velocity-time graph of the motion is as shown in the figure.
    Describe the motion of particle. What is the significance of shaded area?

    Solution

    From the graph given, we can infer that the particle starts from rest at t = 0 and accelerates uniformly, acquiring a velocity of 10 m/s in 5 sec.
    Thereafter, the particle moves with a constant velocity.

    The shaded area represents the displacement undergone by the particle during accelerated motion. 

    Question 143
    CBSEENPH11017185

    A car starts from rest and accelerates uniformly. After n seconds, it acquires the speed v. Find the distance travelled by the car in last 2s.

    Solution

    Given that,
    Initial velocity, u = 0
    Acceleration, a = ?
    At  t = n, the velocity of particle is v is given by the first equation of motion. 
    therefore space straight nu space equals space 0 space plus space an
    rightwards double arrow   space space straight a space equals space straight v over straight n
    Velocity of car at t = n - 2 is given by,
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#6 {main}</pre>
    Now, the distance tracvelled by the car in last two second is,
    straight S equals 2 straight V plus 1 half straight alpha open parentheses 2 close parentheses squared
space space space equals 2 straight v over straight n left parenthesis straight n minus 2 right parenthesis plus 1 half straight v over straight n 4

space space space space equals space fraction numerator 2 left parenthesis straight n minus 1 right parenthesis straight v over denominator straight n end fraction

    Question 144
    CBSEENPH11017187

    A car moves with uniform acceleration along a straight line ABC. Its speed at A and C is 5 m/s and 25 m/s respectively. If AB = BC, then find the velocity at B.

    Solution

    Speed of car A, ua = 5 m/s
    Speed of car C, uc= 25 m/s
    Given, AB = BC
    That is B is the mid-point of AC.
    Therefore, velocity at B is given by, 
    straight v space equals space square root of fraction numerator straight u subscript straight A squared space plus space straight u subscript straight c squared over denominator 2 end fraction end root space equals space square root of fraction numerator 25 plus 625 over denominator 2 end fraction end root space equals space square root of 325 space straight m divided by straight s

    v = 18 m/s is the velocity of particle at B. 

    Question 145
    CBSEENPH11017189

    A particle is moving in a straight line with constant acceleration. It travels a distance of straight alpha in first n seconds, straight beta in next n seconds and y in next n seconds.

    Prove that,

    α + γ = 2β.

    Solution
    Consider,
    'u' be the initial velocity and 'a' be the acceleration of particle.
    Particle travels straight alpha distance in first n seconds, β in next n seconds and gammain next n seconds.
    Thus, particle travels α distance in first n seconds, α + β in first 2n seconds and α+β+γ in first 3n seconds.
    Therefore, equation of kinematics can be written as
    
straight alpha equals u n plus 1 half a n to the power of 2 space space space space space 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... left parenthesis 1 right parenthesis
straight alpha plus straight beta equals 2 u u n plus 1 half straight a left parenthesis 2 straight n right parenthesis squared space space space space space space space space space space space space space space space space space space space... left parenthesis 2 right parenthesis
straight alpha plus straight beta plus straight gamma space equals space 3 u n plus 1 half straight alpha left parenthesis 3 straight n right parenthesis squared space space space space space space space space space space space space space space... left parenthesis 3 right parenthesis
    Subtracting (1)  from (2),
             straight beta equals un plus 3 over 2 an squared
    For equation (3),
    straight alpha plus straight beta plus straight gamma equals 3 open parentheses un plus 3 over 2 an squared close parentheses equals 3 straight beta
    That is,
                    straight alpha plus straight nu equals 2 straight beta
    Question 146
    CBSEENPH11017198

    A piece of paper and stone are dropped simultaneously from same point in vacuum. Which of the two reaches the ground earlier?

    Solution

    Both reach the  ground simultaneously as the acceleration due to gravity is same for both the objects. 

    Question 147
    CBSEENPH11017200

    The acceleration time graph of a particle moving along a straight line is as shown in figure. At what time the particle acquires its initial velocity?


    Solution

    The area under velocity time graph represents the change in velocity.
    Let at time tthe particle acquire the initial velocity.
    Therefore the change in velocity of the particle in time t0 is zero. i.e. Area under a v/s t graph is zero

    Thus area of ∆ AOB = Area of ∆ BCD
    or space space space space space space 1 half cross times OA cross times OB equals 1 half cross times BC cross times CD
or space space space space space OB space tan space straight theta space cross times space OB space equals space CD space tan space straight theta space cross times space CD
or space space space space space left parenthesis OB right parenthesis squared space equals space left parenthesis CD right parenthesis squared
or space space space space space OB equals CD
or space space space space space 4 space equals space straight t subscript 0 minus 4
or space space space space space straight t subscript 0 equals equals 8
    So, at time t=8 sec the particle acquires it's initial velocity. 

    Question 148
    CBSEENPH11017202

    A bullet fired into a very thick wall loses one fourth of its velocity after penetrating 4m. How much further will it penetrate?

    Solution
    Let, u be the velcoity with which the bullet enters the wall.
    At A and after penetrating AB = 4m, it loses one fourth of its velocity, therefore its velocity at B reduces to vB = 3u/4.
    Let further it travel BC = x to come to rest.
    Let a be the retardation offered by the wall.

    For motion of AB:
    Using third equation of motion, 
       v- u= 2as
    open parentheses fraction numerator 3 straight u over denominator 4 end fraction close parentheses minus straight u squared equals negative 2 straight a cross times 4
rightwards double arrow space space space space space space space 7 over 16 straight u squared equals 8 straight a space space space space space space space space space space space space space space space space space space space... left parenthesis 1 right parenthesis
    For motion of BC:
    Again using third equation of motion, 
    v- u= 2as
    
open parentheses 0 close parentheses squared minus open parentheses fraction numerator 3 straight u over denominator 4 end fraction close parentheses squared equals negative 2 a x
rightwards double arrow space space space fraction numerator 9 straight u squared over denominator 16 end fraction equals 2 a x space space space space space space space space space space space space space space space space space space space space space space space space space space space space space... left parenthesis 2 right parenthesis 
    Dividing (2) by (1), we get 
    The distance of penetration, straight x space equals space 36 over 7 equals 5.143 space straight m

    Question 149
    CBSEENPH11017203

    What is motion? Is absolute rest or absolute motion possible?

    Solution
    Whe the position of the object changes with respect to surroundings with time, the particle is in motion. 
    Absolute rest or motion is not possible. Every body in the universe is in relative motion or rest with respect to some other object which is taken as the reference. 
    Question 150
    CBSEENPH11017204

    Two trains on the same line are approaching each other with velocity u1, and u2When there is distance of x between them, both apply brakes and brakes give the retardation of at1 and a2. Prove that it is just possible to avoid the collision if

    space space space straight u subscript 1 squared straight a subscript 2 plus straight u subscript 2 squared straight alpha subscript 1 equals 2 straight alpha subscript 1 straight alpha subscript 2 straight x

    Solution
    Let u1, be velocity of train A and a1, be retardation.
    Let x, be the distance travelled by the train A before it comes at rest.


    Using the third equation of motion, 
     space space space space space space space space straight s equals fraction numerator straight v squared minus straight u squared over denominator 2 straight a end fraction w e space g e t comma

space space space space space space space space straight x subscript 1 equals fraction numerator 0 squared minus straight u subscript 1 squared over denominator negative 2 straight alpha subscript 1 end fraction equals fraction numerator straight u subscript 1 squared over denominator 2 straight alpha subscript 1 end fraction
    Similarly, if x2 is the distance travelled by train B before it comes at rest then
    straight x subscript 2 equals fraction numerator 0 squared minus straight u subscript 2 squared over denominator negative 2 straight a subscript 2 end fraction equals fraction numerator straight u subscript 2 squared over denominator 2 straight a subscript 2 end fraction
    Collision can be avoided, if the total distance travelled by both the trains is x.
    i.e.   straight x equals straight x subscript 1 plus straight x subscript 2

space equals space fraction numerator straight u subscript 1 squared over denominator 2 straight alpha subscript 1 end fraction plus fraction numerator straight u subscript 2 squared over denominator 2 straight alpha subscript 2 end fraction
equals fraction numerator straight u subscript 1 squared straight alpha subscript 2 plus straight u subscript 2 squared straight a subscript 1 over denominator 2 straight alpha subscript 1 straight alpha subscript 2 end fraction

rightwards double arrow space straight u subscript 1 squared straight alpha subscript 2 plus straight u subscript 2 squared straight alpha subscript 1 equals 2 straight a subscript 1 straight a subscript 2 straight x
    Question 151
    CBSEENPH11017205

    A body starts from rest and accelerates uniformly at rate α for time tand then retards instantaneously at the rate β for time t2 to come to rest. Show that the ratio of the time of accelerated motion to the time of retarded motion is equal to ratio of retardation to the acceleration.

    Solution

    For accelerated motion:
    Initial velocity, u = 0 m/s
    Acceleration =straight alpha m/s
    Time taken = t1 sec 
    Let, v be the velocity acquired by the body at the end of accelerated motion.
    By using the relation,
    v = u + at,
    we have 
    V = 0 + alpha t subscript 1 space equals space alpha t subscript 1       ...(1) 
    For retarded motion, 
    straight u space equals space straight V space space space space space space space straight alpha space equals space minus straight beta
straight t space equals space straight t subscript 2 space space space space space space space space space straight v space equals space 0 
    Now, using this relation, v= u + at, we have
    0 space equals space straight V space minus space beta t subscript 2 space
space space space straight V space equals space space beta t subscript 2 space space space space space space space space space space space space space... space left parenthesis 2 right parenthesis

So space from space equations space left parenthesis 1 right parenthesis space and space left parenthesis 2 right parenthesis comma space we space have

alpha t subscript 1 space equals space beta t subscript 2 space

therefore space space alpha t subscript 1 space equals space beta t subscript 2 space
space space space space space space space space space
space space space space straight t subscript 1 over straight t subscript 2 space equals space straight beta over straight alpha

    Question 152
    CBSEENPH11017206

    Show that acceleration is directly proportional to time. 

    Solution

    Since it given that,
                              straight nu proportional to space straight t squared
    i.e.           space space space space space space space space space space space space space
space space space space space space space space space space space space space straight nu equals kt squared
therefore space space space space space space space space space straight alpha equals dν over dt equals 2 kt
straight i. straight e. space space space space space space space space straight alpha space space proportional to space space straight t
    Acceleration of the particle is directly proportional to the time, i.e. acceleration increases linearly with time.

    Question 154
    CBSEENPH11017208

    The position of particle at any instant t is given by relation t = αx2 + βx + γ. What is the acceleration of the particle?

    Solution

    Given,
    Position of particle at any insatnt t is,
                         straight t equals αx squared plus βx plus straight capital upsilon
    Differentiating both sides w.r.t. tme,
    
1 equals 2 ax dx over dt plus straight beta dx over dt
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#6 {main}</pre>
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#6 {main}</pre>
    Now,
    straight alpha equals straight v fraction numerator d v over denominator d x end fraction
space space space equals straight v fraction numerator straight d over denominator d x end fraction left parenthesis 2 alpha x plus straight beta right parenthesis to the power of negative 1 end exponent
rightwards double arrow space space space

A c c e l e r a t i o n comma space straight alpha equals negative straight nu 2 straight alpha left parenthesis 2 alpha x plus straight beta right parenthesis 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 equals space minus 2 alpha nu nu 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 equals negative 2 a nu cubed            

    Question 155
    CBSEENPH11017209

    The position of the particle at any instant is given by the equation

    x = 3t2- 2.5t2 - 4.

    what is the velocity and acceleration of particle at 3s?

    Solution

    The position of the particle is given by,
                    x=3t-2.5t-4
    Velocity of the particle at any instant is,
    straight v equals dx over dt equals straight d over dt left parenthesis 3 straight t cubed minus 2.5 space straight t squared minus 4 right parenthesis
equals 9 straight t squared minus 5 straight t
    space space therefore Velocity  at  t = 3s is, 
    v(3) =  9(3)- 5 x 3 = 81 - 15 = 66 units 
    Acceleration of the particle at any instant is, 
    straight v equals dx over dt equals straight d over dt left parenthesis 9 straight t squared minus 5 straight t right parenthesis
equals space 18 straight t minus 5 
    space space therefore  Acceleration at t = 3 s is,
    a(3) = 18 x 3-5 = 54 - 5 = 49 units

    Question 156
    CBSEENPH11017210

    The acceleration of a particle at any instant is given by a = 6t-12. The particle starts from x0 = 4m with a velocity 4 m/s. What is the velocity and position of the particle at any instant?

    Solution

    Given, 
    Acceleration of the particle, a = 6t - 12

    Initial distance, x0 = 4 m
    Initial velocity, u = 4m/s

    Let at any instant the velocity of the particle be v.
    integral subscript straight u superscript straight v d v space equals space integral subscript 0 superscript t a space d t space

rightwards double arrow space integral subscript 4 superscript v d v space equals space integral subscript 0 superscript t left parenthesis 6 t space minus space 12 right parenthesis thin space d t

rightwards double arrow space v vertical line subscript 4 superscript v space equals space left parenthesis 3 t squared space minus space 12 t right parenthesis vertical line subscript 0 superscript t space

rightwards double arrow space v space minus space 4 space equals space 3 t squared space minus space 12 space t space

rightwards double arrow space v space equals space 3 t squared space minus space 12 space t space plus space 4

    Let at any instant the position of the particle be x. 
    Therefore, 
    space space integral subscript straight x subscript straight o end subscript superscript straight x dx space equals space integral subscript 0 superscript straight t space straight v. dt

rightwards double arrow space integral subscript 4 superscript straight x space dx space equals space integral subscript 0 superscript straight t left parenthesis 3 straight t squared space minus space 12 straight t space plus 4 right parenthesis dt space

rightwards double arrow space straight x vertical line subscript 4 superscript straight x space end superscript equals space left parenthesis straight t cubed space minus space 6 straight t squared space plus space 4 straight t right parenthesis vertical line subscript 0 superscript straight t space

rightwards double arrow space straight x space minus space 4 space equals space straight t cubed space minus space 6 straight t squared space plus space 4 straight t space

rightwards double arrow space straight x space equals space straight t cubed space minus space 6 straight t squared space plus space 4 straight t space plus space 4

    Question 157
    CBSEENPH11017211

    The position of the particle moving in one dimension, under the action of constant force is given by equation

    straight t equals square root of straight x plus 3

    where x is in meter and t is in second. Find the displacement of the particle when its velocity is momentarly zero.

    Solution

    We have,
    Position of the particle moving under a constant force, straight t equals square root of straight x plus 3
    rightwards double arrow     square root of straight x equals straight t minus 3
    rightwards double arrow        x = t-6t + 9 
    Differentiating w.r.t. r, we get
         straight v space equals space dx over dt equals straight d over dt left parenthesis straight t squared minus 6 straight t plus 9 right parenthesis 
     
    rightwards double arrow  v = 2t - 6, is the velocity of the particle.
    The time at which the velocity of the particle will be zero is given by, 
     
    space space space space space space 2 straight t minus 6 equals 0 space

rightwards double arrow space space straight t space equals space 3 straight s 
    Now,
    Initial position of the particle is, 
    straight x subscript 0 equals left parenthesis 0 right parenthesis squared minus 6 left parenthesis 0 right parenthesis plus 9 equals 9
    Position of the particle at t = 3 is given by, 
    x=(3)- 6(3) + 9 = 0
    space space therefore Displacement of the particle is, 
    x - x= 0 - 9
              = - 9m

    Question 158
    CBSEENPH11017212

    After the engine of a moving motorboat is cut off, the motorboat retards with retardation directly proportional to v. What is the magnitude of velocity at any instant after the engine is cut off, if the velocity at the time of cut off is v0?

    Solution

    Given, 
    Retardation of the boattau space proportional to space nu
    Therefore, 
     
           r = kv; k is the proportionality constant. 
    i. e. comma space straight a space equals space minus kv
    Now, we know that acceleration is given by rate of change of velocity. 
    Rate space of space change space of space velocity space is space known space as space acceleration. space

rightwards double arrow space dv over straight v space equals space minus space straight k space dt space

Now comma space on space integrating space both space sides comma space

integral subscript straight v subscript straight o end subscript superscript straight v dv over straight v space equals space minus space straight k integral subscript 0 superscript straight t dt space

rightwards double arrow space Ln space straight v vertical line subscript straight v subscript straight o end subscript superscript straight v space equals space minus kt vertical line subscript 0 superscript straight t space

rightwards double arrow Ln space straight v over straight v subscript straight o space equals space minus space kt space

rightwards double arrow space straight v space equals space straight v subscript straight o space straight e to the power of negative kt end exponent

    Question 159
    CBSEENPH11017213

    Two particles A and B are connected by a rigid rod AB as shown in figure. The rod slides along perpendicular rails. Where the rod subtends an angle a = 30° with the horizontal, the velocity of A is 10m/s. Find the velocity of B.

                            

    Solution

    From figure given, we have
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#6 {main}</pre> 
    Differentiating both sides w.r.t. t, we get
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#6 {main}</pre> 
    rightwards double arrow                  2xu + 2yv = 0
    rightwards double arrow            space space space straight v equals negative straight x over straight y straight u equals negative straight u space cot space straight alpha
    Here,
    Initial velocity, u = 10m/s
     alpha space equals space 30 degree
    therefore space space space straight v space equals space minus left parenthesis 10 right parenthesis space c o t space 30 degree space

space space space space space space space space space equals space minus 10 square root of 3 space straight m divided by straight s comma space is space the space velocity space of space straight B.

    Question 160
    CBSEENPH11017214

    A particle is moving along a straight line such that its displacement at any instant is given by,

    s1 = αt1 + 2βt + γ

    where α, β and γ are constants.

    Find the acceleration of the particle.

    Solution

    Given here,
       space space space space space space space space space space space space space space straight s squared equals αt squared plus 2 βt plus straight capital upsilon space space space space space space space space space... left parenthesis 1 right parenthesis 
    Differentiating (1) w.r.t. t,
    
space space space space straight d over dt left parenthesis straight s squared right parenthesis space equals space straight d over dt left parenthesis αt squared plus 2 βt plus straight gamma right parenthesis 
    rightwards double arrow    2 xv equals 2 αt plus 2 straight beta space space space space space space space space space space space space space space space space space space space space space space open square brackets because dx over dt equals straight v close square brackets
    rightwards double arrow        straight v equals fraction numerator αt plus straight beta over denominator straight s end fraction space space space space space space space space space space space space space... left parenthesis 2 right parenthesis
    Differentiating (2) w.r.t. t, 
    straight d over dt left parenthesis straight v right parenthesis equals straight d over dt left parenthesis fraction numerator αt plus straight beta over denominator straight s end fraction right parenthesis 
    That is, acceleration of the particle is given by, 
         space space straight alpha equals fraction numerator straight s begin display style straight d over dt end style left parenthesis straight alpha plus straight beta right parenthesis minus left parenthesis αt plus straight beta right parenthesis begin display style straight d over dt end style left parenthesis straight s right parenthesis over denominator straight s squared end fraction   

         
space space space space space equals space fraction numerator sα minus left parenthesis at plus straight beta right parenthesis straight v over denominator straight s squared end fraction 

    
space space space space space space space space space space space space space equals space fraction numerator sα minus left parenthesis αt plus straight beta right parenthesis straight v over denominator straight s squared end fraction 
       
space space space space space space space space equals fraction numerator sα minus left parenthesis αt plus straight beta right parenthesis begin display style fraction numerator αt plus straight beta over denominator straight s end fraction end style over denominator straight s squared end fraction
             
equals fraction numerator αs minus left parenthesis αt plus straight beta right parenthesis squared over denominator straight s cubed end fraction
             equals fraction numerator straight alpha left parenthesis αt squared plus 2 βt plus straight gamma right parenthesis minus left parenthesis straight alpha squared straight t squared plus 2 αβt plus straight beta squared right parenthesis over denominator straight s cubed end fraction
    This implies,
          straight alpha equals fraction numerator αγ minus straight beta squared over denominator straight s cubed end fraction, is the acceleration of the particle. 

    Question 161
    CBSEENPH11017215

    A particle starts from rest from origin along X-axis and accelerates with acceleration a = 12t + 4. Find displacement of the particle in first 2s.

    Solution

    Acceleration of the particle, a = 12 t + 4
    Acceleration is the rate of change of velocity, a = dv/dt = 12 t + 4
    That is, 
    dv = (12 t+ 4) dt
    On integrating both sides, 
    space space space space space space space space integral subscript 0 superscript straight v dv space equals space integral subscript 0 superscript straight t left parenthesis 12 straight t plus 4 right parenthesis dt space

rightwards double arrow space straight v vertical line subscript 0 superscript straight v space equals space left parenthesis 6 straight t squared space plus space 4 straight t right parenthesis vertical line subscript 0 space end subscript superscript straight t

rightwards double arrow space straight v space equals space 6 straight t squared space plus space 4 straight t space

rightwards double arrow space dx over dt space equals space 6 straight t squared space plus space 4 straight t
rightwards double arrow space dx space equals space space left parenthesis 6 straight t squared space plus space 4 straight t right parenthesis space dt space

Integrating space both space sides comma space

integral subscript 0 superscript straight x dx space equals space integral subscript 0 superscript straight t 6 straight t squared space plus space 4 straight t right parenthesis space dt space

we space get comma space

straight x space equals space 2 straight t cubed space plus space 2 straight t squared space

Position space of space the space particle space at space straight t equals 2 space is comma space

straight x left parenthesis 0 right parenthesis space equals space 0 space

Position space of space particle space at space straight t minus 2 space is comma space

straight x left parenthesis 2 right parenthesis space equals space 2 thin space left parenthesis 2 right parenthesis cubed space plus space 2 left parenthesis 2 right parenthesis squared space equals space 24 space

Therefore comma space displacement space of space particle space in space first space 2 straight s space is comma space

straight S space equals space straight x left parenthesis 2 right parenthesis space minus space straight x left parenthesis 0 right parenthesis space equals space 24 space straight m

    Question 162
    CBSEENPH11017216

    Show that if a body retards at the rate numerically equal to distance travelled, then the distance travelled by the body before comes to rest is numerically equal to velocity at the time at which the body starts retarding.

    Solution

    Here we have, 
    Body retards at a rate of distance travelled. 
    That is, 
             a = -s 
    v dv over dx space equals negative space s
    rightwards double arrow space minus straight v space dv space equals space straight s space ds space space space space space space space space space space space space space space space space space... space left parenthesis 1 right parenthesis
    Let the velocity of the body when it starts retarding be v0 and come to rest by travelling the distance s.

    Integrating both sides of (1) between initial and final limits, 
    space space space space space integral subscript straight v subscript straight o end subscript superscript 0 straight v. space dv space equals space integral subscript 0 superscript straight t straight s space straight s space

rightwards double arrow space space space space right enclose negative straight v squared over 2 end enclose subscript straight v subscript straight o end subscript superscript 0 space equals space right enclose straight s squared over 2 end enclose subscript 0 superscript straight t space

rightwards double arrow space space space space space space space space space straight v subscript straight o squared over 2 space equals space straight s squared over 2

rightwards double arrow space space space space space space space space space space space space space straight v subscript straight o space equals space straight s

    Question 163
    CBSEENPH11017217

    A point moves in a straight line so that its displacement x (m) at time t (s) is given by

    x2 = t2+ 1.

    Find its acceleration at any instant.

    Solution

    The position of the particle at any instant given by,
    x2 = t2 + 1 
    x = (t2 + 1)1/2 
    Differentiating both sides w.r.t. to t, we get
     
    dx over dt space equals space fraction numerator straight d left parenthesis straight t squared plus 1 right parenthesis to the power of bevelled 1 half end exponent over denominator dt end fraction

space space space space space space space space equals space 1 half left parenthesis straight t squared plus 1 right parenthesis to the power of negative begin inline style bevelled 1 half end style end exponent space cross times space 2 straight t space

rightwards double arrow space straight v space equals space straight t space left parenthesis straight t squared space plus space 1 right parenthesis to the power of negative begin inline style bevelled 1 half end style space end exponent
space
space space space space space space space space space equals space open parentheses square root of straight x squared minus 1 end root close parentheses 1 over straight x

space space space space space straight v space equals space fraction numerator square root of straight x squared minus 1 end root over denominator straight x end fraction

Differentiating space both space sides space straight w. straight r. to space straight t comma space we space have

straight a space equals space dv over dt space equals space straight d over dt left square bracket straight t left parenthesis straight t squared plus 1 right parenthesis to the power of negative begin inline style bevelled 1 half end style end exponent space right square bracket space

space space space equals space straight t space straight d over dt space left parenthesis straight t squared plus 1 right parenthesis to the power of negative begin inline style bevelled 1 half end style end exponent space plus space left parenthesis straight t squared space plus space 1 right parenthesis to the power of negative begin inline style bevelled 1 half end style end exponent space straight d over dt left parenthesis straight t right parenthesis space

space space equals space straight t space cross times open parentheses negative 1 half close parentheses left parenthesis straight t squared plus 1 right parenthesis to the power of negative begin inline style bevelled 3 over 2 end style end exponent space cross times space 2 straight t space plus space left parenthesis straight t squared plus 1 right parenthesis to the power of negative begin inline style bevelled 1 half end style end exponent space

space space equals space left parenthesis straight t squared plus 1 right parenthesis to the power of negative begin inline style bevelled 3 over 2 end style end exponent space open square brackets straight t space cross times space open parentheses negative 1 half close parentheses space cross times space 2 straight t space plus space left parenthesis straight t squared plus 1 right parenthesis close square brackets space

space space equals space left parenthesis straight t squared plus 1 right parenthesis to the power of negative begin inline style bevelled 3 over 2 end style end exponent space space space left square bracket space minus straight t squared space plus space left parenthesis straight t squared space plus 1 right parenthesis right square bracket space

space space equals space left parenthesis straight t squared plus 1 right parenthesis right parenthesis to the power of negative begin inline style bevelled 3 over 2 end style end exponent space equals space 1 over straight x cubed

    Question 164
    CBSEENPH11017218

    A particle moves from rest at a distance c from fixed point O with accelerationμ/x3, where μ is constant and x is the distance away from O. What will be the velocity of the particle when it is at a distance 2c from where it started?


    Solution

    Let at any instant the particle be at a distance x from O.
    Acceleration of the particle is given by, 
          astraight v space dv over dx space equals space straight mu over straight x squared 
    rightwards double arrow space straight v. dv space equals space straight mu over straight x squared dx space 

    Integrating both sides, we get
    space space space space space space space integral subscript 0 superscript straight v dv space equals space integral subscript straight c superscript 3 straight c end superscript straight mu over straight x squared dx space

space space space space space space space right enclose straight v squared over 2 end enclose subscript 0 superscript straight v space equals space minus right enclose straight mu over straight x end enclose subscript straight c superscript 3 straight c end superscript space

rightwards double arrow straight v squared over 2 space minus space 0 space equals space minus straight mu space open square brackets fraction numerator 1 over denominator 3 straight c end fraction minus 1 over straight c close square brackets space equals space fraction numerator 2 straight mu over denominator 3 straight c end fraction space

rightwards double arrow space space space space space space space space space space space straight v space equals space square root of fraction numerator 4 straight mu over denominator 3 straight c end fraction end root 

    Question 165
    CBSEENPH11017219

    The acceleration of a particle at any instant is given by a = X + Yv, where X and Y arc constants and v is velocity. Find the velocity of the particle at any instant. Assume that the particle starts from rest.

    Solution

    Given, the particle starts from rest. 

    Acceleration, a = space space space fraction numerator dv over denominator space dt end fraction space equals space straight X space plus space Yv space 
    rightwards double arrow space space space space space space space space space space space space space space space fraction numerator dv over denominator straight D space plus space Yu end fraction space equals space dt space



    At time = 0; velocity = 0 and let v be the velocity at any instant t.
    Therefore,

    space space space space integral subscript 0 superscript straight v fraction numerator dv over denominator straight X space plus space Yv end fraction space equals space integral subscript 0 superscript straight t dt space

rightwards double arrow space 1 over straight Y integral subscript 0 superscript straight v fraction numerator straight Y. straight d over denominator straight X space plus space Yv end fraction space equals space integral subscript 0 superscript straight t dt space

rightwards double arrow 1 over straight Y space Ln space left parenthesis straight X space plus space Yv right parenthesis vertical line subscript 0 superscript straight v space equals space straight t vertical line subscript 0 superscript straight t space

rightwards double arrow space Ln space left parenthesis straight X plus Yv right parenthesis space minus space Ln left parenthesis straight X right parenthesis space equals space straight Y left parenthesis straight t minus 0 right parenthesis space

rightwards double arrow space Ln space open parentheses fraction numerator straight X plus Yv over denominator straight X end fraction close parentheses space equals space Yt space

rightwards double arrow space 1 space plus space Yv over straight X space equals space straight e to the power of straight n

rightwards double arrow space space straight v space equals space straight X over straight Y left parenthesis straight e to the power of straight n space minus space 1 right parenthesis 


    Question 166
    CBSEENPH11017220

    A man of height 1.5m walks away from a lamp hanging at a height 2.7m above ground level. If the man walks with a speed of 7.2km/kr, determine the speed of tip of man's shadow.

    Solution
    The situation is as shown in figure below.
                      
    The tip of the man's shadow is at E.
    From figure, ∆ ABE and ∆ CDE are similar triangles.
    
therefore space space space space space space space space space space space space space AE over CE equals AB over CD

rightwards double arrow space space space fraction numerator AE over denominator AE minus AC end fraction equals AB over CD

rightwards double arrow space space space space space space space space space space fraction numerator straight y over denominator straight y minus straight x end fraction equals straight H over straight h
rightwards double arrow space space space space space space space space space space space space space space space space space straight y space equals space fraction numerator straight H over denominator straight H minus straight x end fraction straight x

rightwards double arrow space space space space space space space space space space space space dy over dt equals fraction numerator straight H over denominator vertical line straight H minus straight h end fraction dx over dt equals fraction numerator 2.7 over denominator 2.7 minus 1.5 end fraction 7.2 space

space space space space space space space space space space space space space space space space space space space space straight v space equals space 16.2 space km divided by hr
    v is the speed of tip of man's shadow. 
    Question 167
    CBSEENPH11017221

    A man standing on the edge of a cliff at some height throws ball straight up with initial velocity u and then throws another ball straight down with same initial speed. Which ball will hit the ground with a greater speed?

    Solution
    Velocity with the ball is thrown upwards = u 
    Velocity with which another ball is thrown down = u
    Both the balls will hit the ground with the same velocity if air resistance is neglected. 
    Ball A:
    Initial space velocity comma space straight u subscript straight A equals straight u

Acceleration comma space straight alpha subscript straight A equals straight g

Distance space travelled comma space straight S subscript straight A equals straight h space

U s i n g space t h e space t h i r 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 straight v squared minus straight u squared equals 2 alpha S space

rightwards double arrow space space space space space space space space straight v space equals space square root of straight u squared plus 2 alpha S end root space

therefore space space space space space space space space straight v subscript straight A equals square root of straight u squared plus 2 g h end root space 
    Ball B:
    Ball B is thrown upward with velocity u, therefore the first the ball rises up, after reaching the maximum height and it starts falling.
    Let ball B rise up by x.
    Initial space velocity comma space straight u subscript straight B equals straight u

Acceleration comma space straight alpha subscript straight B equals negative straight g

Final space velocity comma space straight v space equals space 0

Now comma
space space space space space straight x space equals space fraction numerator straight v squared minus straight u subscript straight B squared over denominator 2 straight g end fraction

space space space space space space space space space equals fraction numerator 0 squared minus straight u squared over denominator negative 2 straight g end fraction

space space space space space space space space space equals fraction numerator straight u squared over denominator 2 straight g end fraction
    That is, the ball B rises by <pre>uncaught exception: <b>mkdir(): Permission denied (errno: 2) in /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/util/sys/Store.class.php at line #56mkdir(): Permission denied</b><br /><br />in file: /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/util/sys/Store.class.php line 56<br />#0 [internal function]: _hx_error_handler(2, 'mkdir(): Permis...', '/home/config_ad...', 56, Array)
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#6 {main}</pre>.
    Now ball B has to fall by a total distance of 
    straight h plus fraction numerator straight u squared over denominator 2 straight g end fraction.
     Let vbe the velocity of ball B, when it reaches the ground. 

    therefore space space space space space space straight v subscript straight B superscript 2 space equals space 2 straight g open parentheses straight h plus fraction numerator straight u squared over denominator 2 straight g end fraction close parentheses equals straight u squared plus 2 g h space

rightwards double arrow space space space space space space straight v subscript straight B equals square root of straight u squared plus 2 g h end root space

H e n c e comma space space space space space space space v subscript straight A equals straight U subscript straight a 
    Question 168
    CBSEENPH11017222

    A body is dropped from top of a cliff 240 m high. At the same time another body is thrown vertically upwards from the ground with a speed 48m/s. Find when and where the two balls meet.

    Solution
    Given a ball is dropped from the top of a cliff. 
    Initial position of ball,  X01 = 240m 
    Initial velocity of ball,  u= 0 
    Acceleration, a= g= -10 m/s
    Therefore, the position of the ball at any instant is,  
    straight x subscript 1 equals straight x subscript 01 plus straight u subscript 1 straight t plus 1 half straight a subscript 1 straight t squared equals 240 minus 5 straight t squared space space space space space space space space space space space... left parenthesis 1 right parenthesis
    When the ball is projected from the ground.
    Initial position of ball, X02 = 0 
    Initial velocity of ball, u= 48m/s 
    Acceleration, a= g= -10 m/s2  
    Therefore, the position of the ball at any instant is given by
    straight x subscript 2 equals straight x subscript 02 plus straight u subscript 2 straight t plus 1 half straight a subscript 2 straight t squared equals 48 straight t minus 5 straight t squared space space space space space space space space space space... left parenthesis 2 right parenthesis 
    When the two balls will meet, the position of both the balls will be the same. 
    i.e.,           x= x

     
    therefore space space space space space space 240 minus 5 straight t squared equals 48 straight t minus 5 straight t squared 
                         t = 5s 
    Putting t = 5 in equation (1), we have
    x=240-5t
        =240-5(5)
        =115m  
    Thus, both the balls meat at t = 5s at a height of 115m from the ground. 
    Question 169
    CBSEENPH11017223

    A body falls freely from rest and covers half of the total height in last second. Find the height from where the body was dropped.

    Solution
    Let a body fall from height 2h and take t seconds to fall.
    therefore space space space space space space space space space space space 2 straight h equals 1 half gt 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... left parenthesis 1 right parenthesis 
    Since the body covers half of total height i.e. h in t seconds,
    therefore space space space space space straight h equals 1 half straight g left parenthesis 2 straight t minus 1 right parenthesis space space space space space space space space space space space space space space space space space space space space space space space space space space space... left parenthesis 2 right parenthesis 
    From (1) and (2), 
              t= 2(2t - 1) 
    rightwards double arrow      t-4t+2=0
     straight t space equals space fraction numerator 4 plus-or-minus square root of 16 minus 8 end root over denominator 2 end fraction

space equals fraction numerator 4 plus-or-minus 2 square root of 2 over denominator 2 end fraction space

space equals space 2 space plus-or-minus space square root of 2 

    Since 2 minus square root of 2 less than 1 comma space therefore  not possible.
    Thus, time taken by the body to reach the ground is open parentheses 2 plus square root of 2 close parentheses s. 
    From equation (2), the height from which the body is dropped is given by, 
    straight h space equals 1 half straight g left parenthesis 2 straight t minus 1 right parenthesis space

space space equals 1 half straight x 10 straight x open square brackets 2 open parentheses 2 plus square root of 2 close parentheses minus 1 close square brackets space

space space equals 5 open square brackets 4 plus 2 square root of 2 minus 1 close square brackets

space space equals 5 open square brackets 3 plus 2 square root of 2 close square brackets 
       = 58.28m
    Question 170
    CBSEENPH11017224

    A body is dropped from height h (top of cliff) and at the same moment another particle is projected upwards from the ground. Both the bodies meet at a height h/n. Find the intial velocity of projection and show that, the velocities of two bodies when they meet are in the ratio (n-2):2(-1). 



    Solution

    Ball A: 
    Initial position of ball, x0A = h 
    Initial velocity of ball, u= 0 
    Acceleration, a= -g 
    Therefore the position of the ball at any instant is, 
    space space space space space space straight x subscript straight A space equals space straight x subscript 0 straight A end subscript plus straight u subscript straight A straight t plus 1 half straight a subscript straight A straight t to the power of 2 space end exponent
space space space space space space space space space space space equals straight h minus 1 half gt squared space space space space space space space space space space space space space space space space space space space space space space... left parenthesis 1 right parenthesis
    Ball B: 
    Initial position of ball, xOB = 0
    Initial velocity of ball, uB = 0
    Acceleration, aB = -g
    Therefore, the position of the ball at any instant is, 
    xB = xOB + uBt + 1 half space straight a subscript straight B straight t squared space equals space ut space minus space 1 half gt squared       ...(3)
    When the balls meet, the position of both the balls is same.
                     XA = XB 
           h - 1 half gt squared space equals space ut space minus space 1 half gt squared           ... (2) 
    rightwards double arrow space space space space space straight t space equals space straight h over straight u 
    Position at which the two balls meet, 
    x = xA = xB = h - 1 half g space open parentheses h over u close parentheses squared 
    But, x = straight h over straight n space space space space space space space space space space space space space space space left square bracket space g i v e n right square bracket
    Therefore, 

    space space space space space straight x space equals space straight x subscript straight A equals space straight x subscript straight B equals space straight h minus 1 half straight g open parentheses straight h over straight u close parentheses squared space
But comma space space space space space space space space space space space space space space space space straight x space equals space straight h over straight n space space space space space space space space space space space space space space space space space space space space space space space space space space space space space open square brackets given close square brackets space

therefore space space space space space space space space space space space space space space space space space straight h over straight n equals straight h minus 1 half straight g open parentheses straight h over straight u close parentheses squared space

rightwards double arrow space space space space space space space space space space space space space space space space space space straight u space equals space square root of open parentheses fraction numerator gmh over denominator 2 left parenthesis straight n minus 1 right parenthesis end fraction close parentheses end root 
    Velocity of balls, when they meet i.e., at t = h/u
    space space space space space space space space space space space space space space straight nu subscript straight A equals straight u subscript straight A plus straight alpha subscript straight A straight t equals negative gh over straight u
and comma space space space space space space straight nu subscript straight B equals straight u subscript straight B plus straight alpha subscript straight B straight t space equals space straight u minus gh over straight u
 
    therefore space space space space space space space space straight nu subscript straight B over straight nu subscript straight A space equals space fraction numerator straight u minus begin display style gh over straight u end style over denominator negative begin display style gh over straight u end style end fraction space

space space space space space space space space space space space space space space space space space space space space equals negative straight u squared over gh plus 1 space

space space space space space space space space space space space space space space space space space space space space equals space minus space fraction numerator gmh over denominator 2 left parenthesis straight n minus 1 right parenthesis end fraction space 1 over gh plus 1 space

space space space space space space space space space space space space space space space space space space space equals space fraction numerator negative straight n plus 2 straight n minus 2 over denominator 2 left parenthesis straight n minus 1 right parenthesis end fraction

space space space space space space space space space space space space space space space space space space space equals space fraction numerator straight n minus 2 over denominator 2 left parenthesis straight n minus 1 right parenthesis end fraction 

    Question 171
    CBSEENPH11017225

    A body is released from rest at t = 0 from a height h from the ground. The body will be at

    x from where it was dropped at two times i.e.  space space space space space space space space space plus square root of fraction numerator 2 straight x over denominator straight g end fraction end root space and space minus fraction numerator square root of 2 straight x end root over denominator straight g end fraction.

    What is the significance of negative value of time?

                       




    Solution
    Using the second equation of motion, 
     straight x space equals ut plus 1 half gt squared
space space space equals space 1 half gt squared space space space space space space space space space space space space space space space space space space space space space open square brackets therefore space straight u equals 0 close square brackets

therefore space space space space space straight t space equals space plus-or-minus square root of fraction numerator 2 straight x over denominator straight g end fraction end root
    't' is positive implies that body will be at  x after time  't' when the particle is dropped. 
    If the particle was projected in upward direction from ground, then the particle would  be at x distance below from the highest point.
    Question 172
    CBSEENPH11017226

    A body is projected vertically upwards. There are two instants when the body will be at same height. Show that the ratio of the two instants when the body will be at height three fourth of its greatest height is 3.

    Solution
    Let a body be projected with velocity u in vertically upward direction from ground and attain the maximum height H.
    Let at any instant t, it is at a height h.
    Thus,
    space space space space space space space space space space space space straight h space equals space ut minus 1 half gt squared space space space space space space left square bracket Using space the space second space equation space of space motion right square bracket
rightwards double arrow space space space space space space space space space straight t space equals space fraction numerator straight u plus-or-minus square root of straight u squared minus 2 gh end root over denominator straight g 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... left parenthesis 1 right parenthesis
Also comma space height space of space the space object space is space given space by comma space

space space space space space space space space space space space space space space space space space space space space straight H space equals space fraction numerator straight u squared over denominator 2 straight g 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... left parenthesis 2 right parenthesis
If comma space space space space space space space space space space space space space space space space straight h space space equals space 3 over 4 straight H space equals space fraction numerator 3 straight u squared over denominator 8 straight g end fraction comma space then space

space space space space space space space space space space space space space space space space space space space space space straight t space equals space fraction numerator straight u over denominator 2 straight g end fraction comma space fraction numerator 3 straight u over denominator 2 straight g end fraction

So comma space the space ratio space of space two space instants space is space given space by comma space

space space space space space space space space space space space space space space space space space straight t subscript 2 over straight t subscript 1 equals fraction numerator 3 straight u divided by 2 straight g over denominator straight u divided by 2 straight g end fraction equals 3
    Question 173
    CBSEENPH11017227

    A particle is projected vertically upwards from ground. At time it is at A at a height h from ground and still going upwards. From A, it takes further t2 time to reach back to ground again. Show that

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    Solution
    Let,
    Velocity with which the particle is projected from the ground be 'u'. 
    The particle takes time t1 to reach A and time t2 to reach the ground. 

    Thus, total time of flight of particle is t1 + t2.
    Also the total time of fight =  fraction numerator 2 straight u over denominator straight g end fraction
    On comparing the two, we get
    space space space space space space space space straight t subscript 1 plus straight t subscript 2 equals fraction numerator 2 straight u over denominator straight g end fraction
rightwards double arrow space space space space space space space space space straight u space equals space 1 half straight g left parenthesis straight t subscript 1 plus straight t subscript 2 right parenthesis
 
    The particle is at a height h from ground at time t1, which is given by

    therefore space space space space space space space straight h space equals space ut subscript 1 minus 1 half gt subscript 1 squared
space space space space space space space space space space space space space space space equals space 1 half straight g left parenthesis straight t subscript 1 plus straight t subscript 2 right parenthesis straight t subscript 1 minus 1 half gt subscript 1 squared
space space space space space space space space space space space space space space space equals 1 half gt subscript 1 straight t subscript 2
    (ii) Maximum height attained is,
    
straight H space equals space fraction numerator straight u squared over denominator 2 straight g end fraction equals fraction numerator open square brackets begin display style 1 half straight g left parenthesis straight t subscript 1 plus straight t subscript 2 end style close square brackets over denominator 2 straight g end fraction

space space space space equals space 1 over 8 straight g left parenthesis straight t subscript 1 plus straight t subscript 2 right parenthesis squared
    Question 174
    CBSEENPH11017366

    What is Geocentric theory?

    Solution
    According to the geocentric theory, all the astronomical bodies like the moon, the sun and stars revolve around the earth, and the earth is at the centre of the universe. 
    Question 175
    CBSEENPH11017367

    What is Heliocentric theory?

    Solution
    According to the Heliocentric theory, the sun is at the centre and various planets revolve around the sun at their axis. 
    Question 176
    CBSEENPH11017368

    The position co-ordinates of two particles of masses m1 and m2are (x1, y1, z1) and (x2, y2, z2) respectively. Find the coordinates of the centre of mass.

    Solution
    The position vectors of masses m1 and m2 are respectively,
    space space space space space space space space space space space space r with rightwards harpoon with barb upwards on top subscript 1 equals space straight x subscript 1 straight i with overbrace on top plus straight y subscript 1 straight j with overbrace on top plus straight z subscript 1 straight k with overbrace on top

a n d space space space space space r with rightwards harpoon with barb upwards on top subscript 2 space equals space straight x subscript 2 straight i with overbrace on top plus straight y subscript 2 straight j with overbrace on top plus straight z subscript 2 straight k with overbrace on top
    Let the position coordinates of the centre of mass be (X, Y, Z).
    Therefore the position vector of centre of mass is,
    R with rightwards harpoon with barb upwards on top equals straight X straight i with overbrace on top plus straight Y straight j with overbrace on top plus straight Z straight k with overbrace on top
    Since,    R with rightwards harpoon with barb upwards on top equals fraction numerator straight m subscript 1 stack straight r subscript 1 with rightwards harpoon with barb upwards on top space plus straight m subscript 2 space straight r with rightwards harpoon with barb upwards on top subscript 2 over denominator straight m subscript 1 plus straight m subscript 2 end fraction space
    therefore space space space straight X straight i with overbrace on top plus straight Y straight j with overbrace on top plus straight Z straight k with overbrace on top 
    equals space space fraction numerator straight m subscript 1 left parenthesis straight x subscript 1 straight i with overbrace on top plus straight y subscript 1 straight j with overbrace on top plus straight z subscript 1 straight k with overbrace on top right parenthesis plus straight m subscript 2 left parenthesis straight x subscript 2 straight i with overbrace on top plus straight y subscript 2 straight j with overbrace on top plus straight z subscript 2 straight k with overbrace on top right parenthesis over denominator straight m subscript 1 plus straight m subscript 2 end fraction
    space equals space space fraction numerator open parentheses straight m subscript 1 straight x subscript 1 plus straight m subscript 2 straight x subscript 2 close parentheses over denominator straight m subscript 1 plus straight m subscript 2 end fraction straight i with overbrace on top plus fraction numerator left parenthesis straight m subscript 1 straight y subscript 1 plus straight m subscript 2 straight y subscript 2 over denominator straight m subscript 1 plus straight m subscript 2 end fraction straight j with overbrace on topplus fraction numerator left parenthesis straight m subscript 1 straight z subscript 1 plus straight m subscript 2 straight z subscript 2 over denominator straight m subscript 1 plus straight m subscript 2 end fraction space straight k with overbrace on top
    Comparing the coefficients of  straight i with overbrace on top comma space straight j with overbrace on top space and space straight k with overbrace on top, we get
    straight X equals fraction numerator left parenthesis straight m subscript 1 straight x subscript 1 plus straight m subscript 2 straight x subscript 2 right parenthesis over denominator straight m subscript 1 space plus space straight m subscript 2 end fraction
straight Y space equals space straight X equals fraction numerator left parenthesis straight m subscript 1 straight y subscript 1 plus straight m subscript 2 straight y subscript 2 right parenthesis over denominator straight m subscript 1 space plus space straight m subscript 2 end fraction
straight z equals fraction numerator left parenthesis straight m subscript 1 straight z subscript 1 plus straight m subscript 2 straight z subscript 2 right parenthesis over denominator straight m subscript 1 space plus space straight m subscript 2 end fraction 

    Question 177
    CBSEENPH11017369

    Is Geodesic always a straight line?

    Solution
    No, Geodesic is a straight line if and only if,  the two points lie on the flat surface. If the two points lie on the curved surface then it is a curved line.
    Question 178
    CBSEENPH11017370

    What is the celestial sphere?

    Solution
    At night, if we see the planets and the stars in the sky, all appear to lie in the hemisphere (rest of the hemisphere we are unable to see because of being on the other side of the earth). This sphere is called the celestial sphere.
    Question 179
    CBSEENPH11017371

    What is meant by gravitation?

    Solution
    Gravitation is the force of attraction between any two bodies in the universe. 
    Question 180
    CBSEENPH11017372

    How does the force of gravitation depend on the masses of two interacting bodies?

    Solution
    The force of gravitation between two masses is directly proportional to the product of their masses.
    Question 181
    CBSEENPH11017373

    Does gravitational force follow Newton’s third law of motion?

    Solution
    Yes, gravitational force follows Newton’s third law of motion. The gravitational force due to the first body on the second body is equal and opposite to the gravitational force due to second body on the first body.
    Question 182
    CBSEENPH11017374

    Is gravitational force central one?

    Solution
    Gravitational force is a central force because it acts along the line joining the two masses.
    Question 183
    CBSEENPH11017375

    Is the gravitational constant a universal constant?

    Solution
    Yes, gravitational constant is a universal constant.
    The value of gravitational constant is 6.674 x 10-11 m3 kg-1 s-2
    Question 184
    CBSEENPH11017376

    Is gravitational constant different for different galaxies?

    Solution
    No, gravitational constant is a universal constant.
    Question 185
    CBSEENPH11017377

    Why is G known as universal gravitational constant?

    Solution
    'G' is known as universal gravitational constant because its value is same everywhere in the universe. 
    Question 186
    CBSEENPH11017379

    Does the value of gravitational constant depend on the system of unit used?

    Solution
    Yes, the values of gravitational constant in SI and CGS are different. 
    In S.I. unit - 
    straight G space equals space 6.67408 space cross times space 10 to the power of negative 11 end exponent space straight m cubed space kg to the power of negative 1 end exponent space straight s to the power of negative 2 end exponent
    In cgs unit, 
    straight G space equals space 6.674 space cross times space 10 to the power of negative 8 end exponent space cm cubed space straight g to the power of negative 1 end exponent space straight s to the power of negative 2 end exponent
    Question 187
    CBSEENPH11017380

    If G is 6.67 x 10-8 dyne cm2/gmon the surface of the moon, then what is the value of G in SI at the centre of the earth?

    Solution
    G is a universal constant. It has the same value everywhere. Therefore, gravitational constant in SI unit at the centre of the earth is also 6.67 x10-11 Nm2/kg2
    Question 188
    CBSEENPH11017381

    What is dimensional formula of gravitational constant?

    Solution
    Dimensional formula of gravitational constant is [M-1L3T-2]. 
    Question 189
    CBSEENPH11017382

    How many dynes are there in one newton?

    Solution
    1 N = 105 dyne.
    Question 190
    CBSEENPH11017383

    Are the inertial and gravitational masses of body equivalent to each other?

    Solution
    Yes. Inertial mass is equal to the gravitational mass. 
    Question 191
    CBSEENPH11017384
    Question 192
    CBSEENPH11017385

    Do the force of friction or other contact forces arise due to gravitational force?

    Solution
    No, both force of friction and contact forces arises because of electric forces.
    Question 193
    CBSEENPH11017387

    Which is greater, the attraction of the earth for 1 kg of iron or attraction of 1 kg of iron for the earth?

    Solution
    Both the attractions are equal in magnitude because mass of the earth is the same for both attractions. 
    Question 194
    CBSEENPH11017388

    What is the sense of rotation of celestial sphere?

    Solution
    The celestial sphere appears to rotate from east to west in the clockwise direction.
    Question 196
    CBSEENPH11017391

    What is gravity?

    Solution
    Gravity is the force between the earth and a body. The attraction of a body towards the earth is known as gravity. 
    Question 197
    CBSEENPH11017392

    What is difference between gravitation and gravity?

    Solution
    Gravitation is the force between any two bodies in the universe.
    Gravity is a special case of gravitation in which one body is the earth. 
    Question 198
    CBSEENPH11017393

    What is acceleration due to gravity?

    Solution
    The acceleration possessed by a body due to the force of gravity is called acceleration due to gravity. 
    Question 199
    CBSEENPH11017394

    Write an expression for acceleration due to gravity at the surface of earth.

    Solution

    Acceleration due to gravity = GM over straight R squared comma
    where M is the mass of the earth and R is the radius of the Earth. 

    Question 200
    CBSEENPH11017395

    What is the value of acceleration due to gravity at the surface of the earth?

    Solution
    The value of acceleration due to gravity at the surface of the Earth = 9.8 m/s2
    Question 201
    CBSEENPH11017396

    What is the value of acceleration due to gravity at the surface of the earth in FPS system?

    Solution
    The value of acceleration due to gravity at the surface of the earth in FPS system = 32 ft/s2.
    Question 202
    CBSEENPH11017397

    Who weighed the earth first?

    Solution
    Henry Cavendish is the first scientist who weighed the earth first. 
    Question 203
    CBSEENPH11017399

    What is the mass of the earth?

    Solution

    Mass of the earth = 6 cross times 10 to the power of 24 kg. 

    Question 204
    CBSEENPH11017400

    What is the average density of the earth?

    Solution
    The average density of the earth is 5497 kg/m3.
    Question 205
    CBSEENPH11017401

    What is the approximate mass of the sun?

    Solution
    The approximate mass of the sun = 2 x 1030 kg.
    Question 206
    CBSEENPH11017402

    Is the acceleration due to gravity same everywhere? 

    Solution
    No, the acceleration due to gravity varies from place to place.
    Acceleration due to gravity at the centre of the earth is zero and 'g' at the poles is the strongest. 
    Question 207
    CBSEENPH11017403

    How does the acceleration due to gravity change with height and depth from the surface of the earth?

    Solution
    Acceleration due to gravity decreases both with height and depth. 
    The value of acceleration due to gravity decreases with increase in height above the surface of the earth. 
    Question 208
    CBSEENPH11017405

    What is acceleration due to gravity at the centre of the earth?

    Solution
    Acceleration due to gravity at the centre of the earth is zero. 
    Question 209
    CBSEENPH11017406

    What is the value of acceleration due to gravity at midway between surface of the earth and its centre?

    Solution
    Acceleration due to gravity at midway between the surface of the earth and it's centre is half of 9.8 m/s2 which is equal to 4.9 m/s2. 
    Question 210
    CBSEENPH11017407

    What is the effect of rotation of the earth on the value of g?

    Solution

    Earth rotates about its own axis as a result of which the objects on the earth experience the centrifugal force. 
    This centrifugal force decreases the acceleration due to gravity and magnitude of change in the value of g due to rotation of the earth depends on the value of the angle of latitude. 

    Question 211
    CBSEENPH11017408

    Write an expression for the acceleration due to gravity near the surface of the earth.

    Solution

    Acceleration due to gravity near the surface of the earth is given by, 
                         straight g subscript straight h equals straight g left parenthesis 1 minus fraction numerator 2 straight h over denominator straight R end fraction right parenthesis 
    where, 
    h is the distance of the place from the surface of the earth. 

    Question 212
    CBSEENPH11017409

    Is the acceleration due to gravity same everywhere on the surface of the earth?

    Solution
    No, the acceleration due to gravity at a point on the earth is dependent on the angle of latitude on the earth.
    Question 213
    CBSEENPH11017410

    How does acceleration due to gravity depend on the distance from the center of the earth inside the earth? 

    Solution
    Acceleration due to gravity is directly proportional to the distance from the centre of the earth inside the earth. 
    Question 214
    CBSEENPH11017411

    What is acceleration due to gravity at the surface of the earth at angle of latitude λ?

    Solution
    Acceleration due to gravity at the surface of the earth at angle of latitude λ, gλ = g - Rω2 cos2 λ
    Question 215
    CBSEENPH11017412

    Where is the effect of rotation at the surface of earth maximum?

    Solution

    The effect of rotation is maximum at the equator of the Earth. 

    Question 216
    CBSEENPH11017413

    At what place of the earth, the value of g does not change due to its rotation motion?

    Solution
    At poles, there is no effect on 'g' due to rotation of poles, attributing to the minimum rotation caused at poles. 
    Question 217
    CBSEENPH11017414

    Where does a body weigh more-at the surface of the earth or in a mine?

    Solution
    At the surface of the earth a body weighs more, because as we go into mine the acceleration due to gravity decreases.
    Question 218
    CBSEENPH11017415

    At what point acceleration due to gravity is the largest? (Suppose the earth is a perfect sphere.)

    Solution
    Acceleration due to gravity is maximum at the poles.
    Question 219
    CBSEENPH11017416

    At what point in the universe the mass and weight of the body will be simultaneously zero?

    Solution
    The mass and weight of the body is simultaneously zero nowhere at any point in the universe. 
    Question 220
    CBSEENPH11017417

    If the earth shrinks without changing the mass, what would happen to the value of ‘g’ on the surface of the earth?

    Solution
    The value of 'g' will increase because g is inversely proportional to R2.
    Question 221
    CBSEENPH11017418

    Astronauts feel weightlessness. Why?

    Solution
    An astronaut while orbiting around the planet is in the state of free fall. And, any body in a state of free fall is in the state of weightlessness. 
    Question 222
    CBSEENPH11017419

    Does the acceleration of freely falling body depend upon the mass of body?

    Solution
    No, acceleration does not depend on the mass of the body for a freely falling body. 
    Question 223
    CBSEENPH11017420

    An astronaut inside a small spaceship orbiting around the earth cannot detect gravity. If the space station orbiting around the earth has a large size, can he hope to detect gravity?

    Solution

    Yes, if the size of the space station is large enough, then the astronaut will detect the change in Earth’s gravity (g).

    Question 224
    CBSEENPH11017421

    If a body weighs 100N at the surface of the earth, how much will it weigh at the centre of the earth?

    Solution
    As the body is moved closer to the center of the earth, the mass of the earth between the center of the earth and the body keeps decreasing.
    This causes the force acting from the center of the earth on the body to decrease.
    Hence, the weight of the body at the centre of the earth = zero.
    Question 225
    CBSEENPH11017422

    In the condition of weightless, should the gravity be necessarily zero?

    Solution
    No, a free falling body near the surface of earth weighs zero. 
    Question 226
    CBSEENPH11017423

    A body is lifted from centre of the earth to a point situated at infinite distance away. Where will it possess maximum weight? 

    Solution
    The acceleration due to gravity is maximum at the surface of the earth, therefore the body possesses the maximum weight at the surface of the earth.
    Question 231
    CBSEENPH11017428

    Compare acceleration due to gravity on the surface of the moon with that at the surface of the earth.

    Solution
    The acceleration due to gravity at the surface of the moon is one-sixth the acceleration due to gravity at the surface of earth. 
    Question 232
    CBSEENPH11017429

    A man can jump to a height 1.2m at the surface of the earth. How high he can jump on the surface of the moon?

    Solution
    Acceleration due to gravity at the surface of the moon is g/6.
    That is, for the same energy expended, the person can jump 6 times higher on the moon than on the earth. 
    Therefore, the height to which the person can jump on the moon = 6 x 1.2 = 7.2 m 
    Question 233
    CBSEENPH11017430

    Where does a body weigh more - at the surface of the earth or at the surface of moon? 

    Solution
    The body weighs more on the surface of the earth because acceleration due to gravity at the surface of the earth is greater than that on the surface of the moon. 
     
    Question 234
    CBSEENPH11017431

    What does the inertial mass measure?

    Solution
    Inertia
    Question 235
    CBSEENPH11017435

    State Newton’s law of gravitation.

    Solution

    Law of gravitation states that every body in this universe attracts the other body with a force which is directly proportional to the product of their masses and inversely proportional to the square of the distance between them. 
                 

    i.e.,
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#6 {main}</pre>
                        proportional to 1 over straight r squared
    ∴               straight F proportional to fraction numerator straight m subscript 1 straight m subscript 2 over denominator straight r squared end fraction
    rightwards double arrow            straight F equals straight G fraction numerator straight m subscript 1 straight m subscript 2 over denominator straight r squared end fraction 

    where G is constant of proportionality known as gravitational constant.

    Question 236
    CBSEENPH11017436

    Define gravitational constant.

    Solution

    Force of gravitation, straight F equals straight G fraction numerator straight m subscript 1 straight m subscript 2 over denominator straight r squared end fraction 
    Let,
    Mass, <pre>uncaught exception: <b>mkdir(): Permission denied (errno: 2) in /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/util/sys/Store.class.php at line #56mkdir(): Permission denied</b><br /><br />in file: /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/util/sys/Store.class.php line 56<br />#0 [internal function]: _hx_error_handler(2, 'mkdir(): Permis...', '/home/config_ad...', 56, Array)
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#6 {main}</pre> 
    and               straight r equals 1 straight m 

    Then,          straight F equals straight G fraction numerator 1 cross times 1 over denominator left parenthesis 1 right parenthesis squared end fraction equals straight G 
    Therefore, gravitational constant is equal to the force between two bodies, each of mass 1 kg when their centres of mass are separated 1 metre apart.
    Question 237
    CBSEENPH11017442

    What are the units and dimensions of G?

    Solution
    According to Newton’s law of gravitation, 
                       straight F equals straight G fraction numerator straight M subscript 1 straight M subscript 2 over denominator straight r squared end fraction 
    rightwards double arrow            space space space space space straight G equals fraction numerator Fr squared over denominator straight M subscript 1 straight M subscript 2 end fraction 
    Dimensional formula of G is, 
     straight G equals fraction numerator left square bracket MLT to the power of negative 2 end exponent right square bracket left square bracket straight L squared right square bracket over denominator left square bracket straight M right square bracket left square bracket straight M right square bracket end fraction equals left square bracket straight M to the power of negative 1 end exponent straight L cubed straight T to the power of negative 2 end exponent right square bracket 
    SI units of G is Nm2/kg2 or m3/kg/sec2 
    CGS units of G is dyne cm2/gm2 or cm3/gm/sec2.
    Question 238
    CBSEENPH11017443

    According to Newton’s law of gravitation everybody in this universe attracts every other body. But we do not see bodies on the surface of the earth moving towards one another on account of this interaction. Why?

    Solution
    According to Newton’s law of gravitation, 
                       
    The value of G = 6.67x10-11 Nm2/kg2.
    The masses of two ordinary bodies on the surface of the earth is small i.e. of the order of 102 kg.
    Therefore, the force of attraction between two bodies, each of mass 100 kg separated by 1m, is of the order of 10-6N which is too small to overcome the force of friction between the body and the surface.
    That is why we do not see the bodies on the surface of the earth moving towards each other.
    Question 239
    CBSEENPH11017444

    When the object falls on the ground, the earth moves up. Why the earth’s motion is not noticeable?

                                        OR 

    If the force of gravity between the earth and the body is same and mutual, then does the body fall towards the earth and the earth does not move towards body?

    Solution
    When a body falls on the ground, both the body and the earth move towards each other. But the ratio of distance travelled by each is inversely proportional to the mass.
    i.e., space straight s over straight S equals straight M over straight m 
    where s, S are distances travelled by the body and the earth. 
    Since,               M>>>m 
    ∴                       s>>>S 
    Therefore motion of the earth is not noticeable.
    Question 240
    CBSEENPH11017445

    Any two bodies on the surface of the earth should move towards each other due to force of attraction acting between them but they do not. Why?

    Solution
    The masses of the bodies are very small as compared to the mass of the earth. 
    Hence, the force of attraction between the two bodies is very small as compared to the force of friction between the bodies and the surface of the earth.
    Hence, they do not move towards each other. 
    Question 241
    CBSEENPH11017446

    We cannot move even a finger without disturbing all the stars. Explain.

    Solution
    According to Newton's law of gravitation, the force of attraction between any two bodies in this universe is inversely proportional to the square of the distance between them.
    Therefore, when the finger is moved, the distance between finger and stars changes. Hence, the force of attraction between finger and stars changes. This disturbs all the stars. 
    Question 242
    CBSEENPH11017447

    Among the known types of forces in nature, the gravitational force is the weakest. Why then does it play a dominant role for the motion of bodies on the terrestrial, astronomical and cosmological scale?

    Solution
    Nuclear forces are short-range forces and dominate only over the range of distance of the order of 10-14 m to 10-15 m.
    Therefore, the motion for bodies in the terrestrial, astronomical and cosmological scale is not affected by nuclear forces.
    Electric forces are both attractive as well as repulsive unlike gravitational force, and zero if the bodies are neutral. Therefore, for massive neutral bodies gravitational force plays an important role.
    Question 243
    CBSEENPH11017449

    If you compare the gravitational force on the earth due to the sun to that due to the moon, you would find that the Sun’s pull is greater than the moon’s pull. However, the tidal effect of the moon’s pull is greater than the tidal effect of the sun. Why?

    Solution
    Tidal effect is inversely proportional to the cube of the distance while gravitational force is inversely proportional to the square of the distance. 
    The distance between moon and Earth is less as compared to the distance between Sun and Earth. Greater the distance, less is the pull of the tidal effect.
    Question 244
    CBSEENPH11017450

    What is mass and weight?

    Solution
    Mass is the amount of matter present in the body and weight is the force with which the earth pulls the mass.
    Mass of a body is same at every point and is a scalar quantity.
    The weight of body is different at different places and is a vector quantity. 
    Question 245
    CBSEENPH11017451

    Find the gravitational force between two masses each of 1gm separated by a distance 1cm.

    Solution

    Gravitational Force,space space straight F space equals space fraction numerator Gm subscript 1 straight m subscript 2 over denominator straight r squared end fraction
    Here, we have
    Mass of the body, straight m subscript 1 equals straight m subscript 2 equals 1 gm comma
    Radius, r = 1 cm
    ∴ Gravitational force,
         space space space straight F equals fraction numerator 6.67 cross times 10 to the power of negative 8 end exponent cross times 1 cross times 1 over denominator left parenthesis 1 right parenthesis squared end fraction equals 6.67 cross times 10 to the power of negative 8 end exponent dyne

    Question 246
    CBSEENPH11017452

    A rocket is fired from the earth towards the sun. At what distance from the earth’s
    centre is the gravitational force on the rocket zero ? Mass of the sun = 2× 1030 kg,
    mass of the earth = 6× 1024 kg. Neglect the effect of other planets etc. (orbital radius
    = 1.5 × 1011 m).

    Solution

    Given,
    Mass of the sun, M = 2× 1030 kg
    Mass of the earth, m = 6.0 × 1024 kg 
    Distance from the sun and the earth is, 
                      straight r equals 1.5 cross times 10 to the power of 11 straight m 
            
    Let the force on the rocket be zero at a distance x from the earth. 
    ∴              space space fraction numerator GMm subscript 0 over denominator left parenthesis straight r minus straight x right parenthesis squared end fraction equals Gmm subscript 0 over straight x squared 
    where, straight m subscript 0 is a mass of the rocket. 
    Therefore,
        fraction numerator straight M over denominator left parenthesis straight r minus straight x right parenthesis squared end fraction equals straight m over straight x squared 

    rightwards double arrow   
    rightwards double arrow        straight r over straight x equals 1 plus square root of straight M over straight m end root equals 1 plus square root of fraction numerator 2 cross times 10 to the power of 30 over denominator 60 cross times 10 to the power of 24 end fraction end root 
                     = 578.35 
    Therefore,
    Gravitational force is zero on the rocket at a distance of, 
    space space straight x equals fraction numerator straight r over denominator 587.35 end fraction equals fraction numerator 1.5 cross times 10 to the power of 11 over denominator 587.35 end fraction straight m  
                      space space equals 2.6 cross times 10 to the power of 8 straight m, from the Earth's surface.           

                           

    Question 247
    CBSEENPH11017454

    Find the gravitational force between two neutrons whose centres are 3 x 10-14m apart. Mass of neutron is 1.67 x 10-27kg.

    Solution

    Given,
    Mass of neutron, straight m equals 1.67 cross times 10 to the power of negative 27 end exponent kg 
    Distance between the centres of two neutrons,straight r equals 3 cross times 10 to the power of negative 14 end exponent straight m

    The gravitational force between two neutrons is, 
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#6 {main}</pre> 
                   = 2.07 cross times 10 to the power of negative 37 end exponent straight N 

    Question 248
    CBSEENPH11017455

    Two bodies of masses 200kg and 500kg are placed at a distance 40cm apart in free space. Find the initial acceleration of the masses when released.

    Solution
    Gravitational force between two masses is,
                         space space straight F equals fraction numerator Gm subscript 1 straight m subscript 2 over denominator straight r squared end fraction 
    Here, we have
    straight m subscript 1 equals 300 kg comma     straight m subscript 2 equals 800 space kg comma    space space space straight r equals 0.4 space straight m 
    So, gravitational force is,
             space space space space straight F equals fraction numerator 6.67 cross times 10 to the power of negative 11 end exponent cross times 300 cross times 800 over denominator left parenthesis 0.4 right parenthesis squared end fraction  
                  equals 1.0 cross times 10 to the power of negative 4 end exponent straight N 
    Initial acceleration of mass straight m subscript 1 is, 
                   straight a subscript 1 equals straight F over straight m subscript 1 equals fraction numerator 1.0 cross times 10 to the power of negative 4 end exponent over denominator 300 end fraction 
                      equals 3.33 cross times 10 to the power of negative 7 end exponent straight m divided by straight s squared 
     
    Initial acceleration of mass straight m subscript 2 is, 
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#6 {main}</pre> 
                      equals 2 cross times 10 to the power of negative 7 end exponent straight m divided by straight s squared
    Question 249
    CBSEENPH11017458

    Define inertial mass and gravitational mass.

    Solution

    Inertial mass: It is the ratio of the force applied on a body to acceleration produced in it.
    Gravitational mass: It is the ratio of the gravitational force acting on a body to the acceleration due to gravity at that point. 

    Question 250
    CBSEENPH11017460

    What are the characteristics of inertial mass?

    Solution

    Characteristics of inertial mass are:
    (i) Inertial mass is proportional to the quantity of matter contained in a body.
    (ii) It is independent of the shape of the body.
    (iii) It is additive. 
    (iv) It is a scalar quantity. 
    (v) It is independent of the temperature of the body. 
    (vi) It is independent of the state of the body.
    (vii) When two inertial masses combine physically or chemically, total inertial mass remains conserved.

    Question 251
    CBSEENPH11017461

    What is meant by acceleration due to gravity?

    Derive an expression for it in terms of mass of the earth and gravitation constant.

    Solution

    According to Newton’s law of gravitation, every mass exerts a gravitational force on every other mass in the Universe.
    Thus, the earth exerts a force on every mass in its surroundings. This force of attraction exerted by the earth on a body is called the force of gravity and the acceleration possessed by a body moving under the force of gravity is called acceleration due to gravity. 
    The value of acceleration due to gravity does not depend on the mass of a body but it depends on the mass of the earth and distance of the body from the center of the earth.
    Consider, the earth to be a uniform sphere of mass M and radius R.
    Let a body be placed at a distance r from the surface of the earth.
    The gravitational on the body is, 
                    straight F equals straight G Mm over straight r squared 
    The acceleration produced in the body due to force of gravity i.e. acceleration due to gravity is, 
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#6 {main}</pre> 
    At the surface of the earth, the acceleration due to gravity is, 
                          straight g equals GM over straight R squared

    Question 252
    CBSEENPH11017463

    Cavendish is said to have been the first person to weigh the earth. Comment.

    Solution

    Acceleration due to gravity is given by,
                           straight g equals GM over straight R squared
    rightwards double arrow                  straight M equals gR squared over straight G 
    Since Cavendish was the first scientist who measured the value of G, hence he is said to have been the first person to weigh the earth. 

    Question 253
    CBSEENPH11017464

    Find the acceleration due to gravity at the surface of the earth in terms of density of the earth.

    Solution
    Acceleration due to gravity, g at the surface of the Earth is given by,
                             straight g equals GM over straight R squared 
    Let straight rho be the density of body. 
    The mass of the earth is, 
                            straight M equals fraction numerator 4 straight pi over denominator 3 end fraction straight R cubed straight rho
    Therefore acceleration due to gravity in terms of density of the earth is, 
    straight g equals fraction numerator straight G begin display style fraction numerator 4 straight pi over denominator 3 end fraction end style straight R cubed straight rho over denominator straight R squared end fraction equals fraction numerator 4 straight pi over denominator 3 end fraction ρRG 
    Question 254
    CBSEENPH11017466

    The acceleration due to gravity at the surface of the earth is 9.8 m/s2. Find the density of the earth. (Take R = 6400km and G = 6.67 x 10-11 Nm2/kg2 ).

    Solution
    Acceleration due to gravity at the surface of the earth is,
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#6 {main}</pre>           
    ∴                 straight rho equals fraction numerator 3 straight g over denominator 4 πGR end fraction                   ... (1)
    Here, we have
    straight g equals 9.8 space straight m divided by straight s squared space space space space space space space space space space space straight R space equals space 6400 space km 
    straight G equals 6.67 cross times 10 to the power of negative 11 end exponent Nm squared divided by kg squared 
    So, putting the values in equation (1), we have

    Density of Earth,
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#4 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/plugin/impl/TextServiceImpl.class.php(59): com_wiris_plugin_impl_RenderImpl->computeDigest(NULL, Array)
#5 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/service.php(19): com_wiris_plugin_impl_TextServiceImpl->service('mathml2accessib...', Array)
#6 {main}</pre>  

         equals 5497 space kg divided by straight m cubed 
    Question 255
    CBSEENPH11017467

    If the earth were made of lead of relative density 11.3, then what would be the value of g at the surface of earth? Take radius of the earth to be 6.37 x 106m.

    Solution
    The acceleration due to gravity at the surface of the earth is given by, 
                            <pre>uncaught exception: <b>mkdir(): Permission denied (errno: 2) in /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/util/sys/Store.class.php at line #56mkdir(): Permission denied</b><br /><br />in file: /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/util/sys/Store.class.php line 56<br />#0 [internal function]: _hx_error_handler(2, 'mkdir(): Permis...', '/home/config_ad...', 56, Array)
#1 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/util/sys/Store.class.php(56): mkdir('/home/config_ad...', 493)
#2 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/plugin/impl/FolderTreeStorageAndCache.class.php(110): com_wiris_util_sys_Store->mkdirs()
#3 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/plugin/impl/RenderImpl.class.php(231): com_wiris_plugin_impl_FolderTreeStorageAndCache->codeDigest('mml=<math xmlns...')
#4 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/plugin/impl/TextServiceImpl.class.php(59): com_wiris_plugin_impl_RenderImpl->computeDigest(NULL, Array)
#5 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/service.php(19): com_wiris_plugin_impl_TextServiceImpl->service('mathml2accessib...', Array)
#6 {main}</pre> 
    Here,
    straight rho equals 11.3 gm divided by cc space equals space 11300 space kg divided by straight m cubed 
    straight R equals 6.37 cross times 10 to the power of 6 straight m
                  
    straight G equals 6.67 cross times 10 to the power of negative 11 end exponent Nm squared divided by kg squared 
    Therefore,
    straight g equals fraction numerator 4 cross times straight pi over denominator 3 end fraction cross times 6.67 cross times 10 to the power of negative 11 end exponent cross times 6.37 cross times 10 to the power of 6 cross times 11300 
      equals 20.12 straight m divided by straight s squared
     
    Question 256
    CBSEENPH11017493

    Heavier the body, larger is the acceleration with which it falls towards the earth. Comment on this statement.

    Solution
    The statement is wrong.
    This is because the acceleration with which the body falls is given by GM/R2 which only depends upon the mass of the earth and is independent of the mass of the body. 
    Question 257
    CBSEENPH11017498

    A body of mass 200kg weighs 500N on the surface of some heavenly body. Find the acceleration due to gravity at the surface of heavenly body.

    Solution

    Given,
     Mass, m = 200 kg
    Weight of the body, W = 500N 
    Therefore,
    Acceleration due to gravity, 
    straight g equals straight W over straight m equals 500 over 200 equals 2.5 straight m divided by straight s squared  

    Question 258
    CBSEENPH11017501

    Find the acceleration due to gravity on the surface of satellite of mass 2 x 1011kg and radius 25km.

    Solution

    Acceleration due to gravity, straight g equals GM over straight R squared
    Here,
    Mass, straight M equals 2 cross times 10 to the power of 11 kg
    Radius, straight R equals 25 km equals 25000 straight m
           
    Gravitational constant, straight G equals 6.67 cross times 10 to the power of negative 11 end exponent Nm squared divided by kg squared 

    Acceleration due to gravity, 
     straight g equals fraction numerator 6.67 cross times 10 to the power of negative 11 end exponent cross times 2 cross times 10 to the power of 11 over denominator left parenthesis 25000 right parenthesis squared end fraction 
       equals 2.13 cross times 10 to the power of negative 8 end exponent straight m divided by straight s squared 

    Question 259
    CBSEENPH11017503

    The acceleration due to gravity at the moon’s surface is 1.67 m/s2. If the radius of the moon is 1.74 x 10 m, calculate the mass of the moon.

    Solution

    Given that,
    Acceleration due to gravity, straight g equals GM over straight R squared
    ∴                straight M equals gR squared over straight G 
    Here, we have
    Acceleration, straight g equals 1.67 straight m divided by straight s squared    

    Radius, straight R equals 1.74 cross times 10 to the power of 6 straight m 
    ∴ Mass of the moon, <pre>uncaught exception: <b>mkdir(): Permission denied (errno: 2) in /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/util/sys/Store.class.php at line #56mkdir(): Permission denied</b><br /><br />in file: /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/util/sys/Store.class.php line 56<br />#0 [internal function]: _hx_error_handler(2, 'mkdir(): Permis...', '/home/config_ad...', 56, Array)
#1 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/util/sys/Store.class.php(56): mkdir('/home/config_ad...', 493)
#2 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/plugin/impl/FolderTreeStorageAndCache.class.php(110): com_wiris_util_sys_Store->mkdirs()
#3 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/plugin/impl/RenderImpl.class.php(231): com_wiris_plugin_impl_FolderTreeStorageAndCache->codeDigest('mml=<math xmlns...')
#4 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/plugin/impl/TextServiceImpl.class.php(59): com_wiris_plugin_impl_RenderImpl->computeDigest(NULL, Array)
#5 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/service.php(19): com_wiris_plugin_impl_TextServiceImpl->service('mathml2accessib...', Array)
#6 {main}</pre> 
                                      equals space 7.58 cross times 10 to the power of 22 kg 

    Question 260
    CBSEENPH11017504

    If the radius of the earth were to shrink by one percent, its mass remaining the same, what would happen to the value of g?

    Solution
    The acceleration due to gravity at the surface of the earth is,
                             straight g equals GM over straight R squared 
    The radius of the earth decreases by 1%, 
    ∴          <pre>uncaught exception: <b>mkdir(): Permission denied (errno: 2) in /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/util/sys/Store.class.php at line #56mkdir(): Permission denied</b><br /><br />in file: /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/util/sys/Store.class.php line 56<br />#0 [internal function]: _hx_error_handler(2, 'mkdir(): Permis...', '/home/config_ad...', 56, Array)
#1 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/util/sys/Store.class.php(56): mkdir('/home/config_ad...', 493)
#2 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/plugin/impl/FolderTreeStorageAndCache.class.php(110): com_wiris_util_sys_Store->mkdirs()
#3 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/plugin/impl/RenderImpl.class.php(231): com_wiris_plugin_impl_FolderTreeStorageAndCache->codeDigest('mml=<math xmlns...')
#4 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/plugin/impl/TextServiceImpl.class.php(59): com_wiris_plugin_impl_RenderImpl->computeDigest(NULL, Array)
#5 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/service.php(19): com_wiris_plugin_impl_TextServiceImpl->service('mathml2accessib...', Array)
#6 {main}</pre> 
    Now the acceleration due to gravity will be, 
                    
    Percentage increase in acceleration due to gravity, fraction numerator straight g apostrophe minus straight g over denominator straight g end fraction 100 equals 2 % 

    Question 261
    CBSEENPH11017506

    Two planets having masses M1 and M2 and radii R1 and R2 have equal acceleration due to gravity at the surface of planets. Find the ratio of masses of planets in terms of radii.

    Solution
    Acceleration due to gravity of the first planet of mass M1and radii is,
                             
    Acceleration due to gravity of the second planet of mass M2 and radius R2 is, 
                             straight g subscript 2 equals GM subscript 2 over straight R subscript 2 squared 
    Since straight g subscript 1 equals straight g subscript 2 comma therefore 
                         GM subscript 1 over straight R subscript 1 squared equals GM subscript 2 over straight R subscript 2 squared 
    So, 
                            <pre>uncaught exception: <b>mkdir(): Permission denied (errno: 2) in /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/util/sys/Store.class.php at line #56mkdir(): Permission denied</b><br /><br />in file: /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/util/sys/Store.class.php line 56<br />#0 [internal function]: _hx_error_handler(2, 'mkdir(): Permis...', '/home/config_ad...', 56, Array)
#1 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/util/sys/Store.class.php(56): mkdir('/home/config_ad...', 493)
#2 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/plugin/impl/FolderTreeStorageAndCache.class.php(110): com_wiris_util_sys_Store->mkdirs()
#3 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/plugin/impl/RenderImpl.class.php(231): com_wiris_plugin_impl_FolderTreeStorageAndCache->codeDigest('mml=<math xmlns...')
#4 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/plugin/impl/TextServiceImpl.class.php(59): com_wiris_plugin_impl_RenderImpl->computeDigest(NULL, Array)
#5 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/service.php(19): com_wiris_plugin_impl_TextServiceImpl->service('mathml2accessib...', Array)
#6 {main}</pre> 
    Question 262
    CBSEENPH11017508

    Explain why one can jump higher on a planet whose density is same as that of the earth, but radius is half of the radius of the earth, than on the earth?

    Solution
    The height to which a person can jump depend upon the acceleration due to gravity is,
    i.e.,                   straight h equals fraction numerator straight u squared over denominator 2 straight g end fraction 
     
    The acceleration due to gravity on the surface of planet is given by,
     
                           straight g equals 4 over 3 πG space straight rho space straight R 
    The radius of the planet is half the radius of the earth, therefore acceleration due to gravity on the surface of the planet is half the value on the earth.
    Hence, one can jump on the planet to a height twice that on the earth. 
    Question 263
    CBSEENPH11017513

    If acceleration due to gravity at the surface of the earth is g, what will be the acceleration due to gravity at height h? 

    Solution
    At surface of the earth, the acceleration due to gravity g is given by, 
                              straight g equals GM over straight R squared 
                       
    At height h from surface of the earth, it is given by
    space space space space space space space straight g subscript straight h equals fraction numerator GM over denominator left parenthesis straight R plus straight h right parenthesis squared end fraction equals GM over straight R squared fraction numerator straight R squared over denominator left parenthesis straight R plus straight h right parenthesis squared end fraction 
          space space space equals straight g fraction numerator straight R squared over denominator left parenthesis straight R plus straight h right parenthesis squared end fraction 
        straight g subscript straight h equals straight g fraction numerator straight R squared over denominator straight R squared open parentheses 1 plus begin display style straight h over straight R end style close parentheses squared end fraction equals straight g open parentheses 1 plus straight h over straight R close parentheses to the power of negative 2 end exponent
 
           <pre>uncaught exception: <b>mkdir(): Permission denied (errno: 2) in /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/util/sys/Store.class.php at line #56mkdir(): Permission denied</b><br /><br />in file: /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/util/sys/Store.class.php line 56<br />#0 [internal function]: _hx_error_handler(2, 'mkdir(): Permis...', '/home/config_ad...', 56, Array)
#1 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/util/sys/Store.class.php(56): mkdir('/home/config_ad...', 493)
#2 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/plugin/impl/FolderTreeStorageAndCache.class.php(110): com_wiris_util_sys_Store->mkdirs()
#3 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/plugin/impl/RenderImpl.class.php(231): com_wiris_plugin_impl_FolderTreeStorageAndCache->codeDigest('mml=<math xmlns...')
#4 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/plugin/impl/TextServiceImpl.class.php(59): com_wiris_plugin_impl_RenderImpl->computeDigest(NULL, Array)
#5 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/service.php(19): com_wiris_plugin_impl_TextServiceImpl->service('mathml2accessib...', Array)
#6 {main}</pre>, is the acceleration due to gravity at height h. 
    Question 264
    CBSEENPH11017514

    If acceleration due to gravity at the surface of the earth is g then what is its value at depth d?

    Solution
    Let the earth be a homogeneous sphere of radius R with density ρ.
                       
     
    The acceleration due to gravity at surface of the earth g is, 
       
                     straight g equals GM over straight R squared equals 4 over 3 πGρR
    Let, at deptg 'd' the acceleration due to gravity be gd.
    At a depth d, the gravity is only due to the mass of inner sphere of radius R-d. 
    Therefore,
        straight g subscript straight d equals fraction numerator straight G left square bracket mass space of space sphere space of space radius space left parenthesis straight R minus straight d right parenthesis right square bracket over denominator left parenthesis straight R minus straight d right parenthesis squared end fraction 
           equals fraction numerator straight G begin display style 4 over 3 end style straight pi left parenthesis straight R minus straight d right parenthesis cubed straight rho over denominator left parenthesis straight R minus straight d right parenthesis squared end fraction 
         space space equals 4 over 3 πGρ left parenthesis straight R minus straight d right parenthesis 
       straight g subscript straight d space equals space 4 over 3 πGρR open parentheses fraction numerator straight R minus straight d over denominator straight R end fraction close parentheses 
            = straight g open parentheses 1 minus straight d over straight R close parentheses 
    Question 265
    CBSEENPH11017526

    What is the difference between g and G? Is there any relation between the two?

    Solution

    The value of 'g' is different at different points while G is a universal constant and has the same value everywhere in the universe. 
    The unit of g is m/s2
    The dimensional formula is [M0L1T2].
    SI unit of G is N m2/kg2 and [M-1L3T-2] respectively.
    The relation between g and G, straight g equals GM over straight R squared

    Question 266
    CBSEENPH11017527

    Plot the graph for acceleration due to gravity versus distance from the centre of the earth.

    Solution
    The variation of acceleration due to gravity with distance from centre is given by, 
    straight g equals 4 over 3 πGρ space straight r ;              straight r less or equal than straight R 
      equals GM over straight r squared ;                    straight r less or equal than straight R
    The graph for the same is as given below: 

    Question 267
    CBSEENPH11017554

    Discuss the variation of 'g' with angle of latitude.

    Solution

    Acceleration due to gravity changes with the angle of latitude due to shape and rotation of the earth about its own axis.
    Variation of g due to shape:

    Earth is not a perfect sphere and is flat at the pole, bulges out at the equator. Therefore, the polar radius is smaller than the equatorial radius.
    Acceleration due to gravity is inversely proportional to the square of the radius of the earth, therefore the value of g increases as one moves from equator to pole. 
    Variation of g due to rotation:

    Earth rotates about its own axis as a result of which the objects on the earth experience the centrifugal force.
    This centrifugal force decreases the acceleration due to gravity and magnitude of change in the value of g due to rotation of the earth depends on the value of the angle of latitude. 
    Consider the earth to be a homogenous sphere of mass M and radius R.
    The earth rotates about the polar axis.
    Let, ω be the angular velocity of rotation of the earth.
    All the objects at rest on earth also revolve about its polar axis with same angular velocity ω.
    Consider a body of mass M placed at a point P on the earth at an angle of latitude λ.
                    
    If the earth were at rest, then the body would have been attracted towards the center of the earth with force mg.
    Due to rotation of the earth, the body describes a circle of radius OP = r = <pre>uncaught exception: <b>mkdir(): Permission denied (errno: 2) in /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/util/sys/Store.class.php at line #56mkdir(): Permission denied</b><br /><br />in file: /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/util/sys/Store.class.php line 56<br />#0 [internal function]: _hx_error_handler(2, 'mkdir(): Permis...', '/home/config_ad...', 56, Array)
#1 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/util/sys/Store.class.php(56): mkdir('/home/config_ad...', 493)
#2 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/plugin/impl/FolderTreeStorageAndCache.class.php(110): com_wiris_util_sys_Store->mkdirs()
#3 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/plugin/impl/RenderImpl.class.php(231): com_wiris_plugin_impl_FolderTreeStorageAndCache->codeDigest('mml=<math xmlns...')
#4 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/plugin/impl/TextServiceImpl.class.php(59): com_wiris_plugin_impl_RenderImpl->computeDigest(NULL, Array)
#5 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/service.php(19): com_wiris_plugin_impl_TextServiceImpl->service('mathml2accessib...', Array)
#6 {main}</pre> with center at O.
    In rotating frame of the earth, body experiences pseudo force = mrω squared along space OP with rightwards arrow on top. 
    Now the apparent weight of the body is resultant of force mg directed towards the center and force mrω squared along stack OP. with rightwards arrow on top 
    Applying parallelogram of vector addition, we get 
    <pre>uncaught exception: <b>mkdir(): Permission denied (errno: 2) in /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/util/sys/Store.class.php at line #56mkdir(): Permission denied</b><br /><br />in file: /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/util/sys/Store.class.php line 56<br />#0 [internal function]: _hx_error_handler(2, 'mkdir(): Permis...', '/home/config_ad...', 56, Array)
#1 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/util/sys/Store.class.php(56): mkdir('/home/config_ad...', 493)
#2 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/plugin/impl/FolderTreeStorageAndCache.class.php(110): com_wiris_util_sys_Store->mkdirs()
#3 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/plugin/impl/RenderImpl.class.php(231): com_wiris_plugin_impl_FolderTreeStorageAndCache->codeDigest('mml=<math xmlns...')
#4 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/plugin/impl/TextServiceImpl.class.php(59): com_wiris_plugin_impl_RenderImpl->computeDigest(NULL, Array)
#5 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/service.php(19): com_wiris_plugin_impl_TextServiceImpl->service('mathml2accessib...', Array)
#6 {main}</pre>
       straight g subscript straight lambda equals square root of straight g squared plus left parenthesis rω squared right parenthesis squared minus 2 left parenthesis grω squared right parenthesis space cosλ end root  
          equals straight g square root of 1 plus open parentheses rω squared over straight g close parentheses squared minus 2 rω squared over straight g cosλ end root 
    As  rω squared over straight g less than less than 1 comma thereforespace space space open parentheses rω squared over straight g close parentheses squared can be neglected. 
    Therefore,
           straight g subscript straight lambda equals straight g square root of 1 minus 2 rω squared over straight g cosλ end root 
             equals straight g open parentheses 1 minus rω squared over straight g cosλ close parentheses          [using binomial exp]
    rightwards double arrow    straight g subscript straight lambda equals straight g minus rω squared cosλ 
    Substituting straight r equals Rcosλ comma we get 
    straight g subscript straight lambda equals straight g minus Rω squared cos squared straight lambda is the g. 

    Question 268
    CBSEENPH11017566

    Where the acceleration due to gravity on the surface of earth is (i) maximum (ii) minimum? Find the difference in the maximum and minimum value of g on the surface of the earth.

    Solution

    The acceleration due to gravity on the surface of the earth at angle of latitude space straight lambda is,
     
                     straight g subscript straight lambda equals straight g minus Rω squared cos squared straight lambda 
    (i) The value of g is maximum at points where 
                     cos squared straight lambda equals 0 space or space straight lambda equals 90 degree. 
    Thus, acceleration due to gravity is maximum at poles. 
    ∴  straight g subscript max equals straight g subscript straight p equals straight g minus Rω squared cos squared 90 degree equals straight g 
    (ii) The value of g is minimum at points where
                      cos squared straight lambda equals 1 space or space straight lambda space equals space 0 degree. 
    Thus, acceleration due to gravity is minimum at the equator. 
    i.e.,  <pre>uncaught exception: <b>mkdir(): Permission denied (errno: 2) in /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/util/sys/Store.class.php at line #56mkdir(): Permission denied</b><br /><br />in file: /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/util/sys/Store.class.php line 56<br />#0 [internal function]: _hx_error_handler(2, 'mkdir(): Permis...', '/home/config_ad...', 56, Array)
#1 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/util/sys/Store.class.php(56): mkdir('/home/config_ad...', 493)
#2 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/plugin/impl/FolderTreeStorageAndCache.class.php(110): com_wiris_util_sys_Store->mkdirs()
#3 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/plugin/impl/RenderImpl.class.php(231): com_wiris_plugin_impl_FolderTreeStorageAndCache->codeDigest('mml=<math xmlns...')
#4 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/plugin/impl/TextServiceImpl.class.php(59): com_wiris_plugin_impl_RenderImpl->computeDigest(NULL, Array)
#5 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/service.php(19): com_wiris_plugin_impl_TextServiceImpl->service('mathml2accessib...', Array)
#6 {main}</pre> 
    The difference in the value of g at pole and equator is, 
                <pre>uncaught exception: <b>mkdir(): Permission denied (errno: 2) in /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/util/sys/Store.class.php at line #56mkdir(): Permission denied</b><br /><br />in file: /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/util/sys/Store.class.php line 56<br />#0 [internal function]: _hx_error_handler(2, 'mkdir(): Permis...', '/home/config_ad...', 56, Array)
#1 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/util/sys/Store.class.php(56): mkdir('/home/config_ad...', 493)
#2 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/plugin/impl/FolderTreeStorageAndCache.class.php(110): com_wiris_util_sys_Store->mkdirs()
#3 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/plugin/impl/RenderImpl.class.php(231): com_wiris_plugin_impl_FolderTreeStorageAndCache->codeDigest('mml=<math xmlns...')
#4 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/plugin/impl/TextServiceImpl.class.php(59): com_wiris_plugin_impl_RenderImpl->computeDigest(NULL, Array)
#5 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/service.php(19): com_wiris_plugin_impl_TextServiceImpl->service('mathml2accessib...', Array)
#6 {main}</pre>
                                  equals left parenthesis 6.4 cross times 10 to the power of 6 right parenthesis space open parentheses fraction numerator 2 straight pi over denominator 86400 end fraction close parentheses squared 
                                  <pre>uncaught exception: <b>mkdir(): Permission denied (errno: 2) in /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/util/sys/Store.class.php at line #56mkdir(): Permission denied</b><br /><br />in file: /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/util/sys/Store.class.php line 56<br />#0 [internal function]: _hx_error_handler(2, 'mkdir(): Permis...', '/home/config_ad...', 56, Array)
#1 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/util/sys/Store.class.php(56): mkdir('/home/config_ad...', 493)
#2 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/plugin/impl/FolderTreeStorageAndCache.class.php(110): com_wiris_util_sys_Store->mkdirs()
#3 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/plugin/impl/RenderImpl.class.php(231): com_wiris_plugin_impl_FolderTreeStorageAndCache->codeDigest('mml=<math xmlns...')
#4 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/plugin/impl/TextServiceImpl.class.php(59): com_wiris_plugin_impl_RenderImpl->computeDigest(NULL, Array)
#5 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/service.php(19): com_wiris_plugin_impl_TextServiceImpl->service('mathml2accessib...', Array)
#6 {main}</pre> 

    Question 269
    CBSEENPH11017569

    The weight of a body at depth d is equal to the weight of the body at height h above the surface of the earth. What is the ratio of h to d, for small value of h and d?

                                     OR

    Prove that the distance we have to cover into the earth below its surface is twice the distance we have to cover above the surface of the earth to get the same change in the weight of a body?

    Solution
    For small value of h, the acceleration due to gravity at height h is, 

                   space space space straight g subscript straight h equals straight g open parentheses 1 minus fraction numerator 2 straight h over denominator straight R end fraction close parentheses 
    ∴            "<pre 
    Similarly, at the depth d the weight of body is, 
                      straight W subscript straight d equals mg open parentheses 1 minus straight d over straight R close parentheses 
    Since,   space space space space space space space space straight W subscript straight h equals straight W subscript straight d 
    i.e.,  
        mg open parentheses 1 minus fraction numerator 2 straight h over denominator straight R end fraction close parentheses space equals space mg open parentheses 1 minus straight d over straight R close parentheses 
    ∴                          2 straight h equals straight d 
    rightwards double arrow                       straight h over straight d equals 1 half 
    Question 270
    CBSEENPH11017574

    At what height acceleration due to gravity is three-fourth of that at the surface of the earth?

    Solution

    Acceleration due to gravity at a height from the Earth's surface is given by, 
                         straight g subscript straight h equals straight g over open parentheses 1 plus begin display style straight h over straight R end style close parentheses squared 
    Since,            straight g subscript straight h equals 3 over 4 straight g 
    Therefore,
                       3 over 4 straight g equals straight g over open parentheses 1 plus begin display style straight h over straight R end style close parentheses squared  
    rightwards double arrow        <pre>uncaught exception: <b>mkdir(): Permission denied (errno: 2) in /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/util/sys/Store.class.php at line #56mkdir(): Permission denied</b><br /><br />in file: /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/util/sys/Store.class.php line 56<br />#0 [internal function]: _hx_error_handler(2, 'mkdir(): Permis...', '/home/config_ad...', 56, Array)
#1 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/util/sys/Store.class.php(56): mkdir('/home/config_ad...', 493)
#2 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/plugin/impl/FolderTreeStorageAndCache.class.php(110): com_wiris_util_sys_Store->mkdirs()
#3 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/plugin/impl/RenderImpl.class.php(231): com_wiris_plugin_impl_FolderTreeStorageAndCache->codeDigest('mml=<math xmlns...')
#4 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/plugin/impl/TextServiceImpl.class.php(59): com_wiris_plugin_impl_RenderImpl->computeDigest(NULL, Array)
#5 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/service.php(19): com_wiris_plugin_impl_TextServiceImpl->service('mathml2accessib...', Array)
#6 {main}</pre> 
    On solving we get, 
                  straight h equals open parentheses fraction numerator 2 over denominator square root of 3 end fraction minus 1 close parentheses straight R , is the height at which the acceleration due to gravity is three-fourth of that at the surface of the earth. 

    Question 271
    CBSEENPH11017575

    At what height from the surface of the earth the acceleration due to gravity is 1% less than that on the surface of the earth?

    Solution
    The acceleration due to gravity at the surface of the earth is, 
                    straight g equals GM over straight R squared                         ...(1) 
    Let at height h from the surface of the earth, the acceleration due to gravity be 1% less than that on the surface of the earth. 
    ∴           space space space 0.99 straight g equals fraction numerator GM over denominator left parenthesis straight R plus straight h right parenthesis squared end fraction               ...(2) 
    From (1) and (2), 
               space space space fraction numerator straight g over denominator 0.99 straight g end fraction equals fraction numerator left parenthesis straight R plus straight h right parenthesis squared over denominator straight R squared end fraction 
    rightwards double arrow           fraction numerator straight R plus straight h over denominator straight R end fraction equals 1.005 
    rightwards double arrow                  h = 32km, is the height from the surface of the earth. 
    Question 272
    CBSEENPH11017576

    Mass of a body is 10 kg at the surface of the earth. At what depth does its mass reduce to 5 kg?

    Solution
    Mass of the body is proportional to the quantity of matter contained in the body and independent of the acceleration due to gravity.
    Therefore, the mass of the  body does not change with depth and hence it will not be 5 kg anywhere.  
    Question 273
    CBSEENPH11017579

    A body weighs 63N on the surface of the earth. What is the gravitational force on it due to the earth at a height equal to half the radius of the earth?

    Solution
    Acceleration due to gravity at height h from the surface of the earth is,

                    straight g subscript straight h equals fraction numerator GM over denominator left parenthesis straight R plus straight h right parenthesis squared end fraction 
                       equals GM over straight R squared 1 over open parentheses 1 plus begin display style straight h over straight R end style close parentheses squared equals straight g over open parentheses 1 plus begin display style straight h over straight R end style close parentheses squared 
    where, g is the acceleration due to gravity at the surface of the earth. 
    Here, height is equal to the radiuscomma space straight h equals straight R over 2 
    ∴   straight g subscript straight h equals straight g over open parentheses 1 plus begin display style 1 half end style close parentheses squared equals 4 over 9 straight g 
    Gravitational force at given height is, 
                space space straight F equals mg subscript straight h equals 4 over 9 mg 
    Weight of the body,  mg equals 63 space straight N 
                                      straight F equals 4 over 9 left parenthesis 63 right parenthesis equals 28 straight N 
    Question 274
    CBSEENPH11017580

    Why does a man feel weightlessness while sitting inside a freely falling lift?

    Solution

    The apparent weight of man in a lift falling with acceleration a is given by, 
                        Wa = m(g-a) 
    The acceleration of free-falling lift is g. 
    ∴                   Wa = m(g-g) = 0 
    Hence a man feels weightlessness.

    Question 275
    CBSEENPH11017581

    Explain why does a tennis ball bounce higher on hills than in plains? 

    Solution
    The height to which the ball bounces when thrown with velocity u is given by, 
                               straight h equals fraction numerator straight u squared over denominator 2 straight g end fraction 
    The height to which the ball bounces is inversely proportional to the acceleration due to gravity at that place.
    Acceleration due to gravity on hills is less than that in plains, therefore the tennis ball bounces higher on hills. 
    Question 276
    CBSEENPH11017583

    Assuming the earth to be a sphere of uniform mass density, how much would a body weigh half way down to the centre of the earth if it weighed 250N on the surface?

    Solution
    We know acceleration due to gravity at a depth d from the surface of the earth is, 
                       straight g subscript straight d space equals space straight g space left parenthesis 1 minus straight d over straight R right parenthesis space
    where,
    g is acceleration due to gravity at the surface of the earth.
    The weight of body of mass m at depth d is, 
    Wd = mgd = mg left parenthesis 1 minus straight d over straight R right parenthesis space 
    Given, 
    Weight of the body on the surface = 250 N
    Distance of the body, d = R/2 
    Therefore, 
    straight W subscript straight d space equals space 250 space left parenthesis 1 minus 1 half right parenthesis space

space space space space space space equals space 125 space straight N 
    Question 277
    CBSEENPH11017585

    Where will you weigh more and why: 2km above or 2km below the surface of the earth?

    Solution
    The weight is more at 2km below the surface of the Earth than at 2km above the surface of the earth. This is because the acceleration due to gravity at a depth of 2km is more than that of 2km above the surface of the earth. 
    Question 278
    CBSEENPH11017586

    Why is a body is weightless at the center of the earth?

    Solution

    Weight of the body at any point, W = mg 
    where,
    g is the acceleration due to gravity at that point.
    The acceleration due to gravity at a point inside the earth is, 
                  straight g equals GM over straight R squared open parentheses 1 minus straight d over straight R close parentheses 
    Therefore, 
    Weight,  straight W space equals GMm over straight R squared open parentheses 1 minus straight d over straight R close parentheses 
    At the center of the earth, d = R. 
    So, straight W equals GMm over straight R squared open parentheses 1 minus straight R over straight R close parentheses equals 0
    Thus, the body is weightless at the centre of the earth. 

    Question 279
    CBSEENPH11017587

    Why acceleration due to gravity at the poles is greater than at the equator?

    Solution
    The Earth revolves about its own axis with an angular velocity given by, 
                     straight omega equals fraction numerator 2 straight pi over denominator 86400 end fraction rad divided by sec 
    At the equator, the body will experience the centrifugal force in an outward direction. So, at equator, the apparent weight will decrease and hence the acceleration due to gravity will decrease.
    Acceleration due to gravity at the pole = g
    Acceleration due to gravity at the equator = g - Rω
    where, R is the radius of the earth.
    Question 280
    CBSEENPH11017589

    If the earth stops to revolve about its own axis then how much acceleration will change at equator?

    Solution

    If the earth stops to revolve about it's own axis then the acceleration will increase by Rω squared
    Change in acceleration at equator is, 
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#1 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/util/sys/Store.class.php(56): mkdir('/home/config_ad...', 493)
#2 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/plugin/impl/FolderTreeStorageAndCache.class.php(110): com_wiris_util_sys_Store->mkdirs()
#3 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/plugin/impl/RenderImpl.class.php(231): com_wiris_plugin_impl_FolderTreeStorageAndCache->codeDigest('mml=<math xmlns...')
#4 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/plugin/impl/TextServiceImpl.class.php(59): com_wiris_plugin_impl_RenderImpl->computeDigest(NULL, Array)
#5 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/service.php(19): com_wiris_plugin_impl_TextServiceImpl->service('mathml2accessib...', Array)
#6 {main}</pre>
         <pre>uncaught exception: <b>mkdir(): Permission denied (errno: 2) in /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/util/sys/Store.class.php at line #56mkdir(): Permission denied</b><br /><br />in file: /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/util/sys/Store.class.php line 56<br />#0 [internal function]: _hx_error_handler(2, 'mkdir(): Permis...', '/home/config_ad...', 56, Array)
#1 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/util/sys/Store.class.php(56): mkdir('/home/config_ad...', 493)
#2 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/plugin/impl/FolderTreeStorageAndCache.class.php(110): com_wiris_util_sys_Store->mkdirs()
#3 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/plugin/impl/RenderImpl.class.php(231): com_wiris_plugin_impl_FolderTreeStorageAndCache->codeDigest('mml=<math xmlns...')
#4 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/plugin/impl/TextServiceImpl.class.php(59): com_wiris_plugin_impl_RenderImpl->computeDigest(NULL, Array)
#5 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/service.php(19): com_wiris_plugin_impl_TextServiceImpl->service('mathml2accessib...', Array)
#6 {main}</pre> 
         equals 3.38 space cm divided by sec squared 

    Question 281
    CBSEENPH11017591

    How much weight of a body of mass 25 kg at the equator will change when the earth stops to revolve about its own axis?

    Solution
    Mass of the body = 25 kg
    If the earth stops to revolve, the acceleration due to gravity will increase by Rω2. 
    Therefore, the weight of body will increase by mRω2.
    Thus increase in weight is, 
              increment straight W space equals space mRω squared 
             space space space space space space space space space space equals 25 cross times 6.4 cross times 10 to the power of 6 cross times open parentheses fraction numerator 2 straight pi over denominator 86400 end fraction close parentheses squared 
                    asymptotically equal to space 0.845 space straight N 
    Question 282
    CBSEENPH11017592

    When a body of mass m is taken from latitude λ1 to λthen how much its weight will change? 

    Solution
    The acceleration due to gravity gλ at any angle of latitude is given by, 
                             straight g subscript straight lambda equals straight g minus Rω squared cos squared straight lambda 
    Weight of mass m at latitude straight lambda subscript 1 is m 
                           straight W subscript 1 equals straight m left parenthesis straight g minus Rω squared space cos squared straight lambda subscript 1 right parenthesis 
    At latitude straight lambda subscript 2, weight of mass is, 
                          space space straight W subscript 2 equals straight m left parenthesis straight g minus Rω squared cos squared straight lambda subscript 2 right parenthesis 
    Change in weight is, 
         <pre>uncaught exception: <b>mkdir(): Permission denied (errno: 2) in /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/util/sys/Store.class.php at line #56mkdir(): Permission denied</b><br /><br />in file: /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/util/sys/Store.class.php line 56<br />#0 [internal function]: _hx_error_handler(2, 'mkdir(): Permis...', '/home/config_ad...', 56, Array)
#1 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/util/sys/Store.class.php(56): mkdir('/home/config_ad...', 493)
#2 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/plugin/impl/FolderTreeStorageAndCache.class.php(110): com_wiris_util_sys_Store->mkdirs()
#3 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/plugin/impl/RenderImpl.class.php(231): com_wiris_plugin_impl_FolderTreeStorageAndCache->codeDigest('mml=<math xmlns...')
#4 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/plugin/impl/TextServiceImpl.class.php(59): com_wiris_plugin_impl_RenderImpl->computeDigest(NULL, Array)
#5 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/service.php(19): com_wiris_plugin_impl_TextServiceImpl->service('mathml2accessib...', Array)
#6 {main}</pre> 
       space space space space space space space space space space equals mRω squared left parenthesis cos squared straight lambda subscript 1 minus cos squared straight lambda subscript 2 right parenthesis

    Question 283
    CBSEENPH11017594

    Bodies lying on the surface of the earth rotate with large velocity along with the earth. Why do these not fly off ?  

    Solution
    For bodies which are lying on the surface of the earth, the gravitational pull on the bodies is much greater than the centrifugal force experienced by the bodies rotating along with the earth. That is why the bodies do not fly off. 
    Question 284
    CBSEENPH11017595

    Why does a body weigh more at poles than at equator? 

    Solution
    The body experiences a centrifugal force away from the earth due to rotation of the Earth because of which the weight of the body decreases.
    The centrifugal force is maximum at the equator and minimum at the poles.
    Therefore, the body weighs maximum at poles and minimum at the equator. 
    Question 285
    CBSEENPH11017596

    Two identical masses each of 10 kg are placed 1m apart on a tabletop. How many times larger is the gravitational pull of the earth on one mass than the gravitational attraction of other mass?

    Solution

    The gravitational pull of the earth on one of the mass is, F1=mg 
    The gravitational attraction of other mass on first mass is, 
                      straight F subscript 2 equals straight G straight m squared over straight r squared 
    Now, the ratio of both masses is given by, 
                  <pre>uncaught exception: <b>mkdir(): Permission denied (errno: 2) in /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/util/sys/Store.class.php at line #56mkdir(): Permission denied</b><br /><br />in file: /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/util/sys/Store.class.php line 56<br />#0 [internal function]: _hx_error_handler(2, 'mkdir(): Permis...', '/home/config_ad...', 56, Array)
#1 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/util/sys/Store.class.php(56): mkdir('/home/config_ad...', 493)
#2 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/plugin/impl/FolderTreeStorageAndCache.class.php(110): com_wiris_util_sys_Store->mkdirs()
#3 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/plugin/impl/RenderImpl.class.php(231): com_wiris_plugin_impl_FolderTreeStorageAndCache->codeDigest('mml=<math xmlns...')
#4 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/plugin/impl/TextServiceImpl.class.php(59): com_wiris_plugin_impl_RenderImpl->computeDigest(NULL, Array)
#5 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/service.php(19): com_wiris_plugin_impl_TextServiceImpl->service('mathml2accessib...', Array)
#6 {main}</pre> 
    rightwards double arrow          straight F subscript 1 over straight F subscript 2 equals fraction numerator 9.8 cross times left parenthesis 1 right parenthesis squared over denominator 6.67 cross times 10 to the power of negative 11 end exponent cross times 10 end fraction equals 1.469 cross times 10 to the power of 10
    That is, the gravitational pull on one mass is 1.496 x 1010 times larger than the gravitational attraction of other mass. 

    Question 286
    CBSEENPH11017598

    How much faster than its present rate should the earth rotate about its axis, so that the bodies at the equator experience weightlessness? Also find the duration of day. Take g = 9.8 m/s2.

    Solution

    We know that acceleration due to gravity at equator is, 
                       straight g subscript straight e equals straight g minus Rω subscript 0 squared 
    where,
    straight omega subscript 0 is the angular velocity of the earth.
    Angular velocity of Earth= space space fraction numerator 2 straight pi over denominator straight T end fraction equals fraction numerator 2 straight pi over denominator 86400 end fraction. 
    If the acceleration due to gravity at the equator becomes zero, the body at equator experiences weightlessness.
    Let space straight omega be the angular velocity of rotation at which the acceleration due to gravity at the equator becomes zero. 
    That is,
                    <pre>uncaught exception: <b>mkdir(): Permission denied (errno: 2) in /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/util/sys/Store.class.php at line #56mkdir(): Permission denied</b><br /><br />in file: /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/util/sys/Store.class.php line 56<br />#0 [internal function]: _hx_error_handler(2, 'mkdir(): Permis...', '/home/config_ad...', 56, Array)
#1 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/util/sys/Store.class.php(56): mkdir('/home/config_ad...', 493)
#2 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/plugin/impl/FolderTreeStorageAndCache.class.php(110): com_wiris_util_sys_Store->mkdirs()
#3 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/plugin/impl/RenderImpl.class.php(231): com_wiris_plugin_impl_FolderTreeStorageAndCache->codeDigest('mml=<math xmlns...')
#4 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/plugin/impl/TextServiceImpl.class.php(59): com_wiris_plugin_impl_RenderImpl->computeDigest(NULL, Array)
#5 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/service.php(19): com_wiris_plugin_impl_TextServiceImpl->service('mathml2accessib...', Array)
#6 {main}</pre> 
    rightwards double arrow            <pre>uncaught exception: <b>mkdir(): Permission denied (errno: 2) in /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/util/sys/Store.class.php at line #56mkdir(): Permission denied</b><br /><br />in file: /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/util/sys/Store.class.php line 56<br />#0 [internal function]: _hx_error_handler(2, 'mkdir(): Permis...', '/home/config_ad...', 56, Array)
#1 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/util/sys/Store.class.php(56): mkdir('/home/config_ad...', 493)
#2 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/plugin/impl/FolderTreeStorageAndCache.class.php(110): com_wiris_util_sys_Store->mkdirs()
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#4 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/plugin/impl/TextServiceImpl.class.php(59): com_wiris_plugin_impl_RenderImpl->computeDigest(NULL, Array)
#5 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/service.php(19): com_wiris_plugin_impl_TextServiceImpl->service('mathml2accessib...', Array)
#6 {main}</pre> 
    Now,
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#1 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/util/sys/Store.class.php(56): mkdir('/home/config_ad...', 493)
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#4 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/plugin/impl/TextServiceImpl.class.php(59): com_wiris_plugin_impl_RenderImpl->computeDigest(NULL, Array)
#5 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/service.php(19): com_wiris_plugin_impl_TextServiceImpl->service('mathml2accessib...', Array)
#6 {main}</pre> 
    rightwards double arrow             space space space space straight omega equals 17 straight omega subscript 0 
    Thus, the earth should rotate 17 times faster than the present rate.
    Rotation speed increases by a factor of 17.
    Therefore, the duration of day decreases by factor 17.
    New duration of day, 
                  straight T equals 24 over 17 equals 1.412 hr equals 5082 straight s

    Question 287
    CBSEENPH11017605

    A star 2.5 times the mass of the sun and collapsed to a size of 12 km rotates with a speed of 1.5 rps. (Extremely compact stars of this kind are known as neutron stars. Certain observed stellar objects called pulsars are believed to belong to this category.) Will an object placed on its equator remain stuck to its surface due to gravity? (Mass of the sun = 2 x 1030kg) 

    Solution
    An object placed on the equator will remain stuck to its surface if the gravitational pull at the equator is greater than the outward centrifugal force at the equator. 
                        
    Given,
    Mass of the star is 2.5 times the mass of the sun. 
    That is, 
    space space space space straight M equals 2.5 cross times 2 cross times 10 to the power of 30 kg equals 5 cross times 10 to the power of 30 kg comma
    Distance comma space straight R equals 12 km equals 12000 straight m

Angular space velocity comma space straight omega equals 1.5 space rps space equals space 3 straight pi space rad divided by straight s 
    The gravitational pull on an object is, 
              straight F subscript straight g space equals GMm over straight R squared

space space space space space equals fraction numerator 6.67 cross times 10 to the power of negative 11 end exponent cross times 5 cross times 10 to the power of 30 cross times straight m over denominator left parenthesis 12 cross times 10 cubed right parenthesis squared end fraction 
                  equals space left parenthesis 2.3 cross times 10 to the power of 12 straight m right parenthesis space straight N          ... (1)
    The centrifugal force is, 
     space space space straight F subscript straight c equals mRω squared

space space space space space space space equals straight m cross times 12 cross times 10 cubed cross times left parenthesis 3 straight pi right parenthesis squared 
       space space space space equals left parenthesis 1.1 cross times 10 to the power of 6 straight m right parenthesis straight m divided by straight s squared                ... (2)
    From (1) and (2), we can see that straight F subscript straight g greater than straight F subscript straight c comma 
    Therefore, the object will remain stuck on its surface due to gravity.

    Question 288
    CBSEENPH11017606

    Define gravitational field.

    Solution
    The space around a material in which it's gravitational pull can be experienced is called it's gravitational field. 
    Question 289
    CBSEENPH11017607

    What is the unit and dimensions of gravitational field strength?

    Solution
    SI unit of gravitational field strength = N/kg
    Dimension formula = [M0L1T-2]. 
    Question 290
    CBSEENPH11017608

    Write an expression for gravitational field strength due to mass m at a distance r.

    Solution

    Gravitational field strength = GM over straight r squared
    where, 
    M = mass of Earth, and
    R = radius of Earth

    Question 291
    CBSEENPH11017609

    What is the gravitational field strength due to the earth at the surface of the earth?

    Solution
    Gravitational field strength due to the Earth at the surface = 9.8 N/kg
    Question 292
    CBSEENPH11017610

    What is gravitational field strength due to the earth at the centre of the earth?

    Solution
    Gravitational field strength, I = GM over straight R squared
    At the centre of the Earth, the mass is presumably zero.
    Therefore, gravitational field strength is zero at the centre of the Earth. 
    Question 293
    CBSEENPH11017612

    What is the gravitational field strength due to spherical shell inside the shell?

    Solution
    For a spherical shell, whole of the mass is concentrated at the outer surface of the shell. Hence, inside the spherical shell, there is no mass. Therefore, the gravitational field strength due to spherical shell inside the shell is ZERO. 
    Question 294
    CBSEENPH11017613

    You can shield a charge from electrical forces by putting it inside a hollow conductor. Can you shield a body from the gravitational influence of nearby matter by putting it inside a hollow sphere or by some other means?

    Solution

    We cannot shield a body from the gravitational influence of nearby matter, because the gravitational force on a body due to nearby matter is independent of the presence of other matter, whereas it is not so in the case of electric forces.
    It means the gravitational screens are not possible.

    Question 296
    CBSEENPH11017615

    The acceleration due to gravity due to certain planet at its surface is 1.2m/s2. What is the gravitational field strength at the surface of the planet?

    Solution

    At the surface of a planet, gravitational field strength is equal to acceleration due to gravity.
    At the surface, 
                           I = GM over straight R squared space equals space g
    That is,            I = 1.2N/kg.

    Question 297
    CBSEENPH11017616

    A body of mass 80kg weighs 400N at surface of certain planet. What is the gravitational field strength at the surface of the planet?

    Solution
    Given, 
    Weight of the body, m = 400 N
    Mass of the body = 80 kg
    The gravitational field strength at the surface of planet is, 
    straight E equals straight F over straight m equals 400 over 80 equals 5 straight N divided by kg 
    Question 298
    CBSEENPH11017617

    What is gravitational potential energy?

    Solution
    The gravitational potential energy of a body at a point in a gravitational field of another body is defined as the amount of work done in carrying the mass m from infinity to a given point without acceleration.
    Question 299
    CBSEENPH11017618

    What is gravitational potential?

    Solution
    The gravitational potential at a point in a gravitational field of a body is the amount of work done in carrying the unit mass from infinity to a given point without acceleration.
    Question 300
    CBSEENPH11017619

    What is the relation between gravitational field and gravitational potential?

    Solution
    The gravitational field is negative gradient of gravitational potential.
    Question 301
    CBSEENPH11017620

    What is the relation between gravitational potential energy and gravitational potential?

    Solution
    The gravitational potential is the gravitational potential energy stored per unit mass. 
    Question 302
    CBSEENPH11017621

    What is unit and dimension of gravitational potential?

    Solution
    The SI unit of gravitational potential is J/kg
    Gravitational potential =straight W over straight m subscript straight o space equals space fraction numerator open square brackets straight M space straight L squared space straight T to the power of negative 2 end exponent close square brackets over denominator open square brackets straight M close square brackets end fraction space equals open square brackets space straight M to the power of straight o space straight L squared space straight T to the power of negative 2 end exponent close square brackets
    Question 303
    CBSEENPH11017622

    Is the gradient of the gravitational potential a vector?

    Solution
    Yes, the gradient of the gravitational potential is a vector.
    Question 304
    CBSEENPH11017623

    Is gravitational potential a vector?

    Solution
    No, the gravitational potential is a scalar quantity because work done is a scalar quantity. 
    Question 305
    CBSEENPH11017624

    Write an expression for gravitational potential energy stored in the mass m placed in the field of mass M at a distance r.

    Solution

    Gravitational P.E stored in mass m placed in the field of mass M at a distance r = <pre>uncaught exception: <b>mkdir(): Permission denied (errno: 2) in /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/util/sys/Store.class.php at line #56mkdir(): Permission denied</b><br /><br />in file: /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/util/sys/Store.class.php line 56<br />#0 [internal function]: _hx_error_handler(2, 'mkdir(): Permis...', '/home/config_ad...', 56, Array)
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#6 {main}</pre>

    Question 307
    CBSEENPH11017626

    Does the change in potential energy depend upon the path followed?

    Solution
    No, because gravitational field is conservative. That is, the work done does not depend on the path followed. 
    Question 309
    CBSEENPH11017628

    What is the gravitational potential energy stored in 3kg mass placed in the field where gravitational potential is -10J/kg?

    Solution
    Gravitational potential energy , U = mV
                                                       = 3(-10)
                                                       = -30 J
    Question 311
    CBSEENPH11017631

    Define equipotential surface.

    Solution
    An equipotential surface is that surface at every point of which the gravitational potential is the same.
    Question 312
    CBSEENPH11017632

    Do you think that the gravitational potential at the centre of the earth is zero?

    Solution
    No, gravitational potential at the centre of the earth is not zero. 
    Gravitational potential is zero, when r = infinity
    Question 314
    CBSEENPH11017634

    Define gravitational field and find the strength of gravitational field due to mass M at a distance R?

    Solution

    The space around a material body in which it's gravitational pull can be experienced is called it's gravitational field. 
    The gravitational field strength at any point in the gravitational field is the force experienced by unit mass placed at that point.
    Gravitational field strength is a vector quantity and is directed towards the mass causing the field.
                  
    At a distance R, place the test mass m at point P to find the gravitational field strength.
    Force space straight F with rightwards arrow on top experienced by the test mass m is, 
                straight F with rightwards arrow on top equals GMm over straight R squared, towards the mass M.
    The gravitational field strength due to mass M at observation point P is, 
           space space space space straight E with rightwards arrow on top equals fraction numerator straight F with rightwards arrow on top over denominator straight m end fraction equals GM over straight R squared, towards the mass M. 
        

    Question 315
    CBSEENPH11017635

    Derive an expression for gravitational field strength due to the earth and show that it is equal to acceleration due to gravity at that point.

    Solution
    Consider the earth to be a homogenous sphere of mass M and radius R.
    To find the gravitational field strength at a point P which is at a distance r from center of the earth, place the unit mass at that point.
    The magnitude of the force on the unit mass is gravitational field strength at that point. 
                    
    Gravitational field strength at a point is given by, 
    i.e.          straight E equals fraction numerator GM cross times 1 over denominator straight r squared end fraction equals GM over straight r squared                     ...(1)
    Also, acceleration due to gravity at point P is, 
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#6 {main}</pre>                  ...(2) 
    From (1) and (2), we get, 
                     E = g 
    That is, the gravitational field strength due to the earth is equal to the acceleration due to gravity. 

    Question 316
    CBSEENPH11017636

    If the earth were to contract to half of its present radius without any change in mass, how would the gravitational field strength at the surface be affected?

    Solution
    Gravitational field strength at the surface of the earth is given by, 

                       straight E equals GM over straight R squared 
    If R reduces to 1 half straight R comma then 
    space straight E apostrophe equals fraction numerator GM over denominator left parenthesis straight R divided by 2 right parenthesis squared end fraction equals 4 GM over straight R squared equals 4 straight E 
    Hence the strength of gravitational field increases to four times. 
    Question 317
    CBSEENPH11017637

    Find the gravitational field strength due to the sun at the surface of the earth. Given mass of the sun is 2 x 1030kg.

    Solution

    Here,
    Mass of the sun, straight M equals 2 cross times 10 to the power of 30 kg 
    The distance of the earth from the sun is, straight R equals 1.5 cross times 10 to the power of 11 straight m 

    Therefore, gravitational field strength due to the sun is,
    straight E equals GM over straight R squared equals fraction numerator 6.67 cross times 10 to the power of negative 11 end exponent cross times 2 cross times 10 to the power of 30 over denominator left parenthesis 1.5 cross times 10 to the power of 11 right parenthesis squared end fraction 
      equals 5.93 cross times 10 to the power of negative 3 end exponent straight N divided by kg 

    Question 318
    CBSEENPH11017640

    Two masses of 25 kg and 75kg are separated by a distance of 20m. Find the gravitational field strength at a point at a distance of 5 m from 25kg mass.

    Solution

    Given,

    Mass space of space the space first space body comma space straight M subscript 1 equals 25 kg
Mass space of space the space second space body comma space straight M subscript 2 equals 75 kg 
    Distance between the two bodies = 20 m 
    Distance space of space mass space straight m subscript 1 space end subscript from space 25 space kg comma space straight r subscript 1 equals 5 straight m

Distance space of space mass space straight m subscript 2 space from space 25 kg comma space straight r subscript 2 equals 15 straight m 
    Let space space straight E subscript 1   and space space straight E subscript 2  be the gravitational field strength due to masses 25 kg and 75 kg respectively. 
    Using the formula, we have
     <pre>uncaught exception: <b>mkdir(): Permission denied (errno: 2) in /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/util/sys/Store.class.php at line #56mkdir(): Permission denied</b><br /><br />in file: /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/util/sys/Store.class.php line 56<br />#0 [internal function]: _hx_error_handler(2, 'mkdir(): Permis...', '/home/config_ad...', 56, Array)
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#6 {main}</pre> 
                   space equals 6.66 cross times 10 to the power of negative 11 end exponent straight N divided by kg 

    straight E subscript 2 equals GM subscript 2 over straight r subscript 2 squared equals fraction numerator 6.66 cross times 10 to the power of negative 11 end exponent cross times 75 over denominator 225 end fraction 
                  space equals 2.22 space cross times space 10 to the power of negative 11 end exponent straight N divided by kg 
    Since straight E subscript 1 space and space straight E subscript 2 are in opposite direction, therefore net field strength at given point is, 
               straight E equals straight E subscript 1 minus straight E subscript 2 equals 4.44 cross times 10 to the power of negative 11 end exponent straight N divided by kg

    Question 319
    CBSEENPH11017641

    Two masses m1 and m2 are separated by a distance R. At what distance from m1 the gravitational field will be zero?

    Solution
    Let the gravitational field be zero at P, which is at a distance x from m1.

    Gravitational field at P is zero if the fields due to m1 and m2 are equal and opposite. 

                 
    i.e.           space Gm subscript 1 over straight x squared equals fraction numerator Gm subscript 2 over denominator left parenthesis straight R minus straight x right parenthesis squared end fraction 
    rightwards double arrow        open parentheses fraction numerator straight R minus straight x over denominator straight x end fraction close parentheses squared equals straight m subscript 2 over straight m subscript 1 
    rightwards double arrow            fraction numerator straight R minus straight x over denominator straight x end fraction equals square root of straight m subscript 2 over straight m subscript 1 end root 
                         straight R over straight x equals fraction numerator square root of straight m subscript 2 end root plus square root of straight m subscript 1 end root over denominator square root of straight m subscript 1 end root end fraction 
    i.e.,                 <pre>uncaught exception: <b>mkdir(): Permission denied (errno: 2) in /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/util/sys/Store.class.php at line #56mkdir(): Permission denied</b><br /><br />in file: /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/util/sys/Store.class.php line 56<br />#0 [internal function]: _hx_error_handler(2, 'mkdir(): Permis...', '/home/config_ad...', 56, Array)
#1 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/util/sys/Store.class.php(56): mkdir('/home/config_ad...', 493)
#2 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/plugin/impl/FolderTreeStorageAndCache.class.php(110): com_wiris_util_sys_Store->mkdirs()
#3 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/plugin/impl/RenderImpl.class.php(231): com_wiris_plugin_impl_FolderTreeStorageAndCache->codeDigest('mml=<math xmlns...')
#4 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/plugin/impl/TextServiceImpl.class.php(59): com_wiris_plugin_impl_RenderImpl->computeDigest(NULL, Array)
#5 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/service.php(19): com_wiris_plugin_impl_TextServiceImpl->service('mathml2accessib...', Array)
#6 {main}</pre>

    Question 320
    CBSEENPH11017643

    Three point masses each of mass m are placed at the vertices of an equilateral triangle of side 1. What is the gravitational field due to three masses at the centriod of triangle?

    Solution
    Let three masses each of mass m be placed at the vertices of an equilateral triangle ABC of side 1 unit.

    Let O be the centroid of the triangle.
    The distance of the centroid from each vertex is, 
                 <pre>uncaught exception: <b>mkdir(): Permission denied (errno: 2) in /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/util/sys/Store.class.php at line #56mkdir(): Permission denied</b><br /><br />in file: /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/util/sys/Store.class.php line 56<br />#0 [internal function]: _hx_error_handler(2, 'mkdir(): Permis...', '/home/config_ad...', 56, Array)
#1 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/util/sys/Store.class.php(56): mkdir('/home/config_ad...', 493)
#2 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/plugin/impl/FolderTreeStorageAndCache.class.php(110): com_wiris_util_sys_Store->mkdirs()
#3 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/plugin/impl/RenderImpl.class.php(231): com_wiris_plugin_impl_FolderTreeStorageAndCache->codeDigest('mml=<math xmlns...')
#4 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/plugin/impl/TextServiceImpl.class.php(59): com_wiris_plugin_impl_RenderImpl->computeDigest(NULL, Array)
#5 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/service.php(19): com_wiris_plugin_impl_TextServiceImpl->service('mathml2accessib...', Array)
#6 {main}</pre> 
    The gravitational field at O due to mass m at A is, 
                    straight E with rightwards arrow on top subscript straight A space equals space Gm over straight r squared 
                         equals fraction numerator 3 Gm over denominator ell squared end fraction comma space space space space space space along OA with rightwards arrow on top 
    Similarly,
                      straight E with rightwards arrow on top subscript straight B equals fraction numerator 3 Gm over denominator ell end fraction comma space space space space space space space space space space along space OB with rightwards arrow on top 
    and            straight E with rightwards arrow on top subscript straight C equals fraction numerator 3 Gm over denominator ell squared end fraction comma space space space space space space space space space along OC with rightwards arrow on top 
    Here, the three coplanar vectors straight E with rightwards arrow on top subscript straight A. space straight E with rightwards arrow on top subscript straight B and space straight E with rightwards arrow on top subscript straight C are equal in magnitude and make an angle of 120 degree with the preceding vector.
    Therefore, the resultant field is zero. 
    Question 321
    CBSEENPH11017644

    Why do the stars appear displaced away from the sun?

    Solution
                       

    The rays that are coming from the stars bent towards the Sun because of the higher gravitational field of the Sun. Hence, they appear displaced. 
    Question 322
    CBSEENPH11017645

    What do you mean by gravitational red shift?

    Solution
    The frequency of light emitted by star decreases and shifts towards red end of the spectrum due to the gravitational field of star.
    The decrease in the frequency of light due to gravity is, 
                               straight v minus straight v apostrophe equals gRv over straight c squared 
    where,
    g is the acceleration due to gravity due to star at the surface of star,
    R is the radius of star, and
    c is the velocity of light.
    Question 323
    CBSEENPH11017647

    Define gravitational potential and gravitational potential energy. What are their units and dimensions?

    Solution

    Gravitational potential:
    Gravitational potential at any point is defined as the amount of work done in moving a unit mass from infinity to that point.
    Dimensional formula: [M0L2T2].
    SI unit: J/ kg
    CGS unit: erg/g
    Gravitational potential energy:
    Gravitational potential energy is the amount of work done against the gravitational force to bring the body from infinity to that point.
    Dimensional formula: [M1L2T2].
    SI unit: J
    CGS unit: erg.

    Question 324
    CBSEENPH11017649

    Show that the gravitational potential due to the earth at a height h from the surface of the earth is
    negative fraction numerator GM over denominator straight R plus straight h end fraction

    Solution
    Gravitational potential at any point is defined as the amount of work done in moving a unit mass from infinity to that point.
    Mass, m = 1 kg

     
    Let a unit mass be placed at P at a distance r from the centre of the earth.
    The force acting on unit mass is, 
               straight F equals fraction numerator GM cross times 1 over denominator straight r squared end fraction equals GM over straight r squared 
    If unit mass is moved by a distance dr towards O, then work done by gravitational force is,
    dW equals straight F with rightwards arrow on top. dr with rightwards arrow on top equals straight F space dr 
        equals GM over straight r squared dr 
    Total work done when the unit mass is moved from infinity to a distance R + h from the centre is, 
    space space space straight W equals integral straight d space straight W equals integral subscript infinity superscript straight R plus straight h end superscript GM over straight r squared dr 

        equals open square brackets negative GM over straight r close square brackets subscript infinity superscript straight R plus straight h end superscript equals negative GM open square brackets fraction numerator 1 over denominator straight R plus straight h end fraction minus 1 over infinity close square brackets 

        <pre>uncaught exception: <b>mkdir(): Permission denied (errno: 2) in /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/util/sys/Store.class.php at line #56mkdir(): Permission denied</b><br /><br />in file: /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/util/sys/Store.class.php line 56<br />#0 [internal function]: _hx_error_handler(2, 'mkdir(): Permis...', '/home/config_ad...', 56, Array)
#1 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/util/sys/Store.class.php(56): mkdir('/home/config_ad...', 493)
#2 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/plugin/impl/FolderTreeStorageAndCache.class.php(110): com_wiris_util_sys_Store->mkdirs()
#3 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/plugin/impl/RenderImpl.class.php(231): com_wiris_plugin_impl_FolderTreeStorageAndCache->codeDigest('mml=<math xmlns...')
#4 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/plugin/impl/TextServiceImpl.class.php(59): com_wiris_plugin_impl_RenderImpl->computeDigest(NULL, Array)
#5 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/service.php(19): com_wiris_plugin_impl_TextServiceImpl->service('mathml2accessib...', Array)
#6 {main}</pre> 
    This work done by definition is equal to gravitational potential. 
    ∴            straight V equals negative fraction numerator GM over denominator straight R plus straight h end fraction
    Hence, the result. 
    Question 325
    CBSEENPH11017658

    As you have learnt in the text, a geostationary satellite orbits the earth at a height of nearly 36,000 km from the surface of the earth. What is the potential due to the earth's gravity at the site of this satellite? (Take the potential energy at infinity to be zero), M = 6.0 x l024kg, radius = 6400 km.)

    Solution

    Given here, 
    Mass of the earth, straight M equals 6.0 space cross times space 10 to the power of 24 kg 
    Radius of earth, straight R equals 6400 space km space equals space 6.4 cross times 10 to the power of 6 straight m 
    Height,<pre>uncaught exception: <b>mkdir(): Permission denied (errno: 2) in /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/util/sys/Store.class.php at line #56mkdir(): Permission denied</b><br /><br />in file: /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/util/sys/Store.class.php line 56<br />#0 [internal function]: _hx_error_handler(2, 'mkdir(): Permis...', '/home/config_ad...', 56, Array)
#1 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/util/sys/Store.class.php(56): mkdir('/home/config_ad...', 493)
#2 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/plugin/impl/FolderTreeStorageAndCache.class.php(110): com_wiris_util_sys_Store->mkdirs()
#3 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/plugin/impl/RenderImpl.class.php(231): com_wiris_plugin_impl_FolderTreeStorageAndCache->codeDigest('mml=<math xmlns...')
#4 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/plugin/impl/TextServiceImpl.class.php(59): com_wiris_plugin_impl_RenderImpl->computeDigest(NULL, Array)
#5 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/service.php(19): com_wiris_plugin_impl_TextServiceImpl->service('mathml2accessib...', Array)
#6 {main}</pre>                     


    Gravitational potential at the site of geostationary satellite is, 
                space space space straight V equals negative fraction numerator GM over denominator left parenthesis straight R plus straight h right parenthesis end fraction 
                    equals negative fraction numerator 6.67 cross times 10 to the power of negative 11 end exponent cross times 6.0 cross times 10 to the power of 24 over denominator 42.4 cross times 10 to the power of 6 end fraction straight J divided by kg 
                   space space equals negative 9.44 space cross times space 10 to the power of 6 straight J divided by kg


    V is the required potential due to Earth;s gravity at the site of this satellite. 

    Question 326
    CBSEENPH11017660

    Derive an expression for the potential energy stored in the mass m placed in the gravitational field of mass M.

    Solution
    Gravitational potential energy stored in the mass m is the work done against the gravitational force of mass M to bring it from infinity to that point without acceleration. 

                         
    Let the mass m be at P, at a distance r from the mass M.
    The gravitational force on m exerted by M is, 
                       straight F equals GMm over straight r squared dr 
    If m is moved by the distance 'dr' towards M, then work done by gravitational force is, 
                      <pre>uncaught exception: <b>mkdir(): Permission denied (errno: 2) in /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/util/sys/Store.class.php at line #56mkdir(): Permission denied</b><br /><br />in file: /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/util/sys/Store.class.php line 56<br />#0 [internal function]: _hx_error_handler(2, 'mkdir(): Permis...', '/home/config_ad...', 56, Array)
#1 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/util/sys/Store.class.php(56): mkdir('/home/config_ad...', 493)
#2 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/plugin/impl/FolderTreeStorageAndCache.class.php(110): com_wiris_util_sys_Store->mkdirs()
#3 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/plugin/impl/RenderImpl.class.php(231): com_wiris_plugin_impl_FolderTreeStorageAndCache->codeDigest('mml=<math xmlns...')
#4 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/plugin/impl/TextServiceImpl.class.php(59): com_wiris_plugin_impl_RenderImpl->computeDigest(NULL, Array)
#5 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/service.php(19): com_wiris_plugin_impl_TextServiceImpl->service('mathml2accessib...', Array)
#6 {main}</pre> 

                           equals GMm over straight r squared dr 
    Total work done when the mass m is moved from infinity to a distance r is, 
    <pre>uncaught exception: <b>mkdir(): Permission denied (errno: 2) in /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/util/sys/Store.class.php at line #56mkdir(): Permission denied</b><br /><br />in file: /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/util/sys/Store.class.php line 56<br />#0 [internal function]: _hx_error_handler(2, 'mkdir(): Permis...', '/home/config_ad...', 56, Array)
#1 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/util/sys/Store.class.php(56): mkdir('/home/config_ad...', 493)
#2 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/plugin/impl/FolderTreeStorageAndCache.class.php(110): com_wiris_util_sys_Store->mkdirs()
#3 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/plugin/impl/RenderImpl.class.php(231): com_wiris_plugin_impl_FolderTreeStorageAndCache->codeDigest('mml=<math xmlns...')
#4 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/plugin/impl/TextServiceImpl.class.php(59): com_wiris_plugin_impl_RenderImpl->computeDigest(NULL, Array)
#5 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/service.php(19): com_wiris_plugin_impl_TextServiceImpl->service('mathml2accessib...', Array)
#6 {main}</pre> 
        <pre>uncaught exception: <b>mkdir(): Permission denied (errno: 2) in /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/util/sys/Store.class.php at line #56mkdir(): Permission denied</b><br /><br />in file: /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/util/sys/Store.class.php line 56<br />#0 [internal function]: _hx_error_handler(2, 'mkdir(): Permis...', '/home/config_ad...', 56, Array)
#1 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/util/sys/Store.class.php(56): mkdir('/home/config_ad...', 493)
#2 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/plugin/impl/FolderTreeStorageAndCache.class.php(110): com_wiris_util_sys_Store->mkdirs()
#3 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/plugin/impl/RenderImpl.class.php(231): com_wiris_plugin_impl_FolderTreeStorageAndCache->codeDigest('mml=<math xmlns...')
#4 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/plugin/impl/TextServiceImpl.class.php(59): com_wiris_plugin_impl_RenderImpl->computeDigest(NULL, Array)
#5 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/service.php(19): com_wiris_plugin_impl_TextServiceImpl->service('mathml2accessib...', Array)
#6 {main}</pre> 

        <pre>uncaught exception: <b>mkdir(): Permission denied (errno: 2) in /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/util/sys/Store.class.php at line #56mkdir(): Permission denied</b><br /><br />in file: /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/util/sys/Store.class.php line 56<br />#0 [internal function]: _hx_error_handler(2, 'mkdir(): Permis...', '/home/config_ad...', 56, Array)
#1 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/util/sys/Store.class.php(56): mkdir('/home/config_ad...', 493)
#2 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/plugin/impl/FolderTreeStorageAndCache.class.php(110): com_wiris_util_sys_Store->mkdirs()
#3 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/plugin/impl/RenderImpl.class.php(231): com_wiris_plugin_impl_FolderTreeStorageAndCache->codeDigest('mml=<math xmlns...')
#4 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/plugin/impl/TextServiceImpl.class.php(59): com_wiris_plugin_impl_RenderImpl->computeDigest(NULL, Array)
#5 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/service.php(19): com_wiris_plugin_impl_TextServiceImpl->service('mathml2accessib...', Array)
#6 {main}</pre> 
    This work by definition is equal to gravitational potential energy, 
    ∴        <pre>uncaught exception: <b>mkdir(): Permission denied (errno: 2) in /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/util/sys/Store.class.php at line #56mkdir(): Permission denied</b><br /><br />in file: /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/util/sys/Store.class.php line 56<br />#0 [internal function]: _hx_error_handler(2, 'mkdir(): Permis...', '/home/config_ad...', 56, Array)
#1 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/util/sys/Store.class.php(56): mkdir('/home/config_ad...', 493)
#2 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/plugin/impl/FolderTreeStorageAndCache.class.php(110): com_wiris_util_sys_Store->mkdirs()
#3 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/plugin/impl/RenderImpl.class.php(231): com_wiris_plugin_impl_FolderTreeStorageAndCache->codeDigest('mml=<math xmlns...')
#4 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/plugin/impl/TextServiceImpl.class.php(59): com_wiris_plugin_impl_RenderImpl->computeDigest(NULL, Array)
#5 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/service.php(19): com_wiris_plugin_impl_TextServiceImpl->service('mathml2accessib...', Array)
#6 {main}</pre> 
    where, V is the gravitational potential due to mass M at a distance r. 
    Question 327
    CBSEENPH11017661

    Is the gravitational potential energy of our galaxy milky way positive or negative? Explain.

    Solution
    The gravitational potential energy of milky way is negative because different stars in a milky way are bounded by gravitational force of attraction. Therefore, the potential energy of milky way is negative. 
    Question 328
    CBSEENPH11017662

    Can the gravitational potential have positive value?

    Solution
    Yes, the positive or negative value of potential depends on the reference point chosen.
    For e.g., if the surface of the earth is taken as the reference point for gravitational potential, then the gravitational potential above the surface of the earth is positive. 
    Question 329
    CBSEENPH11017665

    A body of mass m is taken from distance r1 to r2 from the centre of the earth. What is the work required to displace the body?

    Solution
    The force acting on mass m when at a distance r from the centre of the earth is,
                          straight F equals GMm over straight r squared 
    This force must be applied in the upward direction, to displace the body in upward direction.
    If the body is displaced by a small distance 'dr' then work done is,
    space dW equals GMm over straight r squared dr 
    Total work done in displacing the body from r1 to r2 is,
    dW equals integral subscript straight r subscript 1 end subscript superscript straight r subscript 2 end superscript GMm over straight r squared dr equals open square brackets negative GMm over straight r close square brackets subscript straight r subscript 1 end subscript superscript straight r subscript 2 end superscript 
        equals GMm open square brackets 1 over straight r subscript 1 minus 1 over straight r subscript 2 close square brackets, is the amount of work required to displace the body. 
    Question 330
    CBSEENPH11017667

    A body of mass m is taken from the surface of the earth to a height h. What is the change in the potential energy of mass m? Is it equal to mgh?

    Solution
    At a distance r from the centre of the Earth, gravitational potential energy at a point is given by,
                        straight U equals negative GMm over straight r 
    At the surface of the earth, 
                      space straight U subscript 1 equals negative GMm over straight R 
    Gravitational potential energy at height h from the surface of the earth is, 
                       straight U subscript 2 equals negative fraction numerator GMm over denominator straight R plus straight h end fraction 
    Change in potential energy is, 
    increment straight U equals straight U subscript 2 minus straight U subscript 1 equals GMm open square brackets 1 over straight R minus fraction numerator 1 over denominator straight R plus straight h end fraction close square brackets 
                      <pre>uncaught exception: <b>mkdir(): Permission denied (errno: 2) in /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/util/sys/Store.class.php at line #56mkdir(): Permission denied</b><br /><br />in file: /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/util/sys/Store.class.php line 56<br />#0 [internal function]: _hx_error_handler(2, 'mkdir(): Permis...', '/home/config_ad...', 56, Array)
#1 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/util/sys/Store.class.php(56): mkdir('/home/config_ad...', 493)
#2 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/plugin/impl/FolderTreeStorageAndCache.class.php(110): com_wiris_util_sys_Store->mkdirs()
#3 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/plugin/impl/RenderImpl.class.php(231): com_wiris_plugin_impl_FolderTreeStorageAndCache->codeDigest('mml=<math xmlns...')
#4 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/plugin/impl/TextServiceImpl.class.php(59): com_wiris_plugin_impl_RenderImpl->computeDigest(NULL, Array)
#5 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/service.php(19): com_wiris_plugin_impl_TextServiceImpl->service('mathml2accessib...', Array)
#6 {main}</pre> 
                       equals mgh open parentheses fraction numerator straight R over denominator straight R plus straight h end fraction close parentheses 
    Change in potential energy is not equal to mgh.
    But if h<<R, then R + h = h
    ∴ Change in potential energy, increment straight U equals mgh 
    Question 331
    CBSEENPH11017669

    Two masses of 25 kg and 75kg are separated by a distance 20m. Find the gravitational potential at a point at a distance of 5 m from 25kg mass.

    Solution

    Given,
    Mass of first body, m1 = 25 kg
    Mass of second body, m2 = 75 kg
    r1 = 5 m and r2 = 15 m
               
    Let straight V subscript 1 space and space straight V subscript 2 be the gravitational potential due to masses 25 kg and 75 kg respectively. 
    Using the formula for gravitational potential,
      straight V subscript 1 equals negative GM subscript 1 over straight r subscript 1 equals fraction numerator 6.66 cross times 10 to the power of negative 11 end exponent cross times 25 over denominator 5 end fraction 
         equals space 3.33 space cross times space 10 to the power of negative 10 end exponent straight J divided by kg            

    and
    space space space space space space space space straight V subscript 2 equals negative GM subscript 2 over straight r subscript 2 equals fraction numerator 6.66 cross times 10 to the power of negative 11 end exponent cross times 75 over denominator 15 end fraction 
             equals 3.33 cross times 10 to the power of negative 10 end exponent straight J divided by kg 
    The total potential at the given point is,
    straight V equals straight V subscript 1 plus straight V subscript 2 equals 6.66 cross times 10 to the power of negative 10 end exponent straight J divided by kg, is the gravitational potential at a point at a distance of 5m from the 25 kg mass. 

    Question 332
    CBSEENPH11017670

    A body of mass m is moved from surface of the earth to a height equal to radius of the earth. Find the increase in gravitational potential energy of mass.

    Solution
    Gravitational potential energy at the surface of the earth is,
                       straight U subscript 0 equals negative GMm over straight R
    Gravitational potential energy at height R from the surface of the earth is, 
    straight U equals negative fraction numerator GMm over denominator straight R plus straight R end fraction equals negative fraction numerator GMm over denominator 2 straight R end fraction 
    Change in potential energy, 
    <pre>uncaught exception: <b>mkdir(): Permission denied (errno: 2) in /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/util/sys/Store.class.php at line #56mkdir(): Permission denied</b><br /><br />in file: /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/util/sys/Store.class.php line 56<br />#0 [internal function]: _hx_error_handler(2, 'mkdir(): Permis...', '/home/config_ad...', 56, Array)
#1 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/util/sys/Store.class.php(56): mkdir('/home/config_ad...', 493)
#2 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/plugin/impl/FolderTreeStorageAndCache.class.php(110): com_wiris_util_sys_Store->mkdirs()
#3 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/plugin/impl/RenderImpl.class.php(231): com_wiris_plugin_impl_FolderTreeStorageAndCache->codeDigest('mml=<math xmlns...')
#4 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/plugin/impl/TextServiceImpl.class.php(59): com_wiris_plugin_impl_RenderImpl->computeDigest(NULL, Array)
#5 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/service.php(19): com_wiris_plugin_impl_TextServiceImpl->service('mathml2accessib...', Array)
#6 {main}</pre> 
         equals fraction numerator GMm over denominator 2 straight R end fraction equals mgR over 2
    Question 333
    CBSEENPH11017672

    Two masses m1 and m2 are separated by a distance R. What is the gravitational potential at the centre of mass?

    Solution
    Let the centre of mass be at a distance r1, from m1 and r2 from m2.
    Then,
    straight r subscript 1 equals fraction numerator straight R space straight m subscript 2 over denominator straight m subscript 1 plus straight m subscript 2 end fraction     and      straight r subscript 2 equals fraction numerator Rm subscript 1 over denominator straight m subscript 1 plus straight m subscript 2 end fraction 
             
    Gravitational potential at the centre of mass is, 
    straight V equals straight V subscript 1 plus straight V subscript 2 equals negative Gm subscript 1 over straight r subscript 1 minus Gm subscript 2 over straight r subscript 2 
     equals negative straight G open square brackets fraction numerator straight m subscript 1 left parenthesis straight m subscript 1 plus straight m subscript 2 right parenthesis over denominator straight R space straight m subscript 2 end fraction plus fraction numerator straight m subscript 2 left parenthesis straight m subscript 1 plus straight m subscript 2 right parenthesis over denominator straight R space straight m subscript 1 end fraction close square brackets 
     equals negative fraction numerator straight G left parenthesis straight m subscript 1 plus straight m subscript 2 right parenthesis over denominator straight R space end fraction open parentheses straight m subscript 1 over straight m subscript 2 plus straight m subscript 2 over straight m subscript 1 close parentheses 
     equals negative fraction numerator straight G left parenthesis straight m subscript 1 plus straight m subscript 2 right parenthesis left parenthesis straight m subscript 1 squared plus straight m subscript 2 squared right parenthesis over denominator straight R left parenthesis straight m subscript 1 straight m subscript 2 right parenthesis end fraction 
    Question 334
    CBSEENPH11017678

    Three point masses each of mass m are placed at the vertices of an equilateral triangle of side l. What is the gravitational potential due to three masses at the centriod of the triangle?

    Solution
    Let three masses each of mass m be placed at the vertices of an equilateral triangle ABC of side l.
    Let O be the centroid of the triangle.
                  
    The distance of centroid from each vertex is, 
                r = AO = BO = CO = space ell divided by square root of 3 
    The gravitational potential at O due to mass m at A is,
                straight V subscript straight A equals negative Gm over straight r equals negative fraction numerator square root of 3 Gm over denominator ell end fraction 
    Similarly, 
     
                 Error converting from MathML to accessible text. 
    and       Error converting from MathML to accessible text. 
    Therefore total potential due to all the three masses is,
    Error converting from MathML to accessible text.
    Question 335
    CBSEENPH11017680

    Two masses m1 and m2 are separated by a distance R. What is the gravitational potential at a point where gravitational field is zero?

    Solution
    Given, the two masses m1 and m2 are separated by a distance R.
    Let m1 be at a distance of x from L and m2 be at a distance (R-x) from L. 


    Question 336
    CBSEENPH11017681

    What is wrong with the relation,
    GMm open parentheses 1 over straight r subscript 2 minus 1 over straight r subscript 1 close parentheses equals mg left parenthesis straight r subscript 2 minus straight r subscript 1 right parenthesis
    Write the correct relation.

    Solution
    Acceleration due to gravity is not the same everywhere.
    It is different at different points. 
    GMm open parentheses 1 over straight r subscript 2 minus 1 over straight r subscript 1 close parentheses space equals space GMm over straight r subscript 2 minus GMm over straight r subscript 1 
                           space equals GM over straight r subscript 2 squared mr subscript 2 minus GM over straight r subscript 1 squared mr subscript 1 
                            equals straight g subscript 2 mr subscript 2 minus straight g subscript 1 mr subscript 1 
                             equals straight m left parenthesis straight g subscript 2 straight r subscript 2 minus straight g subscript 1 straight r subscript 1 right parenthesis 
    where,
    g1 and g2 are the accelerations due to gravity at r1 and r2 respectively and are not equal. 
                                 
                             
    Question 337
    CBSEENPH11017682

    A rocket is fired vertically with a speed of v km/s from the surface of the earth. Find the maximum height attained by the rocket before it returns to the earth. Take GM =gR2 at the surface of the earth.

    Solution
    Let a rocket of mass m be projected from the surface of the earth with velocity v less than escape velocity of satellite.
    Let h be the maximum height attained by satellite before it returns to the surface of the earth.
    Using law of conservation of the energy,
    Total energy at surface of earth = total energy at height h
    i.e.         1 half mv squared minus GMm over straight R equals 0 minus fraction numerator GMm over denominator straight R plus straight h end fraction 
    rightwards double arrow           space space 1 half mv squared minus mgR equals 0 minus fraction numerator mgR squared over denominator straight R plus straight h end fraction 
                       space space space space fraction numerator straight v squared minus 2 gR over denominator 2 end fraction equals negative fraction numerator gR squared over denominator straight R plus straight h end fraction 
    rightwards double arrow                      space space space space straight R plus straight h equals fraction numerator 2 gR squared over denominator left parenthesis 2 gR minus straight v squared right parenthesis end fraction 
    ∴        straight h equals fraction numerator 2 gR squared over denominator left parenthesis 2 gR minus straight v squared right parenthesis end fraction minus straight R equals fraction numerator Rv squared over denominator 2 gR minus straight v squared end fraction
    'h' is the maximum height attained by the rocket before it returns to the Earth. 
    Question 338
    CBSEENPH11017685

    A rocket is fired vertically with a speed of 5 km/s from the earth’s surface. How far from the earth’s surface does the rocket go before returning to the earth? Mass of the earth = 6.0 x 1024 kg; mean radius of the earth = 6.4 x 106 m; G =6.67 x 10-11 N m2 kg-2.
    Acceleration due to gravity at the surface of the earth = 9.8 m/s2.

    Solution
    When a body is thrown a body in the upward direction, the kinetic energy of the body converts into potential energy.
    Potential energy of the body at ground is, 
             straight U subscript straight o equals negative GMm over straight R 
    If the body is raised to a height h from the earth, then potential energy at height h is, 
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#1 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/util/sys/Store.class.php(56): mkdir('/home/config_ad...', 493)
#2 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/plugin/impl/FolderTreeStorageAndCache.class.php(110): com_wiris_util_sys_Store->mkdirs()
#3 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/plugin/impl/RenderImpl.class.php(231): com_wiris_plugin_impl_FolderTreeStorageAndCache->codeDigest('mml=<math xmlns...')
#4 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/plugin/impl/TextServiceImpl.class.php(59): com_wiris_plugin_impl_RenderImpl->computeDigest(NULL, Array)
#5 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/service.php(19): com_wiris_plugin_impl_TextServiceImpl->service('mathml2accessib...', Array)
#6 {main}</pre> 
    Therefore increase in potential energy is, 
    increment straight U equals GMm open square brackets 1 over straight R minus fraction numerator 1 over denominator straight R plus straight h end fraction close square brackets space

space space space space space space space equals space GMm over straight R fraction numerator straight h over denominator left parenthesis straight R plus straight h right parenthesis end fraction 
          equals GMm over straight R squared fraction numerator Rh over denominator left parenthesis straight R plus straight h right parenthesis end fraction
equals mgh fraction numerator straight R over denominator left parenthesis straight R plus straight h right parenthesis end fraction 
    The increase in potential energy is at the cost of kinetic energy, therefore 
                     space space space space 1 half mv squared equals mgh fraction numerator straight R over denominator straight R plus straight h end fraction 
    rightwards double arrow                 straight R over straight h plus 1 equals fraction numerator 2 gR over denominator straight v squared end fraction 
    rightwards double arrow                     <pre>uncaught exception: <b>mkdir(): Permission denied (errno: 2) in /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/util/sys/Store.class.php at line #56mkdir(): Permission denied</b><br /><br />in file: /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/util/sys/Store.class.php line 56<br />#0 [internal function]: _hx_error_handler(2, 'mkdir(): Permis...', '/home/config_ad...', 56, Array)
#1 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/util/sys/Store.class.php(56): mkdir('/home/config_ad...', 493)
#2 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/plugin/impl/FolderTreeStorageAndCache.class.php(110): com_wiris_util_sys_Store->mkdirs()
#3 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/plugin/impl/RenderImpl.class.php(231): com_wiris_plugin_impl_FolderTreeStorageAndCache->codeDigest('mml=<math xmlns...')
#4 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/plugin/impl/TextServiceImpl.class.php(59): com_wiris_plugin_impl_RenderImpl->computeDigest(NULL, Array)
#5 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/service.php(19): com_wiris_plugin_impl_TextServiceImpl->service('mathml2accessib...', Array)
#6 {main}</pre> 
    rightwards double arrow                      <pre>uncaught exception: <b>mkdir(): Permission denied (errno: 2) in /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/util/sys/Store.class.php at line #56mkdir(): Permission denied</b><br /><br />in file: /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/util/sys/Store.class.php line 56<br />#0 [internal function]: _hx_error_handler(2, 'mkdir(): Permis...', '/home/config_ad...', 56, Array)
#1 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/util/sys/Store.class.php(56): mkdir('/home/config_ad...', 493)
#2 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/plugin/impl/FolderTreeStorageAndCache.class.php(110): com_wiris_util_sys_Store->mkdirs()
#3 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/plugin/impl/RenderImpl.class.php(231): com_wiris_plugin_impl_FolderTreeStorageAndCache->codeDigest('mml=<math xmlns...')
#4 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/plugin/impl/TextServiceImpl.class.php(59): com_wiris_plugin_impl_RenderImpl->computeDigest(NULL, Array)
#5 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/service.php(19): com_wiris_plugin_impl_TextServiceImpl->service('mathml2accessib...', Array)
#6 {main}</pre> 
    Here, we have
    straight g equals 9.8 straight m divided by straight s squared comma space straight R space equals space 6.4 cross times 10 to the power of 6 straight m 
    straight v equals 5000 straight m divided by straight s 
    Therefore, 
    straight h equals fraction numerator 6.4 cross times 10 to the power of 6 cross times left parenthesis 5 cross times 10 cubed right parenthesis squared over denominator 2 cross times 9.8 cross times 6.4 cross times 10 to the power of 6 minus left parenthesis 5 cross times 10 cubed right parenthesis squared end fraction 
      <pre>uncaught exception: <b>file_put_contents(/home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/plugin/web/../../../../../../formulas/f5/b9/a9dde345a6371f63ab6573968dce.ini): failed to open stream: Permission denied (errno: 2) in /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/sys/io/File.class.php at line #12file_put_contents(/home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/plugin/web/../../../../../../formulas/f5/b9/a9dde345a6371f63ab6573968dce.ini): failed to open stream: Permission denied</b><br /><br />in file: /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/sys/io/File.class.php line 12<br />#0 [internal function]: _hx_error_handler(2, 'file_put_conten...', '/home/config_ad...', 12, Array)
#1 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/sys/io/File.class.php(12): file_put_contents('/home/config_ad...', 'mml=<math xmlns...')
#2 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/util/sys/Store.class.php(48): sys_io_File::saveContent('/home/config_ad...', 'mml=<math xmlns...')
#3 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/plugin/impl/FolderTreeStorageAndCache.class.php(112): com_wiris_util_sys_Store->write('mml=<math xmlns...')
#4 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/plugin/impl/RenderImpl.class.php(231): com_wiris_plugin_impl_FolderTreeStorageAndCache->codeDigest('mml=<math xmlns...')
#5 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/plugin/impl/TextServiceImpl.class.php(59): com_wiris_plugin_impl_RenderImpl->computeDigest(NULL, Array)
#6 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/service.php(19): com_wiris_plugin_impl_TextServiceImpl->service('mathml2accessib...', Array)
#7 {main}</pre>

    That is, the rocket travels a distance of 1.6 x 106 m before returning to the Earth's surface. 
    Question 339
    CBSEENPH11017686

    Show that gravitational field due to a point m ass M at any point is equal to negative gradient of gravitational potential at that point.

    Solution
    The gravitational potential due to point mass M at a distance r is, 
                           straight V equals negative GM over straight r 
    Differentiating the above equation w.r.t. r, 
                          dV over dr equals negative GM over straight r             ...(1) 
    The gravitational field due to point mass M at a distance r is, 
                        straight E equals negative GM over straight r squared                   ...(2) 
    The negative sign is due to sign convention that gravitational field is directed towards the mass.
    Therefore, from equation (1) and (2), we get
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#1 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/util/sys/Store.class.php(56): mkdir('/home/config_ad...', 493)
#2 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/plugin/impl/FolderTreeStorageAndCache.class.php(110): com_wiris_util_sys_Store->mkdirs()
#3 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/plugin/impl/RenderImpl.class.php(231): com_wiris_plugin_impl_FolderTreeStorageAndCache->codeDigest('mml=<math xmlns...')
#4 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/plugin/impl/TextServiceImpl.class.php(59): com_wiris_plugin_impl_RenderImpl->computeDigest(NULL, Array)
#5 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/service.php(19): com_wiris_plugin_impl_TextServiceImpl->service('mathml2accessib...', Array)
#6 {main}</pre>
    Hence, the result. 
    Question 340
    CBSEENPH11017688

    The gravitational potential due to some distribution of mass is given by straight V equals negative straight k over straight r to the power of 3 divided by 2 end exponent. Find the gravitational field due to that distribution of mass.

    Solution

    Gravitational field is given by, <pre>uncaught exception: <b>mkdir(): Permission denied (errno: 2) in /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/util/sys/Store.class.php at line #56mkdir(): Permission denied</b><br /><br />in file: /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/util/sys/Store.class.php line 56<br />#0 [internal function]: _hx_error_handler(2, 'mkdir(): Permis...', '/home/config_ad...', 56, Array)
#1 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/util/sys/Store.class.php(56): mkdir('/home/config_ad...', 493)
#2 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/plugin/impl/FolderTreeStorageAndCache.class.php(110): com_wiris_util_sys_Store->mkdirs()
#3 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/plugin/impl/RenderImpl.class.php(231): com_wiris_plugin_impl_FolderTreeStorageAndCache->codeDigest('mml=<math xmlns...')
#4 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/plugin/impl/TextServiceImpl.class.php(59): com_wiris_plugin_impl_RenderImpl->computeDigest(NULL, Array)
#5 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/service.php(19): com_wiris_plugin_impl_TextServiceImpl->service('mathml2accessib...', Array)
#6 {main}</pre> 
    Gravitational P.E is given by, straight V equals negative straight k over straight r to the power of 3 divided by 2 end exponent 
    We know that field the negative gradient of gravitational potential energy. 
    ∴     straight E equals negative dV over dr equals fraction numerator negative straight d over denominator dr end fraction open parentheses negative straight k over straight r to the power of 3 divided by 2 end exponent close parentheses equals negative fraction numerator 3 straight k over denominator 2 straight r to the power of 5 divided by 2 end exponent end fraction 
    This is the required expression for gravitational field due to the distribution of mass. 

    Question 341
    CBSEENPH11017690

    A comet orbits the sun in a highly elliptical orbit. Does the comet have a constant:

    (a) linear speed

    (b) angular speed

    (c) angular momentum

    (d) kinetic energy

    (e) potential energy

    (f) total energy throughout its orbit?

    Neglect any mass loss of the comet when it comes very close to the sun.

    Solution

    The torque exerted by the sun on the comet is zero.
    Therefore, angular momentum of comet remains constant.
    i.e.,           mvr = constant
    rightwards double arrow               vr = constant [because m is constant].
    In an elliptical orbit distance of the comet from the sun changes, therefore, the speed of the comet also changes.
    (a) Linear speed of the comet is not constant.
    (b) Angular speed ω of the comet is not constant.
    (c) Angular momentum of the comet is constant.
    (d) Kinetic energy of comet 1/2 mv2 changes because the speed of the comet changes.
    (e) Potential energy of the comet is <pre>uncaught exception: <b>mkdir(): Permission denied (errno: 2) in /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/util/sys/Store.class.php at line #56mkdir(): Permission denied</b><br /><br />in file: /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/util/sys/Store.class.php line 56<br />#0 [internal function]: _hx_error_handler(2, 'mkdir(): Permis...', '/home/config_ad...', 56, Array)
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#6 {main}</pre>
    (f) Total energy of the comet remains constant. 

     
    Question 346
    CBSEENPH11017696

     A non-homogeneous sphere of radius R has the following density relation: 

    straight rho space equals space straight rho subscript straight o space space space space space space space space space space space space semicolon space 0 space less than space straight r space less or equal than space straight R divided by 2

space space space space equals space 2 straight rho subscript straight o space space space space space space space space space semicolon space straight R over 2 space less than thin space straight r space less or equal than space fraction numerator 3 straight R over denominator 2 end fraction space

space space space space equals space 0 space space space space space space space space space space space space space semicolon space space straight r space greater than thin space fraction numerator 3 straight R over denominator 2 end fraction space

    What is the total mass of sphere and gravitational field at the surface of sphere?



     

    Solution
                    
    According to the figure given above, we have
    Mass of inner sphere of radius R/2 is, 
          straight m subscript 1 space equals 4 over 3 straight pi open parentheses straight R over 2 close parentheses cubed space straight rho subscript straight o

space space space space space space equals fraction numerator pi R cubed straight rho subscript straight o over denominator 6 end fraction
    Mass of outer sphere whose outer radius is fraction numerator 3 straight R over denominator 2 end fraction and inner straight R over 2 is,
     straight m subscript 2 equals 4 over 3 straight pi open square brackets open parentheses fraction numerator 3 straight R over denominator 2 end fraction close parentheses cubed minus open parentheses straight R over 2 close parentheses cubed close square brackets 2 straight rho subscript straight o 
          equals 4 over 3 straight pi fraction numerator 26 straight R cubed over denominator 8 end fraction 2 straight rho subscript straight o equals 52 over 6 straight R cubed pi rho subscript straight o 
    Total mass of the sphere is,
    straight m subscript 1 plus straight m subscript 2 space equals fraction numerator pi R cubed straight rho subscript straight o over denominator 6 end fraction plus 52 over 6 pi R cubed straight rho subscript straight o

space space space space space space space space space space space space space space space equals 53 over 6 pi R cubed straight rho subscript straight o 
    Now gravitational field at the surface of sphere is,
    space straight E subscript straight g equals fraction numerator G M over denominator straight r squared end fraction
space space space space space space equals straight G 53 over 6 fraction numerator pi R cubed straight rho subscript straight o over denominator open parentheses begin display style 3 over 2 straight R end style close parentheses squared end fraction
space space space space space space equals 106 over 27 pi R rho subscript straight o straight G
    Question 347
    CBSEENPH11017700

    Two stars each of one solar mass (= 2 x 1030 kg) are approaching each other for a head-on collision. When they are a distance of 109km, their speeds are negligible. What is the speed with which they collide? The radius of each star is 104 km. Assume the stars to remain undistorted until they collide. (G = 6.67 x 10-11Nm2/kg2)

    Solution

    Let stack straight v subscript 1 with rightwards arrow on top space and space stack straight v subscript 2 with rightwards arrow on top be the velocities of two stars when they collide.
    According to the law of conservation of momentum, 
                   straight M stack straight v subscript 1 with rightwards arrow on top plus straight M stack straight v subscript 2 with rightwards arrow on top equals 0 
    rightwards double arrow                       stack straight v subscript 1 with rightwards arrow on top space equals space minus stack straight v subscript 2 with rightwards arrow on top 
    rightwards double arrow                       space straight v subscript 1 equals straight v subscript 2 equals straight v 
    According to the law of conservation of energy, 
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#6 {main}</pre> 
    We have  
    Mass of the star, straight M equals 2 cross times 10 to the power of 30 kg 
    Distance between the stars, straight r subscript 1 equals 10 to the power of 9 km equals 10 to the power of 12 straight m 
    Radius of star, straight r equals 10 to the power of 4 km equals 10 to the power of 7 straight m 
    The distance between two stars when they collide is,
                   2r = 2 x 107m
    ∴ Speed with which the stars collide is given by, 

    straight v equals square root of GM open square brackets 1 over straight r subscript 2 minus 1 over straight r subscript 1 close square brackets end root 
     equals square root of 6.67 cross times 10 to the power of negative 11 end exponent cross times 2 cross times 10 to the power of 30 cross times open square brackets fraction numerator 1 over denominator 2 cross times 10 to the power of 7 end fraction minus 1 over 10 to the power of 12 close square brackets end root 
     equals 2.58 cross times 10 to the power of 6 space straight m divided by straight s

     

    Question 348
    CBSEENPH11017703

    Two heavy spheres each of mass 100 kg and radius 0.10 m are placed 1.0 m apart on a horizontal table. What is the gravitational force and potential at the mid point of the line joining the centers of the spheres? Is an object placed at that point in equilibrium? If so, is the equilibrium stable or unstable?

    Solution
    We have,
    Mass of the sphere, M1=M2 = 100kg
    Radius of the sphere, r = 0.10 m
    Distance between the spheres, AB = R = 1m
               
    Gravitational field due to mass M1, at P 
    stack straight g subscript 1 with rightwards arrow on top space equals space fraction numerator GM subscript 1 over denominator left parenthesis PA right parenthesis squared end fraction space 
        equals fraction numerator 6.67 cross times 10 to the power of negative 11 end exponent cross times 10 squared over denominator left parenthesis 0.5 right parenthesis squared end fraction space along space PA with rightwards arrow on top
        equals 2.67 cross times 10 to the power of negative 8 end exponent straight N divided by kg space along space PA with rightwards arrow on top 
    Similarly, gravitational field due to mass M2 at P is, 
    stack straight g subscript 2 with rightwards arrow on top equals fraction numerator GM subscript 2 over denominator left parenthesis PB right parenthesis squared end fraction space 
       equals 2.67 cross times 10 to the power of negative 8 end exponent space straight N divided by kg space comma space along space PB with rightwards arrow on top

    The gravitational field due to two masses are equal and opposite, therefore net gravitational field at P is zero.
    The gravitational potential at P is,
    straight V equals negative fraction numerator G M subscript 1 over denominator P A end fraction minus fraction numerator G M subscript 2 over denominator P B end fraction equals negative straight G open square brackets fraction numerator straight M subscript 1 over denominator P A end fraction plus fraction numerator straight M subscript 2 over denominator P B end fraction close square brackets 
      equals negative 6.67 cross times 10 to the power of negative 11 end exponent open square brackets fraction numerator 100 over denominator 0.5 end fraction plus fraction numerator 100 over denominator 0.5 end fraction close square brackets
      equals negative 2.67 cross times 10 to the power of negative 8 end exponent straight J divided by kg 
    The net gravitational field at P is zero, therefore the object placed at P will be in equilibrium.
    The equilibrium is an unstable equilibrium.

    Question 349
    CBSEENPH11017704

    Define escape velocity.

    Solution
    Escape velocity is the minimum velocity with which a body at the surface of the earth has to be projected vertically upwards inorder to escape it from the gravitational field of the earth.
    Question 350
    CBSEENPH11017705

    What is the escape velocity from the surface of the earth?

    Solution
    In the case of Earth, 
    g = 9.8 m/s2
    R = 6.4 x 106 m
    And using the formula of escape velocity, we have
    straight v subscript straight e space equals space square root of 2 gr end root space

So comma space

straight v subscript straight e space equals space square root of 2 space straight x space 9.8 space straight x space 6.4 space straight x space 10 to the power of 6 end root

space space space space equals space 11.2 space straight x space 10 cubed space straight m divided by straight s space

space space space space equals 11.2 space km divided by straight s 
    Question 351
    CBSEENPH11017706

    Does the escape velocity from the surface of the earth depend on the direction of projection?

    Solution
    No, the escape velocity from the surface of the earth does not depend on the direction of projection.
    Escape velocity is independent of the mass and direction of projection of the body from the surface of the earth. 
    Question 352
    CBSEENPH11017708

    Does the escape speed of a body from the earth depend on the mass of the body?

    Solution

    Escape velocity is independent of the mass of the body which is projected from the surface of the earth. 

    Question 353
    CBSEENPH11017709

    Does the escape speed of a body from the earth depend on the height of the location from where the body is launched?

    Solution
    Yes, escape velocity is dependent on the height from which the body is launched. 
    Question 354
    CBSEENPH11017710

    An ant and an elephant are to project out from gravitational field of the earth. Which of the two must be projected with greater velocity?

    Solution
    Both the bodies will be projected with the same velocity.
    Escape velocity is independent of the mass of the body. 
    Question 355
    CBSEENPH11017711

    Is there any hydrogen in the atmosphere of the earth?

    Solution
    No, the atmosphere of the earth does not have any hydrogen.
    Hydrogen being a lighter gas in the atmosphere of earth escapes into the outer space of the planet. 
    Question 356
    CBSEENPH11017712

    Which escapes faster from the earth’s atmospheres - hydrogen or helium?

    Solution
    Hydrogen being a lighter gas than helium, escapes faster than helium from the earth’s atmosphere.
    Question 357
    CBSEENPH11017714

    If the K.E of a satellite revolving in an orbit close to the Earth is doubled, will the satellite escape? 

    Solution
    When the K.E is doubled, the velocity of the satellite will become equal to the escape velocity. Hence, the satellite will be able to escape. 
    Question 358
    CBSEENPH11017715
    Question 359
    CBSEENPH11017716

    Name two factors which determine whether a planet has an atmosphere or not.

    Solution
    Factors which determine whether a planet has an atmosphere or not are the following:
    i) value of acceleration due to gravity of planet, and
    ii) its surface temperature
    Question 360
    CBSEENPH11017717

    What is the shape of orbits in which the planets revolve around the sun?

    Solution
    The planets revolve around the sun in elliptical orbits.
    Question 361
    CBSEENPH11017718

    What is the shape of orbits in which the planets revolve around the sun?

    Solution
    The planets revolve around the sun in elliptical orbits.
    Question 362
    CBSEENPH11017719

    What is apogee and perigee?

    Solution
    The planets move around the sun in an elliptical orbit.
    The farthest point of the planet from the sun is called apogee while its nearest point distance from the sun is called perigee.
    Question 364
    CBSEENPH11017721

    Define orbital velocity.

    Solution
    Orbital velocity is the velocity required to keep a satellite orbiting around the earth. 
    Question 365
    CBSEENPH11017722

    Does the orbital velocity depend upon the mass of satellite?

    Solution
    Orbital velocity does not depend on upon the mass of the satellite. 
    Question 366
    CBSEENPH11017723

    Does the orbital velocity of satellite depend upon the mass of planet?

    Solution
    Yes, orbital velocity depends on upon the mass of the planet.
    Question 367
    CBSEENPH11017724

    What is the orbital velocity of satellite revolving in a circular orbit at a height h from the surface of planet of mass M and radius R?

    Solution

    Orbital velocity of a satellite revolving in a circular orbit at a height h from the surface of the planet is, 
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#6 {main}</pre> 

    Question 368
    CBSEENPH11017725

    If the earth jumps from its present orbit to the orbit of Jupiter, then how will its velocity change?

    Solution
    Orbital velocity is inversely proportional to the distance. 
    Therefore, orbital velocity of the earth will decrease when Earth jumps from it's present orbit to Jupiter's orbit. 
    Question 369
    CBSEENPH11017726

    Is the motion of satellite in its orbit uniform or uniform accelerated?

    Solution
    The motion of a satellite in its orbit is neither uniform nor uniformly accelerated because the direction of acceleration is continuously changing.
    Question 370
    CBSEENPH11017727

    What is the total energy of satellite revolving around the earth?

    Solution

    The total energy of satellite =space space minus GMm over straight r
    where
    m is the mass of satellite,
    M is the mass of the earth and
    r is the radius of orbit. 

    Question 371
    CBSEENPH11017728

    What will happen if the moon stops revolving around the earth?

    Solution
    The moon will fall on the earth if it will stop revolving around the Earth. 
    Question 372
    CBSEENPH11017729

    A satellite is just revolving around the earth. What is the orbital velocity of satellite?

    Solution
    When a satellite is revolving around the earth, h < < R, then
    r = R + h ~ R, and
    v = vo
    Putting the value of g = 9.8 m/s2
    R = 6.4 x 106
    We have, 
     straight v subscript straight o space equals space square root of space 9.8 space straight x space 6.4 space straight x space 10 to the power of 6 end root space

space space space space space equals space 7.92 space straight x space 10 cubed space straight m divided by straight s

space space space space space equals space 7.92 space km space divided by straight s 
    Question 373
    CBSEENPH11017731

    Does a satellite need fuel to orbit around the planet?

    Solution
    No, a satellite does not need fuel once they have been launched into the orbit of the planet. 
    In order to get the satellite into the orbit, the rocket needs a large amount of fuel. 
    Question 374
    CBSEENPH11017732

    What is the sign of total energy of bounded system?

    Solution
    The total energy of bounded system is negative.
    Question 375
    CBSEENPH11017733

    In which direction the artificial satellites are generally fired to put into orbit?

    Solution
    Generally, the satellites are fired in the East direction. 
    Question 376
    CBSEENPH11017734

    What is the ratio of potential energy to kinetic energy of a satellite moving in orbit around the earth?

    Solution
    Ratio of P.E to K.E of a satellite moving in orbit around the earth is 2:1.
    Question 378
    CBSEENPH11017737

    Which is greater - magnitude of potential energy or kinetic energy of an orbiting satellite?

    Solution
    Potential energy is twice of kinetic energy. That is, P.E is greater. 
    Question 379
    CBSEENPH11017739

    What is geostationary satellite?

    Solution
    A satellite which appears stationary to an observer standing on the earth is known as a geostationary satellite.
    Question 380
    CBSEENPH11017740

    What is parking orbit?

    Solution
    The orbit of geostationary satellite is called parking orbit.
    Question 381
    CBSEENPH11017742
    Question 382
    CBSEENPH11017743

    What is the height of geostationary satellite from the surface of the earth?

    Solution
    Height of the geostationary satellite is, 
    h= open parentheses fraction numerator straight T squared space straight R squared space straight g over denominator 4 space straight pi squared end fraction close parentheses to the power of begin inline style 1 third end style end exponent space minus space R
    Put, R = 6.4 x 10m
    g = 9.8 m/s2 
    T = 24 hrs = 24 x 60 x 60 s 
    So, we get
    h = 3.6 x 107 m = 36000 km.
    The height of geostationary satellite from the surface of the earth is 36000 km.
    Question 383
    CBSEENPH11017744

    Can the satellite moon be used for communication?

    Solution
    No, moon cannot be used for communication because the moon is not a geostationary satellite.
    Question 384
    CBSEENPH11017745

    Give one application of geostationary satellite.

    Solution
    Geostationary satellite is used for communication purposes.
    Question 385
    CBSEENPH11017746

    What is polar orbit?

    Solution
    A polar orbit is that orbit whose angle of inclination with the equatorial plane of earth is 90o and a satellite in the polar orbit will pass over both the north and south geographic poles once per orbit. 
    The geographic lies on the polar orbit. 
    Question 386
    CBSEENPH11017747

    What is polar satellite?

    Solution
    A polar satellite revolves in the polar orbit around the Earth. 
    Question 388
    CBSEENPH11017749

    Define escape velocity. Show that the square of escape velocity is equal to the product of diameter of the earth and acceleration due to gravity at the earth.

    Solution
    Escape velocity is the minimum velocity with which a body must be projected vertically upward in order that it may just escape the gravitational pull of the earth.
                              
    Let M be the mass of the earth, and
    R the radius of the earth.
    Let at any instant, mass m be at P which is at a distance r from the centre of the earth.
    The gravitational force on mass m is, 
                            straight F equals GMm over straight r squared 
    The work done by external force to raise the body by small distance dr against gravity is, 
                 dW equals straight F space dr space equals space GMm over straight r squared dr 
    Total work done in taking the body from surface of the earth to infinity is, 
              straight W equals integral subscript straight R superscript infinity straight F space dr space equals space integral subscript straight R superscript infinity GMm over straight r squared dr 
                 equals open square brackets negative GMm over straight r close square brackets subscript straight R superscript infinity equals GMm over straight R 
    If the body is at surface of the earth and if kinetic energy is GMm over straight R comma the mass would escape from gravitational pull of the earth.
    i.e.,           1 half mv subscript straight e superscript 2 equals GMm over straight R 
                           straight v subscript straight e squared space equals space 2 fraction numerator G M over denominator straight R end fraction
space space space space space space equals 2 fraction numerator G M over denominator straight R squared end fraction straight R
space space space space space space space equals 2 straight R space straight g space equals space straight D. straight g 
                            straight v subscript straight e squared equals straight D. straight g 
    That is, escape velocity is the product of diameter of the earth and acceleration due to gravity at the earth. 
    Hence the result.
    Question 389
    CBSEENPH11017752

    Define escape velocity. Derive an expression for escape velocity. Find the value of escape velocity.

    Solution

    Escape velocity is the minimum velocity with which a body must be projected vertically upward so that it may just escape the surface of the Earth. 
    Expression for escape velocity: 
    Let a body of mass m be escaped from the gravitational field of the earth.
    During the course of motion, let at any instant, body be at a distance r from the centre of the earth. 
                             
    The gravitational force between body and the earth is,
                            straight F equals GMm over straight r squared dr  
    Work done to raise the body by distance dr is, 
                         space space dW equals GMm over straight r squared dr 
    Total work done, W in raising the body from the surface of the earth to infinity is, 
                                 straight W equals integral subscript straight R superscript infinity GMm over straight r squared dr equals GMm open square brackets negative 1 over straight r close square brackets subscript straight R superscript infinity 
                             equals GMm open square brackets negative 1 over infinity plus 1 over straight R close square brackets

equals GMm over straight R 
    If we throw the body upward with a velocity ve, then work done to raise the body from surface of the earth to infinity is done by kinetic energy.
    Therefore, 
                       GMm over straight R equals 1 half mv subscript straight e squared 
    rightwards double arrow                       straight v subscript straight e equals square root of fraction numerator 2 GM over denominator straight R end fraction end root space equals space square root of 2 gR end root  
    rightwards double arrow                       straight v subscript straight e equals square root of 2 gR end root               ... (1) 
    Substituting the values in equation (1), we have
    g = 9.81m/sec and R = 6.4 x 106m
    Escape velocity is given by, 
      straight v subscript straight e space equals square root of 9.81 cross times 6.4 cross times 10 to the power of 6 end root space

space space space space space equals space 11200 space straight m divided by straight s 

          equals 11.2 space km divided by straight s

    Question 390
    CBSEENPH11017753

    Why the moon has no atmosphere?

    Solution
    The planet or satellite can bound the atmosphere i.e., mixture of gases if and only if the root mean square velocity of molecules of the gases is less than the escape velocity.
    Escape velocity on the surface of the moon is small i.e. 2.38 km/sec as compared to r.m.s. velocity of gas molecules on the surface of the moon.
    Therefore, all the gases escape from the gravitational field of the moon and hence is not present in the atmosphere.

    Question 391
    CBSEENPH11017754

    Determine the escape velocity of a body from the moon. Take the moon to be a uniform sphere of radius 1.74 x 106 m and mass 7.36 x 1022 kg. Does your answer throw light on why the moon has no atmosphere? (G = 6.67 x 10-11 Nm2/kg2)

    Solution
    Escape velocity from the surface of the moon is, 
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#1 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/util/sys/Store.class.php(56): mkdir('/home/config_ad...', 493)
#2 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/plugin/impl/FolderTreeStorageAndCache.class.php(110): com_wiris_util_sys_Store->mkdirs()
#3 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/plugin/impl/RenderImpl.class.php(231): com_wiris_plugin_impl_FolderTreeStorageAndCache->codeDigest('mml=<math xmlns...')
#4 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/plugin/impl/TextServiceImpl.class.php(59): com_wiris_plugin_impl_RenderImpl->computeDigest(NULL, Array)
#5 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/service.php(19): com_wiris_plugin_impl_TextServiceImpl->service('mathml2accessib...', Array)
#6 {main}</pre>
    where,
    M is the mass of the moon, and
    R is radius of the moon. 
    Here, we have
    straight R equals 1.74 cross times 10 to the power of 6 straight m 
    straight M equals 7.36 cross times 10 to the power of 22 kg
    straight G equals 6.67 cross times 10 to the power of negative 11 end exponent Nm squared divided by kg squared 
    ∴  Escape velocity is,
      straight v subscript straight e equals square root of fraction numerator 2 cross times 6.67 cross times 10 to the power of negative 11 end exponent cross times 7.36 cross times 10 squared over denominator 1.74 cross times 10 to the power of 6 end fraction end root straight m divided by straight s 
          equals 2.3754 space cross times space 10 cubed straight m divided by sec 
           <pre>uncaught exception: <b>mkdir(): Permission denied (errno: 2) in /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/util/sys/Store.class.php at line #56mkdir(): Permission denied</b><br /><br />in file: /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/util/sys/Store.class.php line 56<br />#0 [internal function]: _hx_error_handler(2, 'mkdir(): Permis...', '/home/config_ad...', 56, Array)
#1 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/util/sys/Store.class.php(56): mkdir('/home/config_ad...', 493)
#2 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/plugin/impl/FolderTreeStorageAndCache.class.php(110): com_wiris_util_sys_Store->mkdirs()
#3 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/plugin/impl/RenderImpl.class.php(231): com_wiris_plugin_impl_FolderTreeStorageAndCache->codeDigest('mml=<math xmlns...')
#4 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/plugin/impl/TextServiceImpl.class.php(59): com_wiris_plugin_impl_RenderImpl->computeDigest(NULL, Array)
#5 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/service.php(19): com_wiris_plugin_impl_TextServiceImpl->service('mathml2accessib...', Array)
#6 {main}</pre> 
    Since escape velocity for the moon is 2.38 km/sec which is small as compared to the average speed of gas molecules at the moon, therefore the moon cannot bind the atmosphere. 
                        
    Question 392
    CBSEENPH11017755

    The Jupiter has a mass 318 times that of the earth and its radius is 11.2 times the earth’s radius. Estimate the escape velocity of a body from the Jupiter’s surface, given that the escape velocity from the earth’s surface is 11.2 km/s. Does your answer throw any light on why the atmosphere of the Jupiter contains light gases, mostly hydrogen, whereas the earth’s atmosphere has little of hydrogen gas?

    Solution
    Given,
    Mass of Jupiter is 318 times mass of the Earth.
    i.e.,                    Mj = 318M

    Radius is 11.2 times the radius of the Earth.
                               Rj = 11.2Re

    We know that the escape velocity of body from a planet is given by, 

                          straight v equals square root of fraction numerator 2 GM over denominator straight R end fraction end root 
    space space space space straight v subscript straight j over straight v subscript straight e equals square root of fraction numerator 2 GM subscript straight j over denominator straight R subscript straight j end fraction cross times fraction numerator straight R subscript straight e over denominator 2 GM subscript straight e end fraction end root 
               equals space square root of straight M subscript straight j over straight M subscript straight e cross times straight R subscript straight e over straight R subscript straight j end root space equals space square root of fraction numerator 318 over denominator 11.2 end fraction end root 
    rightwards double arrow   straight v subscript straight j space equals space straight v subscript straight e square root of fraction numerator 318 over denominator 11.2 end fraction end root space equals space 11.2 square root of fraction numerator 318 over denominator 11.2 end fraction end root space km divided by sec 
               = 59.7 km/sec 
    Escape velocity for Jupiter is high as compared to the average speed of light gas molecules like hydrogen.
    Therefore, atmosphere of the Jupiter contains light gases like hydrogen. 
    Question 393
    CBSEENPH11017756

    Does the escape speed of a body from the earth depend on

    (a) the mass of the body,

    (b) the location from where it is projected,

    (c) the direction of projection,

    (d) the height of the location from where the body is launched?

    Solution
    Escape velocity is given by, 
    straight v subscript straight e equals square root of 2 gr end root space equals space square root of fraction numerator 2 GM over denominator straight r end fraction end root
    space space space equals square root of fraction numerator 2 GM over denominator straight R plus straight h end fraction end root
    where,
    g is the acceleration due to gravity at the point from where the body is projected,
    r is the distance from the centre of the earth from where the body is projected, and
    h is the height from the surface of the earth.
    (a) Escape velocity, ve does not depend on the mass of the body.
    (b) Escape velocity, ve depends on upon g and at different locations g is different (gλ = g - Rω2cos2 λ ), therefore, ve slightly depends on upon the angle of latitude ( ∵ Rωcos 2 λ ≤ 3.36 cm/s2).
    (c) Escape velocity, ve does not depend on upon the direction of projection. 
    (d) Escape velocity, ve depends on upon height from the surface of the earth. 
    Question 394
    CBSEENPH11017757

    A body is projected with a velocity 14 km/s from the surface of the earth. Find the velocity of body at infinity. Escape velocity at the surface of the earth is 11.2 km/s.

    Solution

    Let v be the velocity of body at infinity.
    By using law of conservation of energy, we have
              space space space space 1 half mv squared minus GMm over straight R equals 1 half mv subscript infinity superscript 2 plus 0 
    rightwards double arrow      space space space space 1 half mv squared minus 1 half mv subscript straight e superscript 2 equals 1 half mv subscript infinity superscript 2 

    rightwards double arrow                            straight v subscript infinity equals square root of straight v squared minus straight v subscript straight e superscript 2 end root
    Here, we have
    Velocity with which the body is projected,straight v equals 14 space km divided by straight s
    Escape velocity, straight v subscript straight e equals 11.2 space km divided by straight s
    ∴ Velocity of body at infinity, straight v subscript infinity space equals space square root of left parenthesis 14 right parenthesis squared minus left parenthesis 11.2 right parenthesis squared end root space equals space 8.4 space km divided by straight s

     
    Question 395
    CBSEENPH11017758

    Two bodies of masses m1 and m(m1 > m2) are to be projected out of the field of the earth.

    (i) Which of the two must be thrown with greater velocity?

    (ii) Which of the two must be thrown with greater momentum?

    (iii) Which of the two must be thrown with greater kinetic energy?

    Solution

    (i) Escape velocity is independent of the mass of body projected and is same for all the bodies.
    i.e., 11.2 km/s, therefore, both must be projected with equal velocity. 
    (ii) Momentum of body of mass m1 is,
                          straight p subscript 1 equals straight m subscript 1 straight v subscript straight e 
    Momentum of the body of mass straight m subscript 2 is, 
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#1 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/util/sys/Store.class.php(56): mkdir('/home/config_ad...', 493)
#2 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/plugin/impl/FolderTreeStorageAndCache.class.php(110): com_wiris_util_sys_Store->mkdirs()
#3 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/plugin/impl/RenderImpl.class.php(231): com_wiris_plugin_impl_FolderTreeStorageAndCache->codeDigest('mml=<math xmlns...')
#4 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/plugin/impl/TextServiceImpl.class.php(59): com_wiris_plugin_impl_RenderImpl->computeDigest(NULL, Array)
#5 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/service.php(19): com_wiris_plugin_impl_TextServiceImpl->service('mathml2accessib...', Array)
#6 {main}</pre> 
    Since    straight m subscript 1 greater than straight m subscript 2 
    ∴             space space space space space space space straight m subscript 1 straight v subscript straight e greater than straight m subscript 2 straight v subscript straight e 
    That is, a heavy body must be thrown with greater momentum. 
    (iii) Kinetic energy of mass straight m subscript 1 is, 
               space space space space space space straight K subscript 1 equals 1 half straight m subscript 1 straight v subscript straight e squared 
    Kinetic energy of mass straight m subscript 2 is,
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#1 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/util/sys/Store.class.php(56): mkdir('/home/config_ad...', 493)
#2 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/plugin/impl/FolderTreeStorageAndCache.class.php(110): com_wiris_util_sys_Store->mkdirs()
#3 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/plugin/impl/RenderImpl.class.php(231): com_wiris_plugin_impl_FolderTreeStorageAndCache->codeDigest('mml=<math xmlns...')
#4 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/plugin/impl/TextServiceImpl.class.php(59): com_wiris_plugin_impl_RenderImpl->computeDigest(NULL, Array)
#5 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/service.php(19): com_wiris_plugin_impl_TextServiceImpl->service('mathml2accessib...', Array)
#6 {main}</pre> 
    Since m1 > m2, therefore K1 > K2.
    That is, i.e., heavy body must be thrown with greater kinetic energy. 

    Question 396
    CBSEENPH11017760

    Why does hydrogen gas escape faster than oxygen from the earth’s atmosphere?

    Solution
    Root mean square velocity of gas molecule is given by, 
                        straight v subscript straight r equals square root of fraction numerator 3 kT over denominator straight m end fraction end root 
    i.e.               space space space straight v subscript straight r proportional to square root of 1 over straight m end root 
    As,            <pre>uncaught exception: <b>mkdir(): Permission denied (errno: 2) in /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/util/sys/Store.class.php at line #56mkdir(): Permission denied</b><br /><br />in file: /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/util/sys/Store.class.php line 56<br />#0 [internal function]: _hx_error_handler(2, 'mkdir(): Permis...', '/home/config_ad...', 56, Array)
#1 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/util/sys/Store.class.php(56): mkdir('/home/config_ad...', 493)
#2 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/plugin/impl/FolderTreeStorageAndCache.class.php(110): com_wiris_util_sys_Store->mkdirs()
#3 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/plugin/impl/RenderImpl.class.php(231): com_wiris_plugin_impl_FolderTreeStorageAndCache->codeDigest('mml=<math xmlns...')
#4 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/plugin/impl/TextServiceImpl.class.php(59): com_wiris_plugin_impl_RenderImpl->computeDigest(NULL, Array)
#5 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/service.php(19): com_wiris_plugin_impl_TextServiceImpl->service('mathml2accessib...', Array)
#6 {main}</pre> 
    ∴                space space straight v subscript hyd equals 4 space straight v subscript oxy 
    That is, escape velocity of hydrogen is 4 times that of the escape velocity of oxygen gas. 
    Hence, hydrogen gas has more tendency to escape than oxygen from the Earth's atmosphere. 
    Question 397
    CBSEENPH11017762

    The escape speed of a projectile on the surface of earth is 11.2km/s. A body is projected out with thrice this speed. What is the speed of body far away from the earth? Ignore the presence of the sun and other planets.

    Solution
    Let v be the velocity of projection and v' the velocity of the body at infinity.
    According to the principle of law of conservation of energy,
    Total energy at the surface of earth = Total energy at infinity
    i.e.          straight U subscript earth plus straight K subscript earth equals straight U subscript infinity plus straight K subscript infinity  

    rightwards double arrowspace space space space minus GMm over straight R plus 1 half mv squared equals 0 plus 1 half mv apostrophe squared         ...(1)
    If, space straight v equals straight v subscript straight e   
    Then,   straight v apostrophe equals 0 
    Therefore, 
               negative GMm over straight R plus 1 half mv subscript straight e squared equals 0 plus 0 

    rightwards double arrow                  GMm over straight R equals 1 half mv subscript straight e squared                  ...(2)
    From equation (1) and (2), 
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#1 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/util/sys/Store.class.php(56): mkdir('/home/config_ad...', 493)
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#3 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/plugin/impl/RenderImpl.class.php(231): com_wiris_plugin_impl_FolderTreeStorageAndCache->codeDigest('mml=<math xmlns...')
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#6 {main}</pre>  
    rightwards double arrow                          <pre>uncaught exception: <b>mkdir(): Permission denied (errno: 2) in /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/util/sys/Store.class.php at line #56mkdir(): Permission denied</b><br /><br />in file: /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/util/sys/Store.class.php line 56<br />#0 [internal function]: _hx_error_handler(2, 'mkdir(): Permis...', '/home/config_ad...', 56, Array)
#1 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/util/sys/Store.class.php(56): mkdir('/home/config_ad...', 493)
#2 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/plugin/impl/FolderTreeStorageAndCache.class.php(110): com_wiris_util_sys_Store->mkdirs()
#3 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/plugin/impl/RenderImpl.class.php(231): com_wiris_plugin_impl_FolderTreeStorageAndCache->codeDigest('mml=<math xmlns...')
#4 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/plugin/impl/TextServiceImpl.class.php(59): com_wiris_plugin_impl_RenderImpl->computeDigest(NULL, Array)
#5 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/service.php(19): com_wiris_plugin_impl_TextServiceImpl->service('mathml2accessib...', Array)
#6 {main}</pre>           ... (3)
    Here, we have that the body is projected with thrice the speed of escape velocity. 
    i.e.,                       <pre>uncaught exception: <b>mkdir(): Permission denied (errno: 2) in /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/util/sys/Store.class.php at line #56mkdir(): Permission denied</b><br /><br />in file: /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/util/sys/Store.class.php line 56<br />#0 [internal function]: _hx_error_handler(2, 'mkdir(): Permis...', '/home/config_ad...', 56, Array)
#1 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/util/sys/Store.class.php(56): mkdir('/home/config_ad...', 493)
#2 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/plugin/impl/FolderTreeStorageAndCache.class.php(110): com_wiris_util_sys_Store->mkdirs()
#3 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/plugin/impl/RenderImpl.class.php(231): com_wiris_plugin_impl_FolderTreeStorageAndCache->codeDigest('mml=<math xmlns...')
#4 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/plugin/impl/TextServiceImpl.class.php(59): com_wiris_plugin_impl_RenderImpl->computeDigest(NULL, Array)
#5 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/service.php(19): com_wiris_plugin_impl_TextServiceImpl->service('mathml2accessib...', Array)
#6 {main}</pre> 
    Therefore, putting this in equation (3),
    We have,         <pre>uncaught exception: <b>mkdir(): Permission denied (errno: 2) in /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/util/sys/Store.class.php at line #56mkdir(): Permission denied</b><br /><br />in file: /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/util/sys/Store.class.php line 56<br />#0 [internal function]: _hx_error_handler(2, 'mkdir(): Permis...', '/home/config_ad...', 56, Array)
#1 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/util/sys/Store.class.php(56): mkdir('/home/config_ad...', 493)
#2 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/plugin/impl/FolderTreeStorageAndCache.class.php(110): com_wiris_util_sys_Store->mkdirs()
#3 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/plugin/impl/RenderImpl.class.php(231): com_wiris_plugin_impl_FolderTreeStorageAndCache->codeDigest('mml=<math xmlns...')
#4 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/plugin/impl/TextServiceImpl.class.php(59): com_wiris_plugin_impl_RenderImpl->computeDigest(NULL, Array)
#5 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/service.php(19): com_wiris_plugin_impl_TextServiceImpl->service('mathml2accessib...', Array)
#6 {main}</pre> 
    rightwards double arrow                  space space straight v apostrophe space equals space 2 square root of 2 straight v subscript straight e

space space space space space space equals 2 square root of 2 cross times 11.2

space space space space space space equals 31.7 space km divided by straight s
    This is the required speed of the body far away from the earth. 
    Question 398
    CBSEENPH11017764

    Define orbital velocity. Derive the expression for it.

    Solution
    Orbital velocity is the velocity given to artificial satellite so that it may start revolving around the earth. 
                         

    Expression for orbital velocity: 
    Consider, a satellite of mass 'm' is revolving around the earth in a circular orbit of radius 'r', at a height 'h' from the surface of the earth.
    Let 'M' be the mass of the earth and 'R' be the radius of the earth. 
    Therefore,
                      r = R + h
    The centripetal force that is required to revolve the satellite =  <pre>uncaught exception: <b>mkdir(): Permission denied (errno: 2) in /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/util/sys/Store.class.php at line #56mkdir(): Permission denied</b><br /><br />in file: /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/util/sys/Store.class.php line 56<br />#0 [internal function]: _hx_error_handler(2, 'mkdir(): Permis...', '/home/config_ad...', 56, Array)
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#6 {main}</pre> 
    where,
     straight v subscript straight o is orbital velocity.
    Orbital velocity is produced by the gravitational force between the earth and the satellite =  <pre>uncaught exception: <b>mkdir(): Permission denied (errno: 2) in /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/util/sys/Store.class.php at line #56mkdir(): Permission denied</b><br /><br />in file: /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/util/sys/Store.class.php line 56<br />#0 [internal function]: _hx_error_handler(2, 'mkdir(): Permis...', '/home/config_ad...', 56, Array)
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#6 {main}</pre> 

    Therefore, 
                   mv subscript straight o squared over straight r space equals space GMm over straight r squared 

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#6 {main}</pre> 
    rightwards double arrow                 straight v subscript straight o space equals space square root of GM over straight r end root space equals space square root of fraction numerator GM over denominator straight R plus straight h end fraction end root 
    But,
                          GM=gR2 
    where,

    g is the acceleration due to gravity.
    So, 
    straight v subscript straight o space equals space square root of gR squared over straight r end root space equals space square root of fraction numerator gR squared over denominator straight R plus straight h end fraction end root space equals space straight R square root of fraction numerator straight g over denominator straight R plus straight h end fraction end root 
    Let g' be the acceleration due to gravity in the orbit i.e. at a height h from the surface. 
    Therefore,
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#6 {main}</pre>  
    rightwards double arrow         fraction numerator GM over denominator straight R plus straight h end fraction equals straight g apostrophe left parenthesis straight R plus straight h right parenthesis space equals space straight g apostrophe straight r 
    ∴               space space space straight v subscript straight o space equals space square root of straight g apostrophe left parenthesis straight R plus straight h right parenthesis end root space equals space square root of straight g apostrophe straight r end root, is the required orbital velocity. 

    Question 399
    CBSEENPH11017765

    What is the value of orbital velocity when satellite just revolves around the earth?

    Solution
    When the satellite is very close to the earth, then r ~ R.
    ∴ the orbital velocity,

         straight v subscript straight o equals square root of GM over straight R end root equals square root of gR  
            equals space square root of 9.81 cross times 6.4 cross times 10 to the power of 6 end root space straight m divided by straight s 
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#6 {main}</pre> 
            space space equals 7.923 km divided by straight s
    Question 400
    CBSEENPH11017766

    Why does a satellite need no fuel to move round a planet in fixed orbit?

    Solution
    Satellite needs no fuel to move around a planet in a fixed orbit because the gravitational force of attraction between the satellite and the planet provides the necessary centripetal force required to revolve around it. 
    Question 401
    CBSEENPH11017767

    We can make an artificial satellite to revolve around the moon at a height of 50km from its surface but it is not possible to make an artificial satellite to revolve around the earth at a height of 50km from its surface. Explain why.

    Solution
    An artificial satellite can revolve around the moon at a height of 50km from its surface because the moon has no atmosphere.
    The earth binds the atmosphere and its height is about 400km. Therefore, it is not possible to make an artificial satellite revolve around the earth at a height of 50km from its surface.
    Due to air friction, the satellite will lose the energy and falls on the surface of the earth.
    Question 402
    CBSEENPH11017768

    An artificial satellite is generally launched into orbit from the place situated on the equator or near it. Explain why.

    Solution
    The Earth revolves from east to west. However, the speed of the satellite at different places is different and depends on the angle of latitude.
    At equator, the speed is maximum and is equal to 0.465 km/s. When the satellite is fired in an eastern direction, less velocity than the required orbital velocity has to be provided. A velocity equal to the velocity of the place has to be provided and the satellite is fired. Since equator has maximum speed, the satellite is launched from the places situated near the equator. 
    Question 403
    CBSEENPH11017770

    How a rocket fired from the earth can become a satellite of the earth? 

    Solution

    First, the rocket should be fired with such a velocity that it attains the height at which it is required to revolve.
    Orbital velocity rovided to the rocket = square root of fraction numerator GM over denominator straight R plus straight h end fraction end root, in horizontal direction to put into orbit.
    The orbit of the satellite should be above the earth's atomosphere. 

    Question 404
    CBSEENPH11017771

    Show that the satellite revolving around the earth will depart forever if its speed is increased by 41.42%.

    Solution
    Let a satellite of mass m be revolving in circular orbit of radius 'r' around the earth of mass M.
    The orbital velocity of satellite is given by, 
                        straight v subscript straight o equals square root of GM over straight r end root 
    The velocity required by the satellite is obtained from the equation,
                 1 half mv subscript straight e squared space equals space GMm over straight r 
    rightwards double arrow                  straight v subscript straight e equals square root of fraction numerator 2 GM over denominator straight r end fraction end root 
    To escape, the percentage increase in velocity is, 
    fraction numerator straight v subscript straight e minus straight v subscript straight o over denominator straight v subscript straight o end fraction cross times 100 space equals space fraction numerator square root of begin display style fraction numerator 2 GM over denominator straight r end fraction end style end root space minus space square root of begin display style GM over straight r end style end root over denominator square root of begin display style GM over straight r end style end root end fraction cross times 100 
                       space space equals left parenthesis square root of 2 minus 1 right parenthesis cross times 100 space

space space equals space 41.42 percent sign
    Question 405
    CBSEENPH11017772

    A satellite of mass m is revolving around the earth in a fixed circular orbit of radius r. Derive an expression for angular momentum.

    Solution
    The orbital speed of a satellite revolving around the earth in circular orbit of radius r is given by, 
                            straight v subscript straight o equals square root of GM over straight r end root 
    Therefore the angular momentum of satellite, 
    straight L equals mv subscript straight o straight r equals mr square root of GM over straight r end root
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#6 {main}</pre>
    Question 406
    CBSEENPH11017773

    A satellite of mass m is revolving in a circular orbit of radius r around the earth of mass M. What is the total energy of the satellite?

    Solution
    The sum of kinetic energy of revolution and potential energy due to the gravitational field of the earth gives us the total energy of the satellite. 
                      
    Satellite is revolving in circular orbit of radius r with an orbital velocity given by, 
    i.e.,               straight v subscript straight o space equals space square root of GM over straight r end root 
    Hence, Kinetic Energy, straight K equals 1 half mv subscript straight o squared space equals fraction numerator GMm over denominator 2 straight r end fraction 
    The satellite is at a distance ‘r’ from the centre of the earth.
    ∴ Potential energy, straight U equals negative GMm over straight r 
    Now the total energy, 
    straight E equals straight K plus straight U equals negative fraction numerator GMm over denominator 2 straight r end fraction
    Hence the result. 
    Question 407
    CBSEENPH11017774

    State Kepler’s laws of planetary motion.

    Solution
    Kepler's Law of planetary motion states that:
    1. First law: All the planets revolve around the sun in an elliptical orbit with the sun at one focus.
                        

    2. Second law: Ariel velocity of the planet is constant.
    3. Third law: Square of the time period of one complete revolution is directly proportional to the cube of the semi-major axis of the elliptical orbit.

    Question 408
    CBSEENPH11017776

    State and prove Kepler’s third law of planetary motion.

    Solution

    Kepler’s third law states that the square of the time period of revolution of a planet is proportional to the cube of the semi-major axis of the elliptic orbit.
    If orbit of the planet is circular, then square of the time period will be proportional to the cube of radius of the orbit.
    Let a planet revolve around the sun in a circular orbit of radius r.
    The necessary centripetal force required is provided by the gravitational pull, 
    i.e.,         mrω squared equals GMm over straight r squared 
    rightwards double arrow         straight r fraction numerator 4 straight pi squared over denominator straight T squared end fraction equals GM over straight r squared 
    rightwards double arrow             space space straight T squared equals fraction numerator 4 straight pi squared over denominator GM end fraction straight r cubed 
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#6 {main}</pre>

    Question 409
    CBSEENPH11017778

    How will you ‘weigh the sun’, that is estimate its mass? You are given the period of one of its planets and the radius of the planetary orbit.

    The mean orbital radius of the earth around the sun is 1.5 × 108 km.


    Solution
    The sun can be weighed by using Kepler’s third law of planetary motion as, 
                     straight T squared equals open parentheses fraction numerator 4 straight pi squared over denominator GM end fraction close parentheses straight r cubed
    This implies,
                      straight M equals open parentheses fraction numerator 4 straight pi squared over denominator GT squared end fraction close parentheses straight r cubed
    On putting the required values in the given formula, the mass of the sun can be found out. 
    Question 410
    CBSEENPH11017780

    Estimate the mass of the sun from the data given below:

    (i) Time period of revolution of the earth is 365 days

    (ii) Radius of the orbit of the earth is 1 A.U.

    Solution
    Earth revolves around the sun.
    The necessary centripetal force required for the orbital motion of Earth around the Sun is provided by gravitational pull of the sun,
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#6 {main}</pre>
    rightwards double arrow              straight M space equals space fraction numerator straight omega squared straight r cubed over denominator straight G end fraction equals fraction numerator 4 straight pi squared straight r cubed over denominator GT squared end fraction 
                        straight M equals fraction numerator 4 straight pi squared left parenthesis 1.5 cross times 10 to the power of 11 right parenthesis squared over denominator left parenthesis 6.67 cross times 10 to the power of negative 11 end exponent right parenthesis space left parenthesis 365 space cross times space 86400 right parenthesis squared end fraction 
                         space space equals 2 cross times 10 to the power of 30 kg
    This is the required mass of the sun. 
    Question 411
    CBSEENPH11017782

    When a satellite suddenly jumps to higher orbit then how does the following change:

    i) angular velocity,

    ii) linear momentum,

    iii) kinetic energy, and

    iv) potential energy?

    Solution

    We have,   straight T squared proportional to space straight r cubed             ... (1)
    So, when a satellite suddenly jumps to the higher orbit, r increases.
    Therefore, from equation (1), T increases.
    i) Angular velocity is given by, space space straight omega equals fraction numerator 2 straight pi over denominator straight T end fraction
    Therefore, as T increases, angular velocity  decreases.
    ii) We know,
    Orbital velocity, straight v subscript straight o space equals space square root of GM over straight r end root
    Therefore, linear momentum of satellite is, 
                      mv subscript straight o equals straight m square root of GM over straight r end root 
    Since r increases, therefore, linear momentum decreases.
    iii) The kinetic energy of satellite is,
                1 half mv subscript straight o squared space equals space fraction numerator GMm over denominator 2 straight r end fraction
    The kinetic energy of satellite decreases with increase in r.
    iv) The potential energy of satellite is,
                           straight U equals negative GMm over straight r 
    The potential energy of satellite increases with increase in r. 

    Question 412
    CBSEENPH11017783

    A satellite revolves in an orbit close to the surface of planet of density 5800 kg/ m3. Find the time period of revolution of the satellite.

    Solution
    The time period of the satellite revolving close to the surface of planet is,
                          straight T equals 2 straight pi square root of straight R cubed over GM end root 
    where M is the mass of the planet.
    The mass of planet, M equals 4 over 3 πR cubed straight rho 
    Therefore,
     Time period, straight T equals 2 straight pi square root of fraction numerator 3 straight R cubed over denominator straight G cross times 4 πρR cubed end fraction end root space equals space square root of fraction numerator 3 straight pi over denominator Gρ end fraction end root
    Here, we have
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#6 {main}</pre> 
    Therefore,
     straight T equals square root of fraction numerator 3 cross times 22 over denominator 7 cross times 6.67 cross times 10 to the power of negative 11 end exponent cross times 5800 end fraction end root space equals space 4937 straight s, is the required time period of the revolution of satellite. 
    Question 413
    CBSEENPH11017784

    An earth satellite moves in a circular orbit with orbital speed of 6km/s. Find the time period of one revolution and radial acceleration of the satellite.

    Solution
    Let r be the radius of the circular orbit.
    The orbital velocity of satellite is given by, 
                        straight v subscript straight o space equals space square root of GM over straight r end root
    i.e.,                  straight r equals GM over straight v subscript straight o squared 
    The time period of one revolution is, 
                      straight T equals square root of fraction numerator 4 straight pi squared over denominator GM end fraction straight r cubed end root 
    rightwards double arrowor           straight T equals square root of fraction numerator 4 straight pi squared over denominator GM end fraction open parentheses GM over straight v subscript straight o squared close parentheses cubed end root
                        equals 2 straight pi GM over straight v subscript straight o cubed equals 2 straight pi gR squared over straight v subscript straight o cubed 
    Here, we have
    Acceleration due to gravity, straight g equals 9.8 space straight m divided by straight s squared
    Radius of the Earth, straight R equals 6.4 cross times 10 to the power of 6 straight m
    Orbital velocity,space space space straight v subscript straight o equals 6000 straight m divided by straight s
    Therefore,
    Time period, straight T equals 2 straight pi fraction numerator 9.8 cross times left parenthesis 6.4 cross times 10 to the power of 6 right parenthesis squared over denominator left parenthesis 6000 right parenthesis cubed end fraction equals 11676 straight s 
    The radial acceleration of satellite is,
                       straight a equals rω squared equals vω equals fraction numerator straight v 2 straight pi over denominator straight T end fraction 
    i.e.,               straight a equals fraction numerator 6000 cross times 2 cross times 22 over denominator 7 cross times 11676 end fraction equals 3.23 straight m divided by straight s squared 
    Question 414
    CBSEENPH11017786

    A satellite orbits the earth at a height of 400 km above the surface. How much energy must be expended to the satellite to rocket out of the earth’s gravitational influence?
    Mass of the satellite = 200 kg, Mass of the earth = 6.0 x 1024 kg, Radius of the earth = 6.4 x l06 m; G = 6.67 x 10-11 Nm2/kg2.

    Solution
    Kinetic energy of the satellite is, 
    1 half mv subscript straight o squared space equals space 1 half straight m open parentheses fraction numerator GM over denominator straight R plus straight h end fraction close parentheses 
                equals 1 half cross times 200 open parentheses fraction numerator 6.67 cross times 10 to the power of negative 11 end exponent cross times 6.0 cross times 10 to the power of 24 over denominator 6.8 cross times 10 to the power of 6 end fraction close parentheses
                equals space 5.89 space cross times space 10 to the power of 9 space straight J 
    Potential energy of the satellite is, 
    space space space space minus fraction numerator GMm over denominator straight R plus straight h end fraction equals negative open parentheses fraction numerator 6.67 cross times 10 to the power of negative 11 end exponent cross times 6.0 cross times 10 to the power of 24 cross times 200 over denominator 6.8 cross times 10 to the power of 6 end fraction close parentheses 
                   equals negative 11.78 space cross times space 10 to the power of 9 straight J 
    Total energy of satellite is,
    E = 5.89 x 109 - 11.78 x l09
       =-5.89 x 109
    The total energy of the satellite is zero, if the satellite just escapes from the gravitational field. Therefore, 5.89 x 109J of energy has to supplied to the satellite to escape the gravitational pull.
    Question 415
    CBSEENPH11017789

    A satellite orbits the earth at a height of 500 km from its surface. Compute its:

    (a) kinetic energy 

    (b) potential energy

    (c) total energy.

    Mass of the satellite = 300 kg,
    Mass of the earth = 60 x 1024kg,
    Radius of the earth = 6.4 x 106m
    G = 6.67 x 10-11 Nm2/kg2. 

    Will your answers alter if the earth shrinks suddenly to half of its present size?

    Solution

    We have,
    Mass of the satellit, m = 300 kg
    Mass of the Earth, M= 6.0 x 1024kg
    Radius of Earth, R = 6.4 x 106m
    Gravitational constant, G = 6.67 x 10-11 Nm2/kg2
    (a) Kinetic energy of satellite is,
    1 half mv subscript straight o squared space space equals space 1 half straight m open parentheses square root of fraction numerator GM over denominator straight R plus straight h end fraction end root close parentheses squared space

space space space space space space space space space space space space space space space space equals space fraction numerator GMm over denominator 2 left parenthesis straight R plus straight h right parenthesis end fraction
                equals fraction numerator 6.67 cross times 10 to the power of negative 11 end exponent cross times 6.0 cross times 10 to the power of 24 cross times 300 over denominator 2 left parenthesis 6 cross times 10 to the power of 6 plus 0.5 cross times 10 to the power of 6 right parenthesis end fraction
                equals space 8.7 cross times 10 to the power of 9 straight J
    (b) Potential energy of satellite is, 
    negative fraction numerator G M m over denominator straight R plus straight h end fraction space equals space minus fraction numerator 6.67 cross times 10 to the power of negative 11 end exponent cross times 6.0 cross times 10 to the power of 24 cross times 300 over denominator 6.9 cross times 10 to the power of 6 end fraction straight J 
                space space equals negative 17.4 cross times 10 to the power of 6 straight J 

    (c)Total energy of satellite is, 
    E = K.E. + P.E.
       = 8.7 x 109 -17.4 x 10
       = -8.7 x 109
    (d) If the earth shrinks to half its size without changing its mass, the answers will remain the same. 

     

    Question 416
    CBSEENPH11017790

    A Saturn year is 29.5 times the earth year. How far is the Saturn from the sun if the earth is 1.50 x 1011m away from the sun?

    Solution
    According to Kepler’s third law of planetary motion,
                    straight T squared proportional to straight R cubed 
    ∴       open parentheses straight R subscript straight s over straight R subscript straight e close parentheses cubed space equals space open parentheses straight T subscript straight s over straight T subscript straight e close parentheses squared 
    rightwards double arrow            straight R subscript straight s equals straight R subscript straight e open parentheses straight T subscript straight s over straight T subscript straight e close parentheses to the power of 2 divided by 3 end exponent                  ... (1)
    Given that, Saturn year is 29.5 times the earth year.
    i.e., Ts = 29.56 TE
    Distance of Earth from Sun = 1.50 x 1011 m
    Therefore, 
                       Rsequals 1.5 cross times 10 to the power of 11 left parenthesis 29.56 right parenthesis to the power of 2 divided by 3 end exponent
                           equals 1.43 cross times 10 to the power of 12 straight m 
    Question 417
    CBSEENPH11017791

    What is the time period of a satellite revolving just near the surface of the earth?

    Solution
    The time period of satellite is given by,
                       straight T equals square root of fraction numerator 4 straight pi squared over denominator GM end fraction straight r cubed end root 
    where, r is the radius of the orbit.
    When the satellite just revolves around the earth, then r = R
    Therefore,
    Time period of satellite near the surface of the earth is, 
    straight T equals square root of fraction numerator 4 straight pi squared over denominator GM end fraction straight R cubed end root space equals space 2 straight pi square root of straight R over straight g end root
      equals 2 straight pi square root of fraction numerator 6.4 cross times 10 to the power of 6 over denominator 9.8 end fraction end root equals 5077 straight s
    space space equals 1.41 space hour 
                                  
    Question 418
    CBSEENPH11017792

    Two satellites are revolving in circular orbits of radius R and 4R around a planet. What is the ratio of the time period?

    Solution
    Given, two satellites are revolving in a circular orbits or radius R and 4R.
    According to Kepler’s law of planetary motion, 
                            straight T squared proportional to straight R cubed 
    Therefore, the ratio of the time period is given by,
                        straight T subscript 1 squared over straight T subscript 2 squared equals fraction numerator straight R cubed over denominator left parenthesis 4 straight R right parenthesis cubed end fraction equals 1 over 64 
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#6 {main}</pre> 
    i.e., T1: T2 is 1: 8
    Question 419
    CBSEENPH11017793

    A Saturn year is 29.5 times the earth year. How far is the saturn from the sun if the earth is 1.5 x 108 km away from the Sun?

    Solution
    According to Kepler’s third law of planetary motion,
                                 straight T squared proportional to straight R cubed
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#6 {main}</pre>
    Here,           straight T subscript straight s equals 29.5 straight T subscript straight e   and    straight R subscript straight e equals 1.5 cross times 10 to the power of 8 km
    ∴          straight R subscript straight s space equals space straight R subscript straight e open parentheses straight T subscript straight s over straight T subscript straight e close parentheses to the power of 2 divided by 3 end exponent
                   space space space equals 1.50 cross times 10 to the power of 8 left parenthesis 29.5 right parenthesis to the power of 2 divided by 3 end exponent space km
                       equals 1.432 space cross times space 10 to the power of 12 km
    Question 420
    CBSEENPH11017794

    Two satellites are revolving around the earth. One which has time period of revolution 24 hours revolves in orbit of radius R. What is the radius of orbit of second satellite whose time period is 12 hours?

    Solution
    Given, two satellites are revolving around the Earth. 
    Time period of revolution of one satellite = 24 hours
    Time  period of the second satellite = 12 hours
    According to Kepler’s third law,
                               T2 ∝ R3

    Let R’ be the radius of the orbit of the second satellite.
    Therefore, the radius of the satellite is,
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#6 {main}</pre> 
                           straight R apostrophe equals open parentheses 1 fourth close parentheses to the power of 1 divided by 3 end exponent space straight R    

    Question 421
    CBSEENPH11017795

    What is geostationary satellite? What are the conditions for satellite to appear stationary?

    Solution

    A satellite which appears to be stationary to an observer standing on the earth is known as a geostationary satellite.
    The conditions for satellite to appear stationary are:
    (i) The time-period should be 24 hours.
    (ii) Its orbit should be in the equatorial plane of the earth.
    (iii) Its direction of motion should be the same as that of the earth about its polar axis.

    Question 422
    CBSEENPH11017796

    What is the height of a geostationary satellite from the surface of the earth?

    Solution
    The time period of the satellite is given by, 
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#6 {main}</pre> 
    rightwards double arrow          space space space space space space space space space space space space straight T squared space equals space fraction numerator 4 straight pi squared over denominator straight g space straight R squared end fraction left parenthesis straight R plus straight h right parenthesis cubed 
    rightwards double arrow            left parenthesis straight R plus straight h right parenthesis cubed space equals space fraction numerator straight g space straight R squared space straight T squared over denominator 4 straight pi squared end fraction 
    rightwards double arrow                      straight h equals open parentheses fraction numerator straight g space straight R squared space straight T squared over denominator 4 straight pi squared end fraction close parentheses to the power of 1 divided by 3 end exponent space minus space straight R 
    We have,
    Acceleration due to gravity, g = 9.8 m/s2
     Raidus space of space the space Earth comma space straight R space equals space 6.4 cross times 10 to the power of 6 space straight m divided by straight s

Time space period comma space straight T space equals space 86400 space straight s 
    Therefore, height of a geostationary satellite from the surface of the earth is, 
    straight h equals open parentheses fraction numerator 9.8 cross times left parenthesis 6.4 cross times 10 to the power of 6 right parenthesis squared space left parenthesis 86400 right parenthesis squared over denominator 4 straight pi squared end fraction close parentheses to the power of 1 divided by 3 end exponent minus left parenthesis 6.4 cross times 10 to the power of 6 right parenthesis 
      equals 4.234 cross times 10 to the power of 6 minus 6.4 cross times 10 to the power of 6 
       equals 3.594 cross times 10 to the power of 7 straight m 
    Question 423
    CBSEENPH11017797

    Can an artificial satellite be put into orbit in such a way that it will always remain directly above Chandigarh?

    Solution
    Chandigarh does not lie on the equatorial line.
    Therefore, no satellite can be made to remain just above Chandigarh. 
    Question 424
    CBSEENPH11017798

    Give four uses of satellites.

    Solution

    Uses of satellites are:
    (i) They are used for telecommunication purposes,
    (ii) They are used to weather forecasting,
    (iii) They are used to study the shape of the earth,
    (iv) They are used in military and country’s safety purposes,
    (v) They are used in studying the environmental conditions of the earth. 

    Question 425
    CBSEENPH11017799

    A sphere of mass 40 kg is attracted by a second sphere of mass 60 kg with a force equal to 4 mgf. If G = 6 x 10-11 N m2 kg-2, calculate the distance between them. Acceleration due to gravity = 10 m/s2

    Solution

    We have, 
    Mass, m1 = 40 kg ; m2 = 60 kg
    Also, F = 4 mgf = 4 x 10-6 x 10
               = 4 x 10-5 N
    G = 6 x 10-11 N mkg-2
    Distance between the masses, r = ?

    Gravitational force is given by, F = fraction numerator straight G space straight m subscript 1 space straight m subscript 2 over denominator straight r squared end fraction
    This implies, 

    r = straight r space equals space square root of fraction numerator straight G space straight m subscript 1 space straight m subscript 2 over denominator straight F end fraction end root

space space equals space square root of fraction numerator 6 space straight x space 10 to the power of negative 11 end exponent space straight x space 40 space straight x space 60 over denominator 4 space straight x space 10 to the power of negative 5 end exponent end fraction end root space

space space equals space 0.06 space straight m space equals space 6 space cm

    Question 426
    CBSEENPH11017800

    At what height above the surface of the earth, the acceleration due to gravity will be of the same as in a mine, 100 km deep? 

    Solution

    Let, the acceleration due to gravity at height h be the same as that at a depth, d, deep into the earth. 
    We have, d = 100 km 
    Therefore, 
    g' = g open parentheses 1 space minus fraction numerator 2 straight h over denominator straight R end fraction close parentheses space equals space straight g space open parentheses 1 space minus straight d over straight R close parentheses
    rightwards double arrow space space space space space space space space fraction numerator 2 straight h over denominator straight R end fraction space equals space straight d over straight R

rightwards double arrow space space space space space space space space space space space space space straight h space equals space straight d over 2 space equals space 100 over 2 space equals space 50 space km space

    Question 427
    CBSEENPH11017802

    What is gravitational field at a point inside the shell of mass M and radius r?

    Solution

    Gravitational field is given by, 
                    straight I thin space equals space GM over straight x squared                         ... (1)
    The mass lies at the periphery of the shell. And at the centre mass is equal to 0. 
    Therefore, according to equation (1), 
    Gravitational field at a point inside the shell = 0

    Question 428
    CBSEENPH11017805

    Which of the following requires greater velocity to escape?

    (i) Spoon

    (ii) Car.

    Solution

    The value of escape velocity does not depend on the mass of the body and angle of projection from the surface of the earth. Therefore, both the objects will escape from the surface at the same velocity. 

    Question 429
    CBSEENPH11017808

    Why gravitational potential energy is negative?

    Solution

    Gravitational P.E is given by, 
                   U = negative space GMm over straight r
    Negative sign implies that the P.E. is due to the attractive gravitational force exerted by the Earth on the body.  In order to bring a body from infinity to a particular point, work has been performed by the gravitational field of the earth. That is, there is an attractive force. 

    Question 430
    CBSEENPH11017811

    A body is placed on smooth horizontal surface. If a force equal to its weight is applied in horizontal direction then with what acceleration it will accelerate?

    Solution

    Given that the force equal to the body's weight is applied in the horizontal direction. Therefore, the body will not accelearte. 

    Question 431
    CBSEENPH11017812

    A body is projected with velocity less than escape velocity. What is the sign of total energy at:

    (i) surface of the earth,

    (ii) the highest point?

    Solution

    (i) Total energy at the surface of the Earth is negative.
    (ii) Total energy at the highest point is negative. 
    Energy is negative because it is due to the attractive gravitational potential force exerted by earth on the body. 

    Question 432
    CBSEENPH11017815

    Why does the moon has no atmosphere? 

    Solution
    The value of acceleration due to gravity on the surface of the moon is small. This implies, the value of escape velocity on the surface of the moon is also small (ve = 2.5 km / s). The molecules of the atmospheric gases on the surface of the moon have thermal velocities grater than the escape velocity. Therefore, the molecules of gases escape and there is no atmosphere on the moon. 
    Question 433
    CBSEENPH11017817

    A body weighs 64 N on the surface of the Earth. What is the gravitational force on it due to the earth, at a height equal to half the radius of the earth? 

    Solution

    Weight of the body = mg = 64 N
    So, mass of the body, m = 64 / g = 64/ 10 = 6.4 kg
    At height h, the value of g' is given by, 
    straight g apostrophe space equals space straight g space straight R squared over open parentheses straight R plus straight h close parentheses squared space equals space 10 open parentheses space fraction numerator straight R over denominator straight R space plus space straight R divided by 2 end fraction close parentheses 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 equals space 10 space straight x space 4 divided by 9

therefore space Weight space at space straight a space height space straight h space equals space mg apostrophe space

space space space space space space space space space space space space space space space space space 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 6.4 space straight x space 10 space straight x space 4 divided by 9 space straight N

space space space space space space space space space space space space space space space space space 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 28.44 space straight N 

    Question 434
    CBSEENPH11017821

    If the Earth stops rotating about its axis, by what value the acceleration due to gravity at equator will change? 

    Solution

    The value of g will increase by Rstraight omega squared
    where, 
    R is the radius of the earth, and 
    straight omega is the angular velocity of the rotation of the Earth. 

    Question 435
    CBSEENPH11017825
    Question 436
    CBSEENPH11017829

    Prove that the value of acceleration due to gravity at the surface of the earth is more than that at points above and below it.

    Solution

    The value of acceleration due to gravity varies with altitude, depth, shape and rotation of earth about it's axis. 
    i) On the surface of the earth, 
    g = GM subscript straight e over straight R squared                                   ... (1)
    ii) At a height h above the surface of the earth, 
    g' = fraction numerator GM subscript straight e over denominator left parenthesis straight R space plus space straight h right parenthesis squared end fraction                            ... (2)
    iii) At a height below the surface of the earth, 
    g ' = straight g space left parenthesis space 1 space minus space fraction numerator 2 straight h over denominator straight R subscript straight e end fraction right parenthesis                        ... (3)
    From (1), (2) and (3), we can say that the value of 'g' decreases with height and is maximum on the surface of the earth. 
    That is why, the value of 'g' is lesser at mountains than in plains. 

     

    Question 437
    CBSEENPH11017831

    Why does a body weigh zero at the centre of the earth though its mass is not zero.

    Solution

    The mass of the body remains the same. But, the weight of the body changes.
    At the centre of the earth, acceleration due to gravity is zero. 
    Weight, W = mg 
    Therefore, 
    Weight of the body = 0

    Question 438
    CBSEENPH11017833

    Why do the bodies possess weight at the surface of moon though moon is satellite of the earth?

    Solution

    All bodies have a definite mass. 
    Acceleration due to gravity on the moon is one-sixth of acceleration due to gravity on earth. 
    Therefore, the body will possess some weight, as weight is the product of mass and acceleration due to gravity. 

    Question 439
    CBSEENPH11017835

    Discuss the effect of rotation of the earth on the value of g.

    Solution

    The value of g decreases, due to rotation of the Earth.

    Question 440
    CBSEENPH11017837

    Compare and differentiate orbital velocity and escape velocity? 

    Solution
    S. No. Orbital Velocity Escape velocity
    1. The minimum velocity required to put the satellite into the Earth's orbit is called orbital velocity.  The minimum velocity required to throw a body out of the gravitational field of earth is called escape velocity. 
    2.  Orbital velocity of a satellite at height, straight v subscript straight o space equals space square root of fraction numerator GM over denominator straight R space plus space straight h end fraction end root Escape velocity of a satellite near the Earth, vesquare root of fraction numerator 2 GM over denominator straight R end fraction end root
    3. When satellite is orbiting close to earth, vosquare root of GM over straight R end root space equals space fraction numerator v subscript o over denominator square root of 2 end fraction Escape velocity f a satellite near the earth, vesquare root of 2 gR end root
    Question 441
    CBSEENPH11017840

    If the diameter of the earth becomes three times its present value but its mass remains unchanged, then how would the weight of the body on the surface of the earth be affected?

    Solution

    If the diameter of the earth becomes three times its present value weight of the body on the surface of the earth decreases to one-ninth. 

    Question 442
    CBSEENPH11017842

    Two bodies of masses 100 kg and 10,000 kg are at a distance 1 m apart. At which point on the line joining them will be the resultant gravitational field intensity zero?

    Solution

    Let, x be the distance of the point P from 100 kg body where the resultant gravitational intensity is zero. 
    Therefore, gravitational intensity at P due to body of mass 100 kg is equal and opposite to that due to body of mass 10, 000 kg. 
    Hence, 

    space space space space space fraction numerator straight G space straight x space 100 over denominator straight x squared end fraction space equals space fraction numerator straight G space straight x space 10000 over denominator left parenthesis 1 minus straight x right parenthesis squared end fraction

rightwards double arrow space space space 100 space straight x squared space equals space left parenthesis 1 minus straight x right parenthesis squared

rightwards double arrow space space space space space space 10 space straight x space equals space 1 space minus space straight x space

rightwards double arrow space space space space space space 11 space straight x space equals space 1 space

straight i. straight e. comma space straight x space equals space 1 over 11 space straight m
    At this point, the gravitational field intensity will be 0. 

    Question 443
    CBSEENPH11017845

    Define gravitational binding energy.
    [Hint : energy required to take the body out of the region of gravitational field]

    Solution

    Gravitational Binding Energy is the minimum energy that is required to free a body from the  gravitational attraction. Binding Energy is the negative of the total energy. 

    Question 445
    CBSEENPH11017851

    An artificial satellite of mass 300 kg is revolving in an orbit at a height equal 500km from the surface of the earth. Find the kinetic energy, potential energy and total energy of satellite. (Take g = 9.8 m/s2 at the surface of the earth)

    Solution

    Mass of the satellite = 300 kg
    Height  = 500 km, from the surface of the Earth
    Here, we have
    r = R + h
       = 6.4 x 106 + 500 x 103
       = 7.0 x 10m
    m = 300 kg;
    M = 6 x 1024 kg;
    G = 6.67 x 10-11 N m2 kg-2

    i) K.E of satellite =
     1 half space mv squared space equals space 1 half space straight m space GM over straight r

space space space space space space space space space space space space space space equals space 1 half cross times 300 straight x open parentheses fraction numerator 6.67 space straight x 10 to the power of negative 11 end exponent space straight x space 6 space straight x space 10 to the power of 24 over denominator 7 space straight x space 10 to the power of 6 end fraction close parentheses

space space space space space space space space space space space space space equals space 8.7 space straight x space 10 to the power of 9 space straight J
    ii right parenthesis space straight P. straight E space equals space minus space fraction numerator straight G space straight M space straight m over denominator straight r end fraction

space space space space space space space space space space space equals space fraction numerator negative left parenthesis 6.67 space straight x space 10 to the power of negative 11 end exponent right parenthesis space straight x space left parenthesis space 6 space straight x 10 to the power of 24 right parenthesis space straight x space 300 over denominator 7 space straight x space 10 to the power of 6 end fraction

space space space space space space space space space space equals space minus 17.4 space straight x space 10 to the power of 9 space straight J

iii right parenthesis straight T. straight E space equals space straight K. straight E space plus space straight P. straight E space

space space space space space space space space space space space equals 8.7 space straight x space 10 to the power of 9 space plus space left parenthesis negative 17.4 space straight x space 10 to the power of 9 right parenthesis space

space space space space space space space space space space space equals space minus 8.7 space straight x space 10 to the power of 9 space straight J

     

    Question 446
    CBSEENPH11017852

    Calculate the velocity of an atmospheric particle 1600 km above the earth's surface, given that the radius of the earth is 6400 km and acceleration due to gravity on the surface of the earth is 9.8 m/s2.

    Solution

    We have, 
    h = 1600 km 
    R = 6400 km = 6.4 x 106
    Therefore, 
    r = R + h = 6400 +1600
      = 8000 m = 8.0 x 106 m
     
    Hence, escape velocity is given by,
    straight v subscript straight e space equals space square root of 2 space straight g apostrophe straight r end root space equals space square root of 2 open parentheses gR squared over straight r squared close parentheses straight r end root space

space space space space space equals space square root of fraction numerator 2 space gR squared over denominator straight r end fraction end root space equals space square root of fraction numerator 2 straight x 9.8 space straight x space left parenthesis 6.4 space straight x space 10 to the power of 6 right parenthesis squared over denominator 8 space straight x space 10 to the power of 6 end fraction end root

space space space space space equals space 10.02 space straight x space 10 cubed space straight m divided by straight s

    Question 447
    CBSEENPH11017853

    Why do different planets have different escape velocities? 

    Solution

    Escape velocity is given by,
    straight v subscript straight e space equals space square root of fraction numerator 2 GM over denominator straight R end fraction end root 
    The mass and size of different planets is different. Therefore, escape velocity is different for different planets. 

    Question 448
    CBSEENPH11017854

    Two artificial satellites, one close to the surface and the other away are revolving around the earth. Which has larger speed? 

    Solution


     The relation for orbital velocity , v = square root of fraction numerator GM over denominator straight R plus straight h end fraction end root
    where, 

    h is the height of the satellite above the Earth's surface. 
    It is evident from the above formula that, smaller the value of h, greater is the value of v and vice versa. 
    Therefore, the satellite which is revolving close to the earth has larger speed. 

    Question 453
    CBSEENPH11019905

    In which of the following examples of motion, can the body be considered approximately a point object: 

    (a) a railway carriage moving without jerks between two stations

    Solution
    The size of a carriage is very small as compared to the distance between two stations. Therefore, the carriage can be treated as a point-sized object.
    Question 454
    CBSEENPH11019906

    In which of the following examples of motion, can the body be considered approximately a point object:

    (b) a monkey sitting on top of a man cycling smoothly on a circular track

    Solution
    The size of a monkey is very small as compared to the size of a circular track. Therefore, the monkey can be considered as a point-sized object on the track.
    Question 455
    CBSEENPH11019907

    In which of the following examples of motion, can the body be considered approximately a point object:

    c) a spinning cricket ball that turns sharply on hitting the ground.

    Solution
    The size of a spinning cricket ball is comparable to the distance through which it turns sharply on hitting the ground. Therefore, the cricket ball cannot be considered as a point object.
    Question 456
    CBSEENPH11019908

    In which of the following examples of motion, can the body be considered approximately a point object:

    (d) a tumbling beaker that has slipped off the edge of a table.

    Solution
    The size of a beaker is comparable to the height of the table from which it slipped. Hence, the beaker cannot be considered as a point object.
    Question 459
    CBSEENPH11019911

    A jet airplane travelling at the speed of 500 km h–1 ejects its products of combustion at the speed of 1500 km h–1 relative to the jet plane. What is the speed of the latter with respect to an observer on the ground ?

    Solution
    Speed space of space jet space airplane comma space straight v subscript jet space equals space 500 space km divided by hr space

relative space spedd space of space product space of space combustion
space straight w. straight r. to space the space plane space is comma space

straight v subscript smoke space equals space minus 1500 space km divided by hr space

Speed space of space the space product space of space combustion space straight w. straight r. to space ground space equals space straight v apostrophe subscript smoke

Relative space speed space of space space products space of space combustion space
with space respect space to space the space airplane comma

space space space space space straight v subscript smoke space equals space straight v prime subscript smoke space – space straight v subscript jet space end subscript

space space space – space 1500 space equals space straight v prime subscript smoke space end subscript – space 500 space straight v prime subscript smoke space end subscript

space space space space space space space space space space space space space space space space space equals space – space 1000 space km divided by straight h space
    Negative sign implies that the direction of its products of combustion is opposite to the direction of motion of the jet airplane. 
    Question 461
    CBSEENPH11019913

    Two trains A and B of length 400 m each are moving on two parallel tracks with a uniform speed of 72 km h–1 in the same direction, with A ahead of B. The driver of B decides to overtake A and accelerates by 1 m s–2. If after 50 s, the guard of B just brushes past the driver of A, what was the original distance between them ?

    Solution

    In case of train A: 
    Initial velocity, u = 72 km/h = 20 m/s
    Time, t = 50 s
    Since the train is moving with uniform velocity, 
    Acceleration, aI = 0
    Now, using second equation of motion, 
    Distance covered by train A is given by, 
    straight s space equals space ut space plus space left parenthesis 1 divided by 2 right parenthesis straight a subscript 1 straight t squared space

space space space equals space 20 cross times 50 plus space 0 space equals space 1000 space straight m 
    For train B, 
    Initial velocity, uB = 72 km/h = 20 m/s
    Acceleration, a = 1 m/s

    Time, t = 50 s
    Now, using second equation of motion, 
    Distance travelled by second train B is, 
    straight s subscript II space equals space ut space plus space 1 half at squared space
space space space space space equals space 20 space straight X space 50 space plus space left parenthesis 1 divided by 2 right parenthesis space cross times space 1 space cross times space left parenthesis 50 right parenthesis squared space

space space space space space equals space 2250 space straight m space
    Length of both trains = 2 cross times 400 space straight m space equals space 800 space straight m
    Therefore, the original distance between the driver of train A and the guard of train B = 2250 - 1000 - 800
    = 450m.

    Question 463
    CBSEENPH11019915

    Two towns A and B are connected by a regular bus service with a bus leaving in either direction every T minutes. A man cycling with a speed of 20 km h-1 in the direction A to B notices that a bus goes past him every 18 min in the direction of his motion, and every 6 min in the opposite direction. What is the period T of the bus service and with what speed (assumed constant) do the buses ply on the road?

    Solution

    Let V be the speed of the bus running between towns A and B.
    Given, 
    Speed of the cyclist  = 20 km/hr 
    Relative speed of bus moving in the direction of the cyclist = V - v = (V-20) km/hr
    Every 18 mins, the bus went past the cyclist. moving in the direction of the bus. 
    That is, 18 over 60 space h r s
    Distance covered by the bus = left parenthesis straight V minus 20 right parenthesis cross times 18 over 60 space km space space space space space space space... space left parenthesis straight i right parenthesis space
    One bus leaves every t minutes. 
    Therefore, 
    Distance travelled by the bus = V cross times space straight T over 60     ... (ii) 
    Equations (i) and (ii) are equal. 
    left parenthesis straight V space minus space 20 right parenthesis space cross times space fraction numerator 18 space over denominator space 60 end fraction space equals space fraction numerator VT space over denominator 60 end fraction space space space space space space space space... left parenthesis iii right parenthesis
    Relative speed of the bus moving in the opposite direction of the cyclist = (V + 20) km/h
    Time taken by the bus to go past the cyclist = 6 min = 6 / 60 h 
    Therefore, 
    left parenthesis straight V plus 20 right parenthesis space cross times space 6 over 60 space equals space VT over 60 space space space space space space... space left parenthesis iv right parenthesis
    Now, from equations (3) and (4), we get
    space space space space space space space space space space left parenthesis straight V space plus space 20 right parenthesis space cross times space 6 space divided by space 60 space equals space left parenthesis straight V space minus space 20 right parenthesis space cross times space 18 space divided by space 60

space rightwards double arrow space space space space space space space space space straight V space plus space 20 space equals space 3 straight V space minus space 60 space

rightwards double arrow space space space space 2 straight V space equals space 80 space straight V space equals space 40 space km divided by straight h
Putting space the space value space of space straight V space in space equation space left parenthesis iv right parenthesis comma space we space get space

left parenthesis 40 space plus space 20 right parenthesis space cross times space 6 space divided by space 60 space equals space 40 straight T space divided by space 60

space space space space space space space space space space space straight T space equals space 360 space divided by space 40 space equals space 9 space min

    Question 464
    CBSEENPH11019916

    A player throws a ball upwards with an initial speed of 29.4 m s-1.

    (a) What is the direction of acceleration during the upward motion of the ball ?

    Solution

    Acceleration due to gravity always acts in the downward direction, irrespective of the direction of motion. 

    Question 465
    CBSEENPH11019917

    A player throws a ball upwards with an initial speed of 29.4 m s-1

    What are the velocity and acceleration of the ball at the highest point of its motion ?

    Solution
    At maximum height, velocity of the ball becomes zero.
    Acceleration due to gravity at a given place is constant and acts on the ball at all points (including the highest point) with a constant value i.e., 9.8 m/s2.
    Question 466
    CBSEENPH11019918

    A player throws a ball upwards with an initial speed of 29.4 m s-1

    Choose the x = 0 m and t = 0 s to be the location and time of the ball at its highest point, vertically downward direction to be the positive direction of x-axis, and give the signs of position, velocity and acceleration of the ball during its upward, and downward motion.

    Solution
    During upward motion, the sign of position is positive, sign of velocity is negative, and sign of acceleration is positive.
    During downward motion, the signs of position, velocity, and acceleration are all positive.
    Question 467
    CBSEENPH11019919

    A player throws a ball upwards with an initial speed of 29.4 m s-1.


    To what height does the ball rise and after how long does the ball return to the player’s hands ?

    (Take g = 9.8 m s-2and neglect air resistance).

    Solution
    Initial velocity of the ball, u = 29.4 m/s 
    At maximum height, the velocity of the ball becomes zero.
    Final velocity of the ball, v = 0 

    Acceleration, a = – g = – 9.8 m/s2  
    Using third equation of motion, 
    Height to which the ball rises is given by, 
    straight v squared space minus space straight u squared space equals space 2 gs

straight s space equals space fraction numerator left parenthesis straight v squared minus straight u squared right parenthesis over denominator 2 straight g end fraction space

space space equals fraction numerator space left square bracket left parenthesis 0 right parenthesis squared space minus space left parenthesis 29.4 right parenthesis squared right square bracket space over denominator 2 cross times left parenthesis negative 9.8 right parenthesis end fraction space space equals 44.1 space straight m
Also comma space

Time space of space ascent space equals space Time space of space desent space

So comma space total space time space taken space by space the space ball space to space return space to space the space
p layer apostrophe straight s space hand space is comma

straight t space equals space 3 space plus space 3 space equals space 6 space sec

    Question 473
    CBSEENPH11019925

    Explain clearly, with examples, the distinction between :

    (a) magnitude of displacement (sometimes called distance) over an interval of time,
    and the total length of path covered by a particle over the same interval

    Solution
    The magnitude of displacement over an interval of time is the shortest distance (which is a straight line) between the initial and final positions of the particle.
    The actual path length covered by the particle in a given interval of time is the total length of the path covered by the particle. 

    Consider, for example, a particle moves from point A to point B and then, comes back to a point, C taking a total time t. 
    Displacement of the particle = AC 
    Total path length = AB + BC
    Magnitude of displacement can never be greater than the total path length. However, in some cases, both quantities are equal to each other.


    Question 476
    CBSEENPH11019928

    In Exercises 3.13 and 3.14, we have carefully distinguished between average speed and magnitude of average velocity. No such distinction is necessary when we consider instantaneous speed and magnitude of velocity. The instantaneous speed is always equal to the magnitude of instantaneous velocity. Why?

    Solution
    Instantaneous velocity is given by the first derivative of distance with respect to time.
    i.e. , vIn = dx / dt 
    Here, the time interval dt is so small that it is assumed that the particle does not change its direction of motion.
    As a result, both the total path length and magnitude of displacement become equal is this interval of time. 
    Therefore, instantaneous speed is always equal to instantaneous velocity.
    Question 478
    CBSEENPH11019930

    Suggest a suitable physical situation for each of the following graphs.






    Solution
    (a) The given position-time graph shows that initially a body was at rest. Gradually, its velocity increases with time and attains an instantaneous constant value. Then, the velocity reduces to zero with an increase in time. Then, its velocity increases with time in the opposite direction and acquires a constant value.
    A physical situation analogous to this graph arises when a football (initially kept at rest) is kicked and gets rebound from a rigid wall so that its speed gets reduced. Then, it passes from the player who has kicked it and ultimately gets stopped after sometime. 
    (b) In the given velocity-time graph, the sign of velocity changes and its magnitude decreases with a passage of time. A similar situation arises when a ball is dropped on the hard floor from a height. It strikes the floor with some velocity and upon rebound, its velocity decreases by a factor. This continues till the velocity of the ball eventually becomes zero.

    (c) The given acceleration-time graph reveals that initially the body is moving with a certain uniform velocity. Its acceleration increases for a short interval of time, which again drops to zero. This indicates that the body again starts moving with the same constant velocity. A similar physical situation arises when a hammer moving with a uniform velocity strikes a nail. 
    Question 479
    CBSEENPH11019931

    Figure 3.23 gives the x-t plot of a particle executing one-dimensional simple harmonic motion. (You will learn about this motion in more detail in Chapter14). Give the signs of position, velocity and acceleration variables of the particle at t = 0.3 s, 1.2 s, - 1.2 s.


    Solution

    At t= 0.3 sec,
    Negative, Negative, Positive.
    At t= 1.2 sec,
    Positive, Positive, Negative
    At t= -1.2 sec
    Negative, Positive, Positive
    For simple harmonic motion (SHM) of a particle, acceleration (a) is given by, 
    a = – ω2x ;  ω → angular frequency       … (i) 
    t = 0.3 s 

    In this time interval, x is negative. Thus, the slope of the position-time graph will also be negative. Therefore, both position and velocity are negative.
    However, using equation (i), acceleration of the particle will be positive.
    t = 1.2 s 

    In this time interval, x is positive. Thus, the slope of the x-graph will also be positive. Therefore, both position and velocity are positive.
    However, using equation (i), acceleration of the particle comes to be negative. 
    t = – 1.2 s 

    In this time interval, x is negative. Thus, the slope of the x-t plot will also be negative. Since both x and t are negative, the velocity comes to be positive.
    From equation (i), it can be inferred that the acceleration of the particle will be positive.

    Question 480
    CBSEENPH11019932

    Figure 3.24 gives the x-t plot of a particle in one-dimensional motion. Three different equal intervals of time are shown. In which interval is the average speed greatest, and in which is it the least? Give the sign of average velocity for each interval.



    Solution
    The average speed of a particle shown in the x-t graph is obtained from the slope of the graph in a particular interval of time.
    It is clear from the graph that the slope is maximum and minimum restively in intervals 3 and 2 respectively. Therefore, the average speed is greatest in interval 3 and least in interval 2. 
    For intervals 1 and 2, average velocity is positive as the slope is positive. For interval 3 slope is negative therefore, the average velocity is negative. 

    Question 481
    CBSEENPH11019933

    Figure 3.25 gives a speed-time graph of a particle in motion along a constant direction. Three equal intervals of time are shown. In which interval is the average acceleration greatest in magnitude ? In which interval is the average speed greatest ? Choosing the positive direction as the constant direction of motion, give the signs of v and a in the three intervals. What are the accelerations at the points A, B, C and D?


    Solution
    Acceleration is given by the slope of the speed-time graph.
    Since the slope of the given speed-time graph is maximum in interval 2, the average acceleration will be the greatest in this interval.
    The height of the curve from the time-axis gives the average speed of the particle.
    In the given graph, it is clear that the height is the greatest in interval 3.
    Hence, the average speed of the particle is the greatest in interval 3.
    In interval 1:

    The slope of the speed-time graph is positive. Hence, acceleration is positive. Similarly, the speed of the particle is positive in this interval.
    In interval 2:
    The slope of the speed-time graph is negative. Hence, acceleration is negative in this interval. However, speed is positive because it is a scalar quantity.
    In interval 3:
    The slope of the speed-time graph is zero. Hence, acceleration is zero in this interval. However, here the particle acquires some uniform speed. It is positive in this interval.
    Points A, B, C, and D are all parallel to the time-axis. Hence, the slope is zero at these points. Therefore, at points A, B, C, and D, acceleration of the particle is zero. 
    Question 482
    CBSEENPH11019934

    A three-wheeler starts from rest, accelerates uniformly with 1 m s-2 on a straight road for 10 s, and then moves with uniform velocity. Plot the distance covered by the vehicle during the nth second (n = 1,2,3….) versus n. What do you expect this plot to be during accelerated motion : a straight line or a parabola ?

    Solution
    Distance space covered space by space straight a space body space in space straight n to the power of th
second space is space given space by space the space relation comma

straight S subscript straight n space end subscript equals space straight u space plus space straight a fraction numerator space left parenthesis 2 straight n space minus space 1 right parenthesis over denominator 2 end fraction space space space space space space space space space... left parenthesis straight i right parenthesis space

where comma space

straight u space equals space Initial space velocity comma
straight a space equals space acceleration comma space
straight n space equals space time space intervals space equals space 1 comma 2 comma 3 comma... straight n

Here space we space have comma space

straight u space equals space 0 space and space straight a space equals space 1 space straight m divided by straight s squared

Therefore comma space

straight S subscript straight n space equals fraction numerator space left parenthesis 2 straight n space minus space 1 right parenthesis space over denominator 2 end fraction space space space space space space space space space space space space space space space space space space space... space left parenthesis ii right parenthesis space

This space relation space shows comma space

straight S subscript straight n space proportional to space straight n space space space space space space space space space space space space space space space... space left parenthesis iii right parenthesis space
    Now, putting different values of n in equation (3), we have

    n 1 2 3 4 5 6 7 8 9 10
    Sn 0.5 1.5 2.5 3.5 4.5 5.5 6.5 7.5 8.5 9.5
     
    The graph between Sn and n will be a straight line as shown in the figure below. 


    Question 484
    CBSEENPH11019936

    On a long horizontally moving belt (Fig. 3.26), a child runs to and fro with a speed 9 km h-1 (with respect to the belt) between his father and mother located 50 m apart on the moving belt. The belt moves with a speed of 4 km h-1. For an observer on a stationary platform outside, what is the

    (a) speed of the child running in the direction of motion of the belt ?.

    (b) speed of the child running opposite to the direction of motion of the belt ?

    (c) time taken by the child in (a) and (b) ?

    Which of the answers alter if motion is viewed by one of the parents ?



    Solution
    a) Speed of the belt, vB = 4 km/h
    Speed of the boy, vb = 9 km/h
    The boy is running in the same direction of the motion of the belt. Therefore, speed of the child is given by, 
    vbB = vb + vB = 9 + 4 = 13 km/h
    b) The boy is running in a direction opposite to the direction of the motion of the belt. 
    Therefore, his speed as observed by the stationary observer is given by, 
    vbB = vb + (– vB) = 9 – 4 = 5 km/h
    c) Distance between child and his parents = 50 m 
    Given that, both parents are standing on the moving belt. Therefore, the speed of the child in either direction as observed by the parents will remain the same i.e., 9 km/h = 2.5 m/s.
    Hence, time taken by the child to move towards one of his parents = fraction numerator 50 over denominator 2.5 end fraction space equals space 20 space s e c
    If the motion is viewed by any one of the parents, answers obtained in (a) and (b) get altered. This is because the child and his parents are standing on the same belt and hence, are equally affected by the motion of the belt. Therefore, for both parents (irrespective of the direction of motion) the speed of the child remains the same i.e., 9 km/h.
    For this reason, it can be concluded that the time taken by the child to reach any one of his parents remains unaltered.
    Question 486
    CBSEENPH11019938

    The speed-time graph of a particle moving along a fixed direction is shown in Fig. 3.28. Obtain the distance traversed by the particle between (a) t = 0 s to 10 s, (b) t = 2 s to 6 s.

    What is the average speed of the particle over the intervals in (a) and (b)?



    Solution
    a)
    Distance travelled by the particle = Area under the given graph.
                                               = (1/2) × (10 - 0) × (12 - 0)
                                               = 60 m

    Average space Speed space equals space Distance over Time equals 60 over 10 space equals space 6 space straight m divided by straight s
    b) 
    Let s1 and s2 be the distances covered by the particle between time = 2 s to 5 s and t = 5s to 6 s respectively.
    Total distance (s) covered by the particle in time t = 2 s to 6 s is, 
                               s = s1 + s2                     … (i) 
    For distance s1
    Let u1 be the velocity of the particle after 2 s, and 
    a1 be the acceleration of the particle in t = 0 to t = 5s. 
    The particle undergoes uniform acceleration in the first 5 seconds. 
    Therefore, using first equation of motion v = u+at, we have
        12 = 0 + a1 × 5
    i.e.,  a1 = 12/5  = 2.4 m/s2
    Again, using the first equation of motion, we have 
        v at 

          = 0 + 2.4 × 2
          = 4.8 m/s
    Distance travelled by the particle between time 2 s and 5 s i.e., in 3 s
    s1 = u1 t + (1/2) a1 t

       = 4.8 × 3 + (1/2) × 2.4 × (3)
     
       = 25.2 m                                   ...(ii) 

    For distance s2
    Let a2 be the acceleration of the particle between time t = 5 s and t = 10 s.
    Using first equation of motion, 
    v = u + at
    0 = 12 + a2 × 5
    a2 = -12 / 5 = - 2.4 ms-2
    Therefore, 
    Distance travelled by the particle in 1s (i.e., between = 5 s and t = 6 s),
    s2u2 t + (1/2)a2 t2
       = 12 × 1 + (1/2) (-2.4) × (1)

       = 12 - 1.2 = 10.8 m                 ...(iii) 
    Now, from equations (i), (ii) and (iii), we have
    s = 25.2 + 10.8 = 36 m
    ∴ Average speed = 36 / 4 = 9 m/s 


    Question 488
    CBSEENPH11020155

    Suppose there existed a planet that went around the sun twice as fast as the earth.What would be its orbital size as compared to that of the earth?

    Solution
    Time taken by the Earth to complete one revolution around the Sun, 
                       Te = 1 year 
    Orbital radius of the Earth in its orbit, R= 1 AU 
    Time taken by the planet to complete one revolution around the Sun, TP = ½Te 
                                                  = ½ year 
    Orbital radius of the planet = Rp
     

    From Kepler’s third law of planetary motion,
     open parentheses straight R subscript straight P over straight R subscript straight e close parentheses cubed space equals space open parentheses T subscript P over T subscript e close parentheses squared space

open parentheses straight R subscript straight P over straight R subscript straight e close parentheses space equals space space open parentheses T subscript P over T subscript e close parentheses to the power of begin inline style bevelled 2 over 3 end style end exponent space

space space space space space space space space space space space space space equals space open parentheses fraction numerator bevelled 1 half over denominator 1 end fraction close parentheses to the power of begin inline style bevelled 2 over 3 end style end exponent space

space space space space space space space space space space space space space equals open parentheses space 0.5 close parentheses to the power of begin inline style bevelled 2 over 3 end style end exponent space

space space space space space space space space space space space space space equals 0.63 space 
    Hence, the orbital radius of the planet will be 0.63 times smaller than that of the Earth.
    Question 489
    CBSEENPH11020156

    Io, one of the satellites of Jupiter, has an orbital period of 1.769 days and the radius of the orbit is 4.22 × 108 m. Show that the mass of Jupiter is about one-thousandth that of the sun.

    Solution
    Orbital period of  I0 , TI0 = 1.769 days 
                                         =  1.769 × 24 × 60 × 60 s 
    Orbital radius of  I0 , RI0 = 4.22 × 108 m 
    Satellite I0 is revolving around the Jupiter 
    Mass of the Jupiter is given by, 
    MJ = 4π2RI03 / GTI02                         ...(i) 
    where, 
    MJ = Mass of Jupiter, 
    G = Universal gravitational constant,
    Orbital period of the earth,
    T= 365.25 days
        = 365.25 × 24 × 60 × 60 s 
    Orbital radius of the Earth,

    R
    = 1 AU = 1.496 × 1011 
    Mass of sun is given as, 
    Ms = 4π2Re3 / GTe                           ...(ii) 
    Therefore,
    Ms / MJ  = open parentheses fraction numerator 4 straight pi squared straight R subscript straight e cubed over denominator GT subscript 2 squared end fraction space close parentheses space x space open parentheses fraction numerator G T subscript 10 squared over denominator 4 pi squared R subscript 10 cubed end fraction close parentheses
                 =  bevelled fraction numerator open parentheses begin display style straight R subscript straight e cubed over straight T subscript 10 squared end style close parentheses over denominator open parentheses begin display style R subscript 10 cubed over T subscript e squared end style close parentheses end fraction
    Substituting the values, 
    = open parentheses fraction numerator 1.769 space straight x space 24 space straight x space 60 straight x 60 over denominator 365.25 space straight x space 24 space straight x 60 straight x 60 end fraction close parentheses squared space x space open parentheses fraction numerator 1.496 space x space 10 to the power of 11 over denominator 4.22 space x space 10 to the power of 8 end fraction close parentheses cubed
     = 1045.04
    ∴ Ms / MJ  ~ 1000 
               M~ 1000 × M
    Hence, it can be inferred that the mass of Jupiter is about one-thousandth that of the Sun.
    Question 490
    CBSEENPH11020157

    Let us assume that our galaxy consists of 2.5 × 1011stars each of one solar mass. How long will a star at a distance of 50,000 ly from the galactic centre take to complete one revolution? Take the diameter of the Milky Way to be 105 ly.

    Solution
    Mass of our galaxy Milky Way, M = 2.5 × 1011 solar mass 
    Solar mass = Mass of Sun = 2.0 × 1036 kg 
    Mass of our galaxy, M = 2.5 × 1011 × 2 × 1036 
                                    = 5 × 1041 kg 
    Diameter of Milky Way, d = 105 ly 
    Radius of Milky Way, r = 5 × 104 ly 
                               1 ly = 9.46 × 1015 m 
    ∴                             r = 5 × 104 × 9.46 × 1015 

                                     = 4.73 ×1020 m
     
    A star revolves around the galactic centre of the Milky Way.
    Time period is given by the relation, 
    = open parentheses fraction numerator 4 straight pi squared straight r cubed over denominator GM end fraction close parentheses to the power of begin inline style bevelled 1 half end style end exponent 

        = square root of fraction numerator 4 space straight x space 3.14 squared straight x space 4.73 cubed space straight x space 10 to the power of 60 over denominator 6.67 space straight x space 10 to the power of negative 11 end exponent straight x space 5 space straight x space 10 to the power of 41 end fraction end root 
       =   square root of fraction numerator 39.48 space straight x space 105.82 space straight x space 10 to the power of 30 over denominator 33.35 end fraction end root

        = 1.12 × 1016 s   
    1 year = 365 × 324 × 60 × 60 s 
           1s = open parentheses fraction numerator 1 over denominator 365 space straight x 24 space straight x 60 straight x 60 end fraction close parentheses years 
    Therefore,
    1.12 × 1016 s = open parentheses fraction numerator 1.12 space cross times space 10 to the power of 16 over denominator 365 space straight x space 24 space straight x 60 space straight x 60 end fraction close parentheses
                          =  3.55 × 108 years. 
    Question 491
    CBSEENPH11020158

     Does the escape speed of a body from the earth depend on: 

    (a) the mass of the body, 

    (b) the location from where it is projected, 

    (c) the direction of projection, 

    (d) the height of the location from where the body is launched?

    Solution
    The escape velocity is independent of the mass of the body and the direction of projection. It depends upon the gravitational potential at the point from where the body is launched. Gravitational potential depends slightly on the latitude and height of the point, therefore, the escape velocity depends on these factors. 
    Question 492
    CBSEENPH11020159

    Which of the following symptoms is likely to afflict an astronaut in space

    (a) swollen feet,

    (b) swollen face,

    (c) headache,

    (d) orientational problem?

    Solution
    (a) When a person is in standing position, legs hold the entire mass of a body due to gravitational pull. While in space, an astronaut feels weightlessness because of the absence of gravity. Therefore, swollen feet of an astronaut do not affect him/her in space.
    (b) The apparent weightlessness in space causes swollen face.
    Sense organs such as eyes, ears nose, and mouth constitute a person’s face. This symptom can affect an astronaut in space.
    (c) Headaches are caused because of mental strain. It can affect the working of an astronaut in space.
    (d) Space has different orientations. Therefore, the orientational problem can affect an astronaut in space.
    Question 493
    CBSEENPH11020160

    Choose the correct answer from among the given ones:The gravitational intensity at the centre of a hemispherical shell of uniform mass density has the direction indicated by the arrow (see Fig 8.12) (i) a, (ii) b, (iii) c, (iv) O.


    Solution
    Gravitational potential (V) is constant at all points in a spherical shell. Hence, the gravitational potential gradient (dV/dR) is zero everywhere inside the spherical shell.
    The gravitational potential gradient is equal to the negative of gravitational intensity. Hence, the intensity is also zero at all points inside the spherical shell. This indicates that gravitational forces acting at a point in a spherical shell are symmetric. 
    If the upper half of a spherical shell is cut out (as shown in the given figure), then the net gravitational force acting on a particle located at centre O will be in the downward direction.
                               
    Since gravitational intensity at a point is defined as the gravitational force per unit mass at that point, it will also act in the downward direction.
    Thus, the gravitational intensity at centre O of the given hemispherical shell has the direction as indicated by arrow c.

    Question 494
    CBSEENPH11020161

    Choose the correct answer from among the given ones: 

    For the problem 8.10, the direction of the gravitational intensity at an arbitrary point P is indicated by the arrow (i) d, (ii) e, (iii) f, (iv) g. 

    Solution
    Hence, the correct answer is (ii).
    Gravitational potential (V) is constant at all points in a spherical shell. Hence, the gravitational potential gradient (dV/dR) is zero everywhere inside the spherical shell. The gravitational potential gradient is equal to the negative of gravitational intensity. Hence, intensity is also zero at all points inside the spherical shell. This indicates that gravitational forces acting at a point in a spherical shell are symmetric. 

    If the upper half of a spherical shell is cut out (as shown in the given figure), then the net gravitational force acting on a particle at an arbitrary point P will be in the downward direction.

    Since gravitational intensity at a point is defined as the gravitational force per unit mass at that point, it will also act in the downward direction. Thus, the gravitational intensity at an arbitrary point P of the hemispherical shell has the direction as indicated by arrow e.
     


    Question 495
    CBSEENPH11020162

    How will you ‘weigh the sun’, that is estimate its mass? The mean orbital radius of the earth around the sun is 1.5 × 108 km.

    Solution
    Orbital radius of the Earth around the Sun, r = 1.5 × 1011 m
    Time taken by the Earth to complete one revolution around the Sun, 
    T = 1 year
      = 365.25 days 
      = 365.25 × 24 × 60 × 60 s 
    Universal gravitational constant, G = 6.67 × 10–11 Nm2 kg–2 

    Thus, mass of the Sun can be calculated using the relation, 
    M = fraction numerator 4 straight pi squared straight r cubed over denominator GT squared end fraction
       = fraction numerator 4 space straight x space 3.14 squared space straight x space left parenthesis 1.5 space straight x space 10 to the power of 11 right parenthesis cubed over denominator left square bracket 6.67 space straight x space 10 to the power of negative 11 end exponent space straight x space left parenthesis 365.25 space straight x space 24 space straight x 60 space straight x 60 right parenthesis squared right square bracket end fraction 
        = 2 × 1030 kg 
    Hence, the mass of the Sun is 2 × 1030 kg.
    Question 496
    CBSEENPH11020163

    A Saturn year is 29.5 times the earth year. How far is the Saturn from the sun if the earth is 1.50 ×108 km away from the sun?

    Solution
    Distance of the Earth from the Sun, re = 1.5 × 108 km 
                                                               = 1.5 × 1011 m  
    Time period of the Earth = 

    Time period of Saturn, Ts = 29. 5 T

    Distance of Saturn from the Sun = r

    From Kepler’s third law of planetary motion, 
     
          T = (4π2r3 / GM)1/2 

    For Saturn and Sun, 
    rs3 / re3  =  Ts2 / Te

             rs = re(Ts / Te)2/3 

                = 1.5 × 1011 (29.5 Te / Te)2/3 

                 = 1.5 × 1011 (29.5)2/3 

                 = 14.32 × 1011 m 
    Hence, the distance between Saturn and the Sun is 1.43 × 1012 m.
    Question 497
    CBSEENPH11020164

    The escape speed of a projectile on the earth’s surface is 11.2 km s–1. A body is projected out with thrice this speed. What is the speed of the body far away from the earth? Ignore the presence of the sun and other planets. 

    Solution
    Escape velocity of a projectile from the Earth, vesc = 11.2 km/s 
    Projection velocity of the projectile, vp = 3vesc 

    Mass of the projectile = 

    Velocity of the projectile far away from the Earth = v

    Total energy of the projectile on the Earth = open parentheses 1 half close parenthesesmvp2 - open parentheses 1 half close parenthesesmvesc

    Gravitational potential energy of the projectile far away from the Earth is zero. 
    Total energy of the projectile far away from the Earth = open parentheses 1 half close parenthesesmvf

    From the law of conservation of energy, we have 
    open parentheses 1 half close parenthesesmvp2 - open parentheses 1 half close parenthesesmvesc2  =  open parentheses 1 half close parentheses mvf
                                   vf = ( vp2 - vesc)1/2 

                                      = [ (3vesc)2 - vesc2 ]1/2 

                                      = √8 vesc 

                                      = √8 × 11.2 
                                      =  31.68 km/s.
    Question 498
    CBSEENPH11020165

    A spaceship is stationed on Mars. How much energy must be expended on the spaceship to launch it out of the solar system? Mass of the space ship = 1000 kg; mass of the Sun = 2 × 1030 kg; mass of mars = 6.4 × 1023 kg; radius of mars = 3395 km; radius of the orbit of mars = 2.28 × 108kg; G= 6.67 × 10–11 m2kg–2.

    Solution
    Mass of the spaceship, m= 1000 kg
    Mass of the Sun, M = 2 × 1030 kg 
    Mass of Mars, mm = 6.4 × 10 23 kg 
    Orbital radius of Mars, R = 2.28 × 10kg
                                        =2.28 × 1011
    Radius of Mars, = 3395 km
                              = 3.395 × 106 m 
    Universal gravitational constant, G = 6.67 × 10–11 m2kg–2 

    Potential energy of the spaceship due to the gravitational attraction of the Sun = negative GMm subscript straight s over straight R

    Potential energy of the spaceship due to the gravitational attraction of Mars = -fraction numerator GM subscript straight m straight m subscript straight s over denominator straight r end fraction 

    Since the spaceship is stationed on Mars, its velocity and hence, its kinetic energy will be zero. 
    Total energy of the spaceship =  fraction numerator negative G M m subscript straight s over denominator straight R end fraction space minus space fraction numerator G M subscript straight m straight m subscript straight s over denominator straight r end fraction
                                                = negative space Gm subscript straight s space open square brackets straight M over straight R plus straight m subscript straight m over straight r close square brackets 
    The negative sign indicates that the system is in bound state. 
    Energy required for launching the spaceship out of the solar system, 
    = – (Total energy of the spaceship) 
    = space Gm subscript straight s space open square brackets straight M over straight R plus straight m subscript straight m over straight r close square brackets 
    = 6.67 × 1011 × 103 ×open square brackets open parentheses fraction numerator 2 space straight x space 10 to the power of 30 over denominator 2.28 space straight x space 10 to the power of 11 end fraction close parentheses space plus space open parentheses fraction numerator 6.4 space x space 10 to the power of 23 over denominator 3.395 space x space 10 to the power of 6 end fraction close parentheses close square brackets
    = 596.97 × 109 
    =  6 × 1011 J.
    Question 499
    CBSEENPH11020166

    A rocket is fired ‘vertically’ from the surface of mars with a speed of 2 km s–1. If 20% of its initial energy is lost due to Martian atmospheric resistance, how far will the rocket go from the surface of mars before returning to it? Mass of mars = 6.4× 1023 kg; radius of mars = 3395 km; G = 6.67× 10-11 N m2 kg–2.

    Solution
    Initial velocity of the rocket, v = 2 km/s = 2 × 103 m/s
    Mass of Mars, M = 6.4 × 1023 kg 
    Radius of Mars, R = 3395 km = 3.395 × 106 m 
    Universal gravitational constant, G = 6.67× 10–11 N m2 kg–2 

    Mass of the rocket = 

    Initial kinetic energy of the rocket = open parentheses 1 half close parenthesesmv

    Initial potential energy of the rocket = fraction numerator negative space GMm over denominator straight R end fraction 

    Total initial energy = open parentheses 1 half close parenthesesmv2 - fraction numerator negative space GMm over denominator straight R end fraction 

    If 20 % of initial kinetic energy is lost due to Martian atmospheric resistance, then only 80 % of its kinetic energy helps in reaching a height. 
    Total initial energy available = open parentheses 80 over 10 close parentheses x open parentheses 1 half close parentheses m v squared space minus space fraction numerator G M m over denominator R end fraction
    = (80/100) × (1/2) mv2 - GMm over straight R 
                                            =  0.4mv2 - GMm over straight R 

    Maximum height reached by the rocket = 

    At this height, the velocity and hence, the kinetic energy of the rocket will become zero. 
    Total energy of the rocket at height h = fraction numerator GMm over denominator left parenthesis straight R plus straight h right parenthesis end fraction

    Applying the law of conservation of energy for the rocket,
    0.4mv2 - GMm over straight R  =  fraction numerator GMm over denominator left parenthesis straight R plus straight h right parenthesis end fraction
                     0.4v2 =GM over straight R space minus space fraction numerator G M over denominator left parenthesis R plus h right parenthesis end fraction 
                              = fraction numerator GMh over denominator straight R left parenthesis straight R plus straight h right parenthesis end fraction 
                   fraction numerator straight R space plus straight h over denominator straight h end fraction space equals space fraction numerator GM over denominator 0.4 space straight v squared straight R end fraction  

                       straight R over straight h  = open parentheses fraction numerator GM over denominator 0.4 space straight v squared space straight R end fraction close parentheses space minus space 1
                          h = fraction numerator straight R over denominator open square brackets begin display style bevelled fraction numerator GM over denominator 0.4 space straight v squared straight R end fraction end style space minus space 1 space close square brackets end fraction 
                             = open parentheses fraction numerator 0.4 space straight R squared straight v squared over denominator GM space minus space 0.4 space straight v squared straight R end fraction close parentheses
    =fraction numerator 0.4 straight x space left parenthesis 3.395 space straight x space 10 to the power of 6 right parenthesis squared space straight x space left parenthesis 2 space straight x 10 cubed right parenthesis squared over denominator left square bracket space 6.67 space straight x space 10 to the power of 11 space straight x space 6.4 space straight x space 10 to the power of 23 space minus space 0.4 space straight x space left parenthesis 2 space straight x 10 cubed right parenthesis squared space straight x space left parenthesis 3.395 space straight x space 10 to the power of 6 right parenthesis right square bracket end fraction 
      = fraction numerator 18.442 space straight x space 10 to the power of 18 over denominator left square bracket 42.688 space straight x space 10 to the power of 12 space minus space 5.432 space straight x space 10 to the power of 12 right square bracket end fraction
      = fraction numerator 18.422 space straight x space 10 to the power of 6 over denominator 37.256 end fraction 
      = 495 × 103 m
      = 495 km.
    Question 500
    CBSEENPH11020254

    A satellite is revolving in a circular orbit at a height ‘h’ from the earth’s surface (radius of earth R ; h<<R). The minimum increase in its orbital velocity required, so that the satellite could escape from the earth’s gravitational field, is close to: (Neglect the effect of atmosphere.)

    • square root of 2 gR end root
    • square root of gR
    • square root of gR divided by 2 end root
    • square root of gR space left parenthesis square root of 2 space minus 1 right parenthesis

    Solution

    D.

    square root of gR space left parenthesis square root of 2 space minus 1 right parenthesis

    Given, a satellite is revolving in a circular orbit at a height h from the Earth's surface having radius of earth R, i.e h <<R
    Orbit velocity of a satellite
    straight v equals space square root of fraction numerator GM over denominator straight R plus straight h end fraction end root space equals space square root of GM over straight R end root space left parenthesis as space straight h less than less than straight R right parenthesis
Velocity space required space to space escape
1 half space mv squared space equals space fraction numerator GMn over denominator straight R plus straight h end fraction

straight v apostrophe space equals space square root of fraction numerator 2 GM over denominator straight R plus straight h end fraction end root space equals space square root of fraction numerator 2 GM over denominator straight R end fraction end root space left parenthesis straight h less than less than straight R right parenthesis
    therefore, the minimum increase in its orbital velocity required to escape from the Earth's Gravitational Field.
    straight v apostrophe minus straight v space equals space square root of fraction numerator 2 GM over denominator straight R end fraction end root space minus square root of GM over straight R end root
space equals space square root of 2 gR end root minus square root of gR space equals space square root of gR space left parenthesis square root of 2 minus 1 right parenthesis

    Question 501
    CBSEENPH11020258

    A uniform string of length 20 m is suspended from a rigid support. A short wave pulse is introduced at its lowest end. It starts moving up the string. The time taken to reach the support is: (take g = 10 ms−2 )

    • 2 straight pi square root of 2 straight s
    • 2s

    • 2 square root of 2 straight s
    • square root of 2 straight s

    Solution

    C.

    2 square root of 2 straight s

    A uniform string of length 20 m is suspended from a rigid support. Such that the time taken to reach the support.
    straight T equals mgx over straight l
So comma space velocity space at space point space straight P space equals space square root of fraction numerator begin display style mgx over straight l end style over denominator straight m divided by straight l end fraction end root
straight v equals space square root of gx
dx over dt space equals space square root of gx
integral subscript 0 superscript 20 fraction numerator dxt over denominator square root of straight x end fraction space equals space integral subscript 0 superscript straight t square root of straight g space dt
left square bracket 2 square root of straight x right square bracket subscript 0 superscript 20 space equals space square root of 10 straight t
square root of 20 space equals space square root of 10 straight t
straight t space equals space 2 square root of 2 space straight s 

    Question 502
    CBSEENPH11020260

    Two stones are thrown up simultaneously from the edge of a cliff 240 m high with an initial speed of 10 m/s and 40 m/s respectively. Which of the following graph best represents the time variation of relative position of the second stone with respect to the first? (Assume stones do not rebound after hitting the ground and neglect air resistance, take g = 10 m/s2 )

    Solution

    C.

    The concept of relative motion can be applied to predict the nature of motion of one particle with respect to the other.
    Consider the stones thrown up simultaneously as shown in the diagram below.
    Considering the motion of the second particle with respect to the first we have relative acceleration
    |a21| = |a2-a1| = g-g = 0

    Thus, motion of the first particle is straight line with respect to the second particle till the first particle strikes ground at a time given by
    negative 240 space equals space 10 straight t space minus space 1 half space straight x space 10 space straight x space straight t squared
or space straight t squared minus 2 straight t minus 48 space equals space 0
or space straight t squared space minus 8 straight t space plus space 6 straight t space minus 48 space equals space 0
or space space straight t squared space equals space 8 comma negative 6 space left parenthesis not space possible right parenthesis
    Thus, distance covered by the second particle with respect to the first particle in 8s is
    S12 = (v21) t = (40-10)(8s)
     = 30 x 8 = 240m
    Similarly, time taken by the second particle to strike the ground is given by
    negative 240 space equals space 40 straight t space minus space 1 half space straight x space 10 space straight x space straight t squared
minus 240 space equals space 40 space straight t space minus 5 straight t squared
5 straight t squared minus 40 straight t minus 240 space equals 0
straight t squared minus 8 straight t minus 48 space equals space 0
straight t squared minus 12 straight t space plus space 4 straight t minus 48 space equals 0
straight t left parenthesis straight t minus 12 right parenthesis plus 4 left parenthesis straight t minus 12 right parenthesis space equals 0
straight t space equals space 12 comma negative 4 space left parenthesis not space possible right parenthesis
    Thus, after the 8s magnitude of relative velocity will increases up to 12 s when the second particle strikes the ground.

    Question 504
    CBSEENPH11020290

    What is the minimum energy required to launch a satellite of mass m from the surface of a planet of mass M and radius R in a circular orbit at an altitude of 2R?

    • 5GmM/6R

    • 2GmM/3R

    • GmM/2R

    • GmM/3R

    Solution

    A.

    5GmM/6R

    From conservation of energy,
    Total energy at the planet = Total energy at the altitude
    negative GMm over straight R space plus space left parenthesis KE right parenthesis subscript surface space equals space minus fraction numerator GMm over denominator 3 straight R end fraction space plus space 1 half mv subscript straight A superscript 2 space... space left parenthesis straight i right parenthesis
    In its orbit the necessary centripetal force provided by gravitational force.
    ∴ therefore space fraction numerator mv subscript straight A superscript 2 over denominator left parenthesis straight R space plus 2 straight R right parenthesis end fraction space equals space fraction numerator GMm over denominator left parenthesis straight R space plus 2 straight R right parenthesis squared end fraction
straight v subscript straight A superscript 2 space equals space fraction numerator GM over denominator 3 straight R end fraction space... space left parenthesis ii right parenthesis
From space eq space left parenthesis straight i right parenthesis space and space left parenthesis ii right parenthesis space we space get
left parenthesis KE right parenthesis subscript surface space equals space 5 over 6 GMm over straight R

    Question 505
    CBSEENPH11020306

    The mass of a spaceship is 1000 kg. It is to be launched from the earth's surface out into free space. The value of 'g'and 'R'(radius of earth) are 10 m/s2 and 6400km respectively. The required energy for this work will be;

    • 6.4 x1011 J

    • 6.4 x108 J

    • 6.4 x109 J

    • 6.4 x109 J

    Solution

    D.

    6.4 x109 J

    Potential energy on the earth's surface is -mgR while in free space, it is zero. So, to free the spaceship minimum required energy is
    E =mgR
     = 103 x 10 x 6400 x 103 J
     = 6.4 x 1010 J

    Question 509
    CBSEENPH11020462

    A particle moves in a straight line with retardation proportional to its displacement. Its loss of kinetic energy for any displacement x is proportional to

    • x2

    • ex

    • x

    • logex

    Solution

    A.

    x2

    In this problem acceleration (a) is given in terms of displacement (x)  to determine the velocity with respect to position or displacement we have to apply integration method.
    From given information a =-kx, where a is acceleration, x is displacement and k is proportionality constant. 
    vdx over dx space equals space minus space kx space open square brackets therefore space dv over dt space equals space fraction numerator begin display style dv end style over denominator dx end fraction open parentheses dx over dt close parentheses space equals space straight v dv over dx close square brackets
straight v space dv space equals space minus kx space dx
    Let for any displacement from 0 to x , the velocity changes from vo to v
    integral subscript straight v subscript straight o end subscript superscript straight v space vdv space equals space minus integral subscript 0 superscript straight x space kx space dx
rightwards double arrow space integral subscript 0 superscript straight x space kx space dx
rightwards double arrow space fraction numerator straight v squared minus straight v subscript 0 superscript 2 over denominator 2 end fraction space equals space minus space kx squared over 2
rightwards double arrow space straight m space open parentheses fraction numerator straight v squared minus space straight v subscript 0 superscript 2 over denominator 2 end fraction close parentheses space equals space fraction numerator negative mkx squared over denominator 2 end fraction
increment straight K space proportional to space straight x squared

    Question 512
    CBSEENPH11020511

    At what height from the surface of earth the gravitation potential and the value of g are -5.4 x 107 J kg-2 and 6.0 m/s2 respectively? (Take the radius of the earth as 6400 km)

    • 1600 km

    • 1400 km

    • 2000 km

    • 2600 km

    Solution

    D.

    2600 km

    Gravitational potential at some height h from the surface of the earth is given by,
    V = fraction numerator negative GM over denominator straight R plus straight h end fraction    ... (i)
    Acceleration due to gravity at some height h from the earth surface can be given as,

    straight g apostrophe space equals space fraction numerator GM over denominator left parenthesis straight R plus straight h right parenthesis squared end fraction  ... (ii)
    From equation (i) and (ii), we get

    space space space space space fraction numerator open vertical bar straight V close vertical bar over denominator straight g apostrophe end fraction equals fraction numerator G M over denominator R plus h end fraction space x space fraction numerator left parenthesis R plus h right parenthesis squared over denominator G M end fraction

rightwards double arrow space fraction numerator open vertical bar straight V close vertical bar over denominator straight g apostrophe end fraction space equals space R plus h    ... (iii)
    because V = -54 x 107 J kg-2
    g' = 6.0 m/s2
    Radius of Earth, R = 6400 km


    Putting the values in Eqn. (iii), we get
    fraction numerator 5.4 space straight x space 10 to the power of 7 over denominator 6.0 end fraction = R+h
    rightwards double arrow space 9 space straight x space 10 to the power of 6 space equals space straight R plus straight h
    rightwards double arrowh = (9-6.4)x106 = 2.6 x 106 m
    rightwards double arrow h = 2600 km
    Question 513
    CBSEENPH11020518

    If the magnitude of sum of two vectors is equal to the magnitude of difference of the two vectors, the angle between these vectors is,

    • 90o

    • 45o

    • 180o

    • 0o

    Solution

    A.

    90o

    There are two vectors P and Q.
    It is given that,
    open vertical bar straight P plus straight Q close vertical bar space equals open vertical bar P minus Q close vertical bar
    Let, angle between P and Q is straight ϕ,
    straight P squared plus straight Q squared space plus space 2 PQ space cosϕ space equals space straight P squared space plus space straight Q squared space minus space 2 space PQ space cos space straight ϕ
    rightwards double arrow space 4 PQ space cos space straight ϕ space equals space 0

rightwards double arrow space cos space straight ϕ space equals space 0 space left square bracket because space straight P comma space straight Q space not equal to space 0 right square bracket

rightwards double arrow space straight ϕ space equals space straight pi over 2 equals space 90 to the power of straight o

    Question 514
    CBSEENPH11020522

    Two identical charged spheres suspended froma  common point by two massless strings of lengths l, are initially at a distance d (d<< l) apart because of their mutual repulsion. The charges begin to leak from both the spheres at a constant rate. As a result, the spheres approach each other with a velcoity v. Then, v varies as a function of the distance x between the sphere, as

    • vproportional tox

    • straight v space proportional to space straight x to the power of begin inline style bevelled fraction numerator negative 1 over denominator 2 end fraction end style end exponent
    • vproportional tox-1

    • straight v space proportional to space straight x space to the power of begin inline style bevelled 1 half end style end exponent

    Solution

    B.

    straight v space proportional to space straight x to the power of begin inline style bevelled fraction numerator negative 1 over denominator 2 end fraction end style end exponent

    Two identical charged spheres suspended from a common point by two massless strings of length L.

    In increment space ABC,
    tan space theta space equals space fraction numerator F over denominator m g end fraction
    rightwards double arrow space straight F over mg space equals space tan space straight theta space... space left parenthesis straight i right parenthesis
    The charge begins to leak from both the sphere at a constant rate. As a result, the spheres approach each other with velocity v.
    Therefore, equation (i) can be rewritten as,
    fraction numerator Kq squared over denominator straight X squared mg end fraction space equals space fraction numerator X divided by 2 over denominator square root of l squared minus begin display style x squared over 4 end style end root end fraction
    rightwards double arrow space fraction numerator Kq squared over denominator straight X squared mg end fraction space equals space fraction numerator straight x over denominator 2 straight l end fraction
rightwards double arrow space straight q squared space proportional to space straight x cubed
rightwards double arrow space straight q space proportional to space straight x to the power of begin inline style bevelled 3 over 2 end style end exponent
rightwards double arrow space dq over dt space proportional to space fraction numerator straight d space left parenthesis straight x to the power of bevelled 3 over 2 end exponent right parenthesis over denominator dx end fraction. space dx over dt
rightwards double arrow space dq over dt space proportional to space straight x to the power of begin inline style bevelled 1 half end style end exponent
rightwards double arrow space straight v space proportional to space 1 over straight x to the power of bevelled 1 half end exponent
rightwards double arrow space straight v space proportional to space straight x to the power of negative 1 divided by 2 end exponent

    Question 515
    CBSEENPH11020537
    Question 520
    CBSEENPH11020566

    The force F acting on a particle of mass m is indicated by the force-time graph as shown below. The climate change in momentum of the particle over the time interval from zero to 8 s is,

    • 24 Ns

    • 20 Ns

    • 12 Ns

    • 6 Ns

    Solution

    C.

    12 Ns

    The area under F-t graph gives change in momentum.
    For 0 to 2 s,
    increment straight p subscript 1 space equals space 1 half straight x 2 straight x 6 space equals space 6 space kg minus straight m divided by straight s
    For 2 to 4 sec,
    increment straight p subscript 2 space equals space 2 space straight x space left parenthesis negative 3 right parenthesis space equals space minus 6 space kg minus straight m divided by straight s
    For 4 to 8 sec,
    increment straight p subscript 3 space equals space left parenthesis 4 right parenthesis straight x 3 space equals space plus 12 space kg minus straight m divided by straight s
    So, total change in momentum for 0 to 8 sec is,
    increment straight p subscript net space equals space increment straight p subscript 1 plus increment straight p subscript 2 plus increment straight p subscript 3

space space space space space space space space space space space equals left parenthesis plus 6 space minus 6 space plus 12 right parenthesis

space space space space space space space space space space space equals space 12 space kg minus straight m divided by straight s

space space space space space space space space space space space equals space 12 space Ns

    Question 529
    CBSEENPH11020606

    A uniform force of (3 j) N acts on a particle of mass 2 kg. Hence the particle is displaced from position (2 k) m to position ( 4 + 3 k) m. The work done by the force on the particle is,

    • 9 J

    • 6 J

    • 13 J

    • 15 J

    Solution

    A.

    9 J

    Given,
    Force, F = 3 i + j
    r1 = ( 2 i + k) m
    r2 = (4 i + 3 j - k) m
    Therefore,
    S = r2 - r1
      equals space left parenthesis 4 space straight i space plus space 3 space straight j space minus space straight k right parenthesis space minus space left parenthesis space 2 straight i space plus space straight k right parenthesis
equals space left parenthesis 2 space straight i space plus space 3 space straight j space minus space 2 space straight k right parenthesis straight m
    Therefore,
    W = F. s
        = (3i + j). (2 i +3j - 2k)
        = 3 x2 + 3x0
        = 6 + 3 = 9 J

    Question 533
    CBSEENPH11020645

    A satellite S is moving in an elliptical orbit around the earth. The mass of the satellite is very small as compared to the mass of the earth. Then,

    • the angular momentum of S about the centre of the earth changes in direction, but its magnitude remains constant

    • the total mechanical energy of S varies periodically with time

    • the linear momentum of S remains constant in magnitude

    • the acceleration of S is always directed towards the centre of the earth

    Solution

    D.

    the acceleration of S is always directed towards the centre of the earth

    As we know that, the force on satellite is an only gravitational force which will always be towards the centre of the earth. Thus, the acceleration is S is always directed towards the centre of the earth

    Question 534
    CBSEENPH11020658

    A plank with a box on it at one end is gradually raised about the other end. As the angle of inclination with the horizontal reaches 30o, the box starts to slip and slides 4.0 m down the plank in 4.0 s. The coefficients of static and kinetic friction between the box and the plank will be respectively

    • 0.6 and 0.6

    • 06 and 0.5

    • 0.5 and 0.6 

    • 0.4 and 0.3

    Solution

    B.

    06 and 0.5

    Given a plank with a box on it one end is gradually raised about the end having angle of inclination is 300, the box starts to slip and slides down  4 m the plank in 4 s as shown in a figure.

    The coefficient of static friction.
    straight mu subscript straight s space equals space tan space 30 to the power of straight o space equals space fraction numerator 1 over denominator square root of 3 space end fraction space equals space 0.6
    So, distance covered by a plank,
    straight s space equals space ut space plus space 1 half at squared
Here comma space straight u space equals space 0 space and space straight a space equals space straight g space left parenthesis sinθ space minus space straight mu space cos space straight theta right parenthesis
therefore space 4 space equals space 1 half space straight g space left parenthesis sin space 30 space space minus space straight mu subscript straight k space cos space 30 right parenthesis space left parenthesis 4 right parenthesis squared

rightwards double arrow space 0.5 space straight x space 10 space straight x space 1 half straight x space minus straight mu subscript straight k space straight x space 10 space straight x fraction numerator square root of 3 over denominator 2 end fraction
rightwards double arrow space 5 square root of 3 space straight mu subscript straight k space equals space 4.5 space rightwards double arrow space straight mu subscript straight k space equals space 0.51
    Thus, coefficient of kinetic friction between the box and the plank is 0.51

    Question 535
    CBSEENPH11020662

    A remote sensing satellite of earth revolves in a circular orbit at a height of 0.25 x 106 m above the surface of the earth. if earth 's radius is 6.38 x 106 m and g = 9.8 ms-2 then the orbital speed of the  satellite is 

    • 7.76 kms-1

    • 8.56 kms-1

    • 9.13 kms-1

    • 6.67 kms-1

    Solution

    A.

    7.76 kms-1

    Given, height of a satellite
    h = 0.25 x 106
    Earth's radius, Re = 6.38 x 106
    For the satellite revolving around the  earth, orbital velocity of the satellite

    straight v subscript straight o space equals space square root of GM subscript straight e over straight R subscript straight e end root space equals square root of space fraction numerator GM over denominator straight R subscript straight e open square brackets 1 plus space begin display style straight h over straight R subscript straight e end style close square brackets end fraction end root

straight v subscript straight o space equals space square root of fraction numerator gR over denominator 1 plus begin display style straight h over straight R subscript straight e end style end fraction end root

substitutes space the space values space of space straight g comma space straight R subscript straight e space and space straight h comma space we space get
straight v subscript straight o space equals space square root of 60 space straight x space 10 to the power of 6 space end root space straight m divided by straight s
straight v subscript straight o space equals space 7.76 space straight x space 10 cubed space straight m divided by straight s space equals space 7.76 space km divided by straight s 

    Question 546
    CBSEENPH11020721

    A body mass 1 kg is thrown upwards with velocity 20 ms-1. It momentarily comes to rest after attaining a height of 18 m. How much energy is lost due to air friction?  (g = 10 ms-2)

    • 20 J

    • 30 J

    • 40 J

    • 10 J

    Solution

    A.

    20 J

    The energy lost due to air friction is equal to the difference of initial kinetic energy and final potential energy.
    Initially, body possess only kinetic energy and after attaining a height the kinetic energy is zero.
    Therefore, loss of energy = KE - PE
    =1 half mv squared minus mgh

equals space 1 half space straight x 1 space straight x space 1400 space minus 1 space straight x space 18 space straight x space 10

space equals space 200 minus 80 space straight J space equals space 20 space straight J

1 half mv squared minus mgh

equals space 1 half space straight x 1 space straight x space 1400 space minus 1 space straight x space 18 space straight x space 10

space equals space 200 minus 80 space straight J space equals space 20 space straight J

    Question 547
    CBSEENPH11020735

    The figure shows elliptical orbit of a planet m about the sun S. The shaded are SCD is twice the shaded area SAB. If t1 is the time for the planet to move from C to D and t2 is the time to move from A to B then,

    • t1 > t2

    • t1 =4 t2

    • t1 = 2t2

    • t1 = t2

    Solution

    C.

    t1 = 2t2

    Apply Kepler's law of area fo planetary motion.
    The line joining the sun to the planet sweeps out equal areas time interval ie, areal velocity is constant.
    dA over dt equals space constant

or space straight A subscript 1 over straight t subscript 1 equals straight A subscript 2 over straight t subscript 2
rightwards double arrow space straight t subscript 1 space equals space straight A subscript 1 over straight A subscript 2 straight t subscript 1
Given comma
straight A subscript 1 space equals space 2 straight A subscript 2
therefore comma space straight t subscript 1 space equals 2 straight t subscript 2

    Question 548
    CBSEENPH11020742

    A roller coaster is designed such that riders experience " weightlessness" as they go round the top of a hill whose radius of curvature is 20 m.  The speed of the car at the top of the hill is between

    • 14 m/s and 15 m/s

    • 15m/s and 16 m/s

    • 16 m/s and 17 m/s

    • 13m/s and 17 m/s

    Solution

    A.

    14 m/s and 15 m/s


    Balancing the forces, we get
    Mg-N = Mv2/R
    For weightlessness, N = 0 
    therefore, MV2/R = Mg
    where R is the radius of curvature and v is the speeds of the car.
    Therefore,
    straight v space equals square root of Rg
putting space the space values comma space straight R space equals space 20 straight m comma space straight g space equals space 10.0 space straight m divided by straight s squared
So comma space straight v equals space square root of 20 space straight x space 10.0 end root space equals 14.14 space straight m divided by straight s squared
    Thus, the speed of the car at the top of the hill is between 14 m/s and 15 m/s
    Question 549
    CBSEENPH11020766

    Assuming the sun to have a spherical outer surface of radius r, radiating like a black body at a temperature  to C, the power received by a unit surface (normal to the incident rays) at a distance R from the centre of the sun is :

    • fraction numerator 4 πr squared σt to the power of 4 over denominator straight R squared end fraction
    • fraction numerator straight r squared straight sigma left parenthesis straight t plus 273 right parenthesis to the power of 4 over denominator 4 πR squared end fraction
    • fraction numerator 16 space straight pi squared straight r squared σt to the power of 4 over denominator straight R squared end fraction
    • fraction numerator straight r squared straight sigma left parenthesis straight t plus 273 right parenthesis to the power of 4 over denominator straight R squared end fraction

    Solution

    D.

    fraction numerator straight r squared straight sigma left parenthesis straight t plus 273 right parenthesis to the power of 4 over denominator straight R squared end fraction

    From Stefan's law, the rate at which energy is radiated by sun at its surface si 
    P = σ x 4 πr2T4

    [Sun is a perfectly black body as it emits radiations of all wavelengths and so for it e =1]
    The intensity of this power at earth's surface (under the assumption R >>ro) is
    straight I space equals space fraction numerator straight P over denominator 4 space πR squared end fraction
space equals space fraction numerator straight sigma space straight x space 4 space πr squared straight T to the power of 4 over denominator 4 πR squared end fraction
space equals space fraction numerator σr squared straight T to the power of 4 over denominator straight R squared end fraction
equals space fraction numerator σr squared left parenthesis straight t space plus 273 right parenthesis squared over denominator straight R squared end fraction
    Question 550
    CBSEENPH11020776

    Two satellites of earth S1 and S2, are moving in the same orbit. The mass of S1 is four times the mass of S2. Which one of the following statement is true?

    • The time period of S1 is four times that of S2

    • The potential energies of earth and satellite in the two cases are equal 

    • S1 and S2 are moving with the same speed

    • The kinetic energies of the two satellites are equal

    Solution

    C.

    S1 and S2 are moving with the same speed

    In same orbit, orbtial speed of satellites remains same. When two satellites of earth are moving in same orbit, then time period of both are equal.
    From Kepler's third law
    straight T squared space proportional to space straight r cubed
    Time period is independent of mass, hence their t ime periods will be equal.
    The potential energy and kinetic energy are mass dependent, hence the PE and KE of satellites are not equal.
    But, if they are orbiting in a same orbit, then they have equal orbital speed.

    Question 551
    CBSEENPH11020792

    A body of mass 3 kg is under a constant force which causes a displacement s in metres in it, given by the relation straight s equals 1 third straight t squared, where t is in s. Work done by the force in 2 s is:

    • 5 over 19 straight J
    • 3 over 8 straight J
    • 8 over 3 straight J
    • 19 over 5 straight J

    Solution

    C.

    8 over 3 straight J

    If a constant force is applied on the object causing a displacement in it, then it is said that work has been done on the body to displace it.
      Work done by the force  = Force x Displacement
    or          W = F x s                            ...(i)
    But from Newton's 2nd law, we have
      Force  =  Mass x Acceleration
    i.e.,       F = ma                                ...(ii)
    Hence, from Eqs. (i) and (ii), we get
    straight W space equals space mas space equals space straight m open parentheses fraction numerator straight d squared straight s over denominator dt squared end fraction close parentheses straight s space space space space space space... left parenthesis iii right parenthesis space space space open parentheses because space straight a space equals space fraction numerator straight d squared straight s over denominator dt squared end fraction close parentheses
Now comma space we space have comma space space straight s space equals space 1 third straight t squared
therefore space space space space space space space space space fraction numerator straight d squared straight s over denominator dt squared end fraction space equals space straight d over dt open square brackets straight d over dt open parentheses 1 third straight t squared close parentheses close square brackets
space space space space space space space space space space space space space space space space space space space equals straight d over dt cross times open parentheses 2 over 3 straight t close parentheses
space space space space space space space space space space space space space space space space space space equals 2 over 3 dt over dt 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 2 over 3 space space space space space space space
space space space space space space space space space space space space space space space
    Hence, Eq (ii) becomes
                    straight W space equals 2 over 3 ms space equals space 2 over 3 straight m cross times 1 third straight t squared
                       equals 2 over 9 mt squared
    We have given
                       straight m equals 3 space kg comma space straight t space equals space 2 straight s
therefore space space space space straight w space equals space 2 over 9 cross times 3 cross times left parenthesis 2 right parenthesis squared space equals space 8 over 3 straight J
     

    Question 552
    CBSEENPH11020793

    A particle moves along a straight line OX. At a time t (in seconds) the distance x (in metres) of the particle from O is given by
            straight x equals 40 straight t space plus 12 straight t minus straight t cubed
    How long would the particle travel before coming to rest?

    • 24m

    • 40m

    • 56m

    • 16m

    Solution

    C.

    56m

    Velocity is rate of change of distance or displacement.
    Distance travelled by the particle is
                          straight x equals space 40 space plus 12 straight t minus straight t cubed
    We know that, velocity is rate of change of distance i.e., straight v equals dx over dt.
    therefore space space space straight v space equals space straight d over dt left parenthesis 40 plus 12 straight t minus straight t cubed right parenthesis
space space space space space space equals space 0 plus 12 minus 3 straight t squared
    but final velocity v = 0
    therefore space space 12 minus 3 straight t squared space equals space 0
or space space space space space space space space straight t squared space equals space 12 over 3 space equals 4
or space space space space space space straight t space equals space 2 straight s
    Hence, distance travelled by the particle before coming to rest is given by
                       straight x equals 40 plus 12 left parenthesis 2 right parenthesis minus left parenthesis 2 right parenthesis cubed
space space equals 40 plus 24 minus 8 space equals space 64 minus 8
space space equals space 56 space straight m
                

    Question 553
    CBSEENPH11020799

    The earth is assumed to be a sphere of radius R. A platform is arranged at a height R from the surface of the earth. The escape velocity of a body from this platform is fve, where ve is its escape velocity from the surface of the earth. The value of f is

    • square root of 2
    • fraction numerator 1 over denominator square root of 2 end fraction
    • 1 third
    • 1 half

    Solution

    B.

    fraction numerator 1 over denominator square root of 2 end fraction

    If energy in the form of kinetic energy which is equal to binding energy, is supplied to the sphere, it leaves the gravitational field of earth.
    At a platform at a height h, 
    escape energy = binding energy of sphere
    or space space 1 half mv subscript straight e to the power of apostrophe 2 end exponent space equals space fraction numerator GMm over denominator straight R plus straight h end fraction
    or       straight v subscript straight e apostrophe space equals space square root of fraction numerator 2 GM over denominator straight R plus straight h end fraction end root space equals space square root of fraction numerator 2 GM over denominator 2 straight R end fraction end root space space left parenthesis because straight h equals straight R right parenthesis
    But at surface of earth, 
                           straight v subscript straight e space equals space square root of fraction numerator 2 GM over denominator straight R end fraction end root
    As given,       straight v subscript straight e apostrophe space equals space fv subscript straight e
    Hence,  square root of fraction numerator 2 GM over denominator 2 straight R end fraction end root space equals space straight f square root of fraction numerator 2 GM over denominator straight R end fraction end root
        or space space space space space fraction numerator 1 over denominator 2 straight R end fraction space equals space straight f squared over straight R
    therefore space space space space space straight f space equals space fraction numerator 1 over denominator square root of 2 end fraction

    Question 554
    CBSEENPH11020808

    Two bodies, A (of mass 1 kg) and B (of mass 3 kg) are dropped from heights of 16 m and 25 m, respectively. The ratio of the time taken by them to reach the ground is

    • 5/4

    • 12/5

    • 5/12

    • 4/5

    Solution

    D.

    4/5

    As bodies are dropped from a certain height, their initial velocities are zero i.e., u = 0.
    For free fall from a height u = 0 (initial velocity).
    From second equation of motion
                           straight h equals ut plus 1 half gt squared
    or                    straight h equals 0 plus 1 half gt squared
    therefore space space space space space space space space space space space space space space straight h subscript 1 over straight h subscript 2 space equals space open parentheses straight t subscript 1 over straight t subscript 2 close parentheses squared
Given comma space space space space space space space space space straight h subscript 1 space equals space 16 space straight m comma space space space straight h subscript 2 space equals space 25 space straight m
therefore space space space space space space space space straight t subscript 1 over straight t subscript 2 space equals square root of straight h subscript 1 over straight h subscript 2 end root space equals square root of 16 over 25 end root space equals 4 over 5

    Question 556
    CBSEENPH11020854

    A toy car with charge q moves on a frictionless horizontal plane surface under the influence of a uniform electric field E . Due to the force q E, its velocity increases from 0 to 6 m/s in one-second duration. At that instant, the direction of the field is reversed. The car continues to move for two more seconds under the influence of this field. The average velocity and the average speed of the toy car between 0 to 3 seconds are respectively

    • 2 m/s, 4 m/s

    • 1 m/s, 3 m/s

    • 1.5 m/s, 3 m/s

    • 1 m/s, 3.5 m/s

    Solution

    B.

    1 m/s, 3 m/s

    v = - 6ms-1

    Acceleration, 

    a = v-ut = 6-01 = 6 ms-2For t = 0 to t = 1sS1 = 12 x 6 (1)2 = 3m ... (i)For t = 1s = to  t = 2s,S2 = 6.1 -12 x 6 (1)2 = 3m .... (ii)For t = 2s to t = 3sS3 = 0 - 12 x 6 (1)2 = - 3m (iii)Total displacementS = S1 + S2 + S3  = 3m vaverage = 3/3 = 1 ms-1Total  distance travelled = 9 mAverage  speed  = 9/3 =3 ms-1

    Question 557
    CBSEENPH11024912

    What do you mean by mean free path and write its formula.

    Solution

    In kinetic theory the mean free path of a particle, such as a molecule, is the average distance the particle travels between collisions with other moving particles. The formula still holds for a particle with a high velocity relative to the velocities of an ensemble of identical particles with random locations.

     

    Question 558
    CBSEENPH11026128

    A police jeep is chasing with velocity of 45 km/h. A thief in another jeep moving with velocity 153 km/h. Police fires a bullet with muzzle velocity of 180 m/s. The velocity it will strike the car of the thief is

    • 150 m/s

    • 27 m/s

    • 450 m/s

    • 250 m/s

    Solution

    A.

    150 m/s

    Relative velocity is the vector difference between the velocities of two bodies, the velocity of a body with respect to another regarded as being at rest.

    Here, we calculate the relative velocity of the jeep:

     153 -45 =108 km/h

    km/hour =100060×60=518= 108 ×518 =30 m/s

    Therefore, striking speed = relative speed of a bullet with respect to the thief's car:

     180 - 30 = 150 m/s 

    Question 559
    CBSEENPH11026132

    Two similar charges having mass m and 2m are placed in an electric field. The ratio of their kinetic energy is :

    • 4:1

    • 1:1

    • 2:1

    • 1:2

    Solution

    C.

    2:1

    Force on charged particle is electric field F = q E

     Acceleration a = F m                          =q EmVelocity v=u + a t                = 0+ q Em t  u - initial velocity , v -final velocity        So kinetic energy K = 12mv2                            K  =12m q Em t2Ratio of kinetic energies      =12m1 q Em1 t2 12m2 q Em2 t2     = m2m1      =2 mm     =21Ratio of kinetic energies      = 2:1

    Question 560
    CBSEENPH11026136

    A car starts from rest, moves with an acceleration a and then decelerates at a constant rate b for sometimes to come to rest. If the total time taken is t. The maximum velocity of car is given by :

    • a b ta + b 

    • a2 ta + b

    • a ta+ b

    • b2 ta+b

    Solution

    A.

    a b ta + b 

    Let car accelerates for time t1 and decelerates for time t2 then

    t1 + t2 = t

    From  ν = u + at

    v = u + at1

    at1 = bt2

     t2 =at1b  t1 + at1b =1t11 +ab =tt1 = bta+b  t1 =bta + bMaximum velocity of carv = at1 =a b ta+ b  

    Question 561
    CBSEENPH11026150

    A river is flowing from west to east with a speed of 5 m/min. A man can swim in still water with a velocity 10 m/min. In which direction should the man swim so, as to take the shortest possible path to go to the south?

    • 30o east of south

    • 60o east of south

    • 60o west of south

    • 30o west of north

    Solution

    A.

    30o east of south

    Let the swimmer swims at an angle θ with the vertical.

    sin θ = νrνs      =510=12=sin 30oθ = 30o

    The swimmer should swim 30o east of south to take the shortest possible path to go to the south.

    Question 562
    CBSEENPH11026165

    Two boys are standing at the ends A and B of a ground, where AB = a. The boy at B starts running in a direction perpendicular to AB with velocity v1 The boy at A starts running simultaneously with velocity v and catches the other boy in a time t, where t is 

    • av2 + v12

    • a2ν2  - ν12

    • aν -  ν1

    • aν + ν1

    Solution

    B.

    a2ν2  - ν12

    Distance covered by boy A in time t:-  AC= vt

    Distance covered by boy B in time t
    BC = ν1 t
    Using Pythagoras theorem

    Ac2  =  AB2 + BC2

    ⇒ ( νt )2  =  a2 +(  ν1t )2

    ⇒    a2   =   ν2 t2  - (ν1t)2

    t2 ν2 - ν12 = a2            t = a2ν2 - ν12         

    Question 563
    CBSEENPH11026169

    A ball is thrown vertically upward. It has a speed of 10 m/s when it has reached one-half of its maximum height. How high does the ball rise? ( Take, g = 10 m/s2 )

    • 15 m

    • 10 m

    • 20 m

    • 5 m

    Solution

    B.

    10 m

    The problem can be solved using third equation of motion at A and O'.

          

    Let maximum height attained by the ball be H. Third equation of motion gives

      v2 = u2 - 2gh

    v2 - final velocity ; u2 = initial velocity

    At  A,

    102 = u2 - 2 × 10 × H2

    ⇒ u = 100 + 10 H                          ....(i)

    At  'O' 

    ( 0)2 = u2 - 2 × 10 × H                     ....(ii)

    Thus, from Eqs. (i) and (ii), we get

     20 H = 100 + 10 H

    ⇒ 10 H = 100

    ∴  H = 10 m

    Question 564
    CBSEENPH11026174

    A car moves from X to Y with a uniform speed vu and returns to Y with a uniform speed vd The average speed for this round trip is

    • 2 νd νuνd + νu

    • vu vd

    • vd vuvd + vu

    • vu + vd2

    Solution

    A.

    2 νd νuνd + νu

    Average velocity is defined as the change in position or displacement divided by the time intervals ( Δt ) in which displacement occurs.

    v = xt

    Let t1 and t2 times taken by the car to go from X to Y and then from Y to X respectively.

    t1 + t2 = XYvu + XYvdt1 + t2 = XY vu + vdvu vd

    Total distance travelled = XY + XY

                                     = 2XY

    Therefore, average speed of the car for this round trip is

    vav = 2 XYXY vu + vdvu vd vav = 2 vu vdvu + vd

    Question 565
    CBSEENPH11026198

    A particle moves along a straight line OX. At a time t (in second) the distance x ( in metre ) of the particle from O is given by x = 40 + 12 t - t3 . How long would the particle travel before coming to rest?

    • 24 m

    • 40 m

    • 56 m

    • 16 m

    Solution

    C.

    56 m

    Speed is rate of change of distance.

    Distance travelled by the particle is

                   x = 40 + 12 t - t3

    We know the speed is rate of change of distance

           i.e       ν = dxdt

                      ν = ddt40 + 12t - t3

                         = 0 + 12 - 3t2

    but final velocity ν = 0

    ∴               12 - 3t2 = 0

    ⇒               t2 = 123   t2 = 4

     ⇒              t = 2s

    Hence, distance travelled by the particle before coming to rest is given by
                 x = 40 + 12(2)-(2)3

                 x = 64 - 8

                  x = 56 m

    Question 566
    CBSEENPH11026201

    A small disc of radius 2 cm is cut from a disc of radius 6 cm. If the distance between their centres is 3.2 cm, what is the shift in the centre of mass of the disc?

    • 0.4 cm

    • 2.4 cm

    • 1.8 cm

    • 1.2 cm

    Solution

    A.

    0.4 cm

    The situation can be shown as: Let radius of complete disc is a and that of small disc is b. Also let centre of mass now shifts to 02 at a distance X2 from original centre

                           

    The position of new centre of mass is given by

                          XCM = -σ. πb2. x1σ.π a2 - σ.π b2

    Here, a = 6 cm, b = 2 cm, x1 = 3.2 cm

      XCM = -σ × π 22 ×3.2σ × π ×62 - σ ×π ×22                = 12.832π

    ⇒    XCM = 0.4 cm

    Question 567
    CBSEENPH11026203

    The displacement of particle is given by x = αo+ α1t2 - α2t23 .What is its acceleration?

    • 2α23

    • -2α23

    • α2

    • Zero

    Solution

    B.

    -2α23

    Acceleration is the rate of change of velocity and velocity is the rate of change of displacement.
    The displacement equation is given by

                  x = αo + α1 t2 - α2 t23

    Veiocity = rate of change of displacement

                      v = dxdt     = ddtαo + α1 t2 - α2 t23 

                        ν  =  0 + α12-2 α2t3v   =  α1 2 - 2 α2t3

    Acceleration = rate of change of velocity

                          α = dvdtα = ddtα12 - 2α23tα = 0 - 2α23α = - 2α23

    Question 568
    CBSEENPH11026204

    The area of the acceleration-displacement curve of a body gives

    • impulse

    • change in momentum per unit mass

    • change in KE per unit mass

    • total change in energy

    Solution

    C.

    change in KE per unit mass

    Area of the acceleration-displacement curve gives change in KE per unit mass.

           12m v2 - u2 = F s = mdvdt × s change in KEMass  = mdvdt × sm       change in KEMass = dvdt ×s

    Question 569
    CBSEENPH11026231

    Velocity-time curve for a body projected vertically upwards is

    • parabola

    • ellipse

    • hyperbola

    • straight line

    Solution

    D.

    straight line

    When the ball goes upwards the direction of velocity of the ball and the direction of acceleration of gravity are in the opposite directions of acceleration of gravity are in the opposite directions so the velocity of the ball decrease until it reaches the top and stops momentarily.

     v = u - gt          ... (i)

    Then from the peak position, the ball will start to descend downwards due to the acceleration due to gravity that is g.

     v = -gt               ....(ii)

    ( negative sign is due to motion in downward direction)   

    From both equations, the velocity and time are linearly proportional to each other which is the same as the equation of a straight line. So this graph will be a straight line starting from initial velocity (u) on the y-axis and slope g with time as the x-axis.

    Question 570
    CBSEENPH11026232

    The numerical ratio of displacement to the distance covered is always

    • less than one

    • equal to one

    • equal to or less than one

    • equal to or greater than one

    Solution

    C.

    equal to or less than one

    Distance is scalar, it is never negative. Therefore distance can be equal or greater than displacement. That implies the ratio of displacemrnt to distance is always equal to or less than one.

    Question 571
    CBSEENPH11026234

    A train is moving towards east and a car is along north, both with same speed. The observed direction of car to the passenger in the train is

    • east-north direction

    • west-north direction

    • south-east direction

    • None of the above

    Solution

    B.

    west-north direction

    Let O be the origin, then

        

    passenger in the train at P observes the car at Q along direction PQ, where direction PQ is west north direction.

    Question 572
    CBSEENPH11026243

    A metro train starts from rest and in 5 s achieves 108 km/h. After that it moves with constant velocity and comes to rest after travelling 45 m with uniform retardation. If total distance travelled is 395 m, find total time of travelling.

    • 12.2 s

    • 15.3 s

    • 9 s

    • 17.2 s

    Solution

    D.

    17.2 s

    Given:- v= 108 km/h = 30 m/s

    For first equation of motion

              v = u +  α t

             30 = 0 + α × 5                   (  u = 0)

    ⇒          a = 6 m/s2

    So, distance travelled by metro train in 5s

                 s1 = 12α t2  

                     = 12 × 6 ×52

                 s1 = 75 m

    Distance travelled before coming to rest = 45 m

    So, from third equation of motion

                02 = 302 - 2 α'  × 45

    ⇒          α' = 30 × 302 × 45

    ⇒           α' = 10 m/s2

    Time taken in travelling 45 m is

                 t3 = 3010

                   t3 = 3s

    Now, total distance = 395 m

    i.e       75 + s' + 45 = 395 m

    i.e        s'  = 395 - (75 + 45)

    ⇒          s' = 275 m             

    ∴           t2 =  27530

                 t2  = 9.2 s

    Hence, total time taken in whole journey

                  = t1 + t2 + t3

                  = 5 + 9.2 + 3

                   = 17.2 s 

    Question 573
    CBSEENPH11026258

    A police party is moving in a jeep at a constant speed v. They saw a thief at a distance x on a motorcycle which is at rest. The moment the police saw the thief, the thief started at constant acceleration a. Which of the following relations is true if the police is able to catch the thief?

    • v2 < α x

    • v2 < 2α x

    • v2  2 α x

    • v2 =  αx

    Solution

    C.

    v2  2 α x

    Let the police party catch the thief after t 

    Distance travelled by police party in t  = v t

    and distance travelled by thief    =   x + 12 α t2

    ⇒               x + 12α t2  v t

    ⇒               α t2 2 - vt + x = 0

    Since by using the Quadratic equation formula

                    x = -b ± b2 - 4 ac2 a

    ⇒              t = --v ± v2 - 2 α2× x2 × α2                

    ⇒               t = v ± v2 -2 α xα

    Question 574
    CBSEENPH11026276

    The engine of a jet aircraft applies a thrown force of 105 N during take off and causes the plane to attain a velocity of 1 km/s in 10 s. The mass of the plane is

    • 102 kg

    • 103 kg

    • 104 kg

    • 105 kg

    Solution

    B.

    103 kg

    Given:- Force F = 105 N

      Velocity during take off = 1 km/s = 1000 m

          i.e       initial velocity = 0

    Hence Change in velocity = (103 - 0) 

     And time t = 1

    The acceleration produced in jet

                 α =Change in velocitytime

                 α = 103 - 010

                     = 100 m/s2

    ∴ Mass = Forceacceleration

              = 105102

    ⇒  Mass = 103 kg

    Question 575
    CBSEENPH11026277

    A stone is weighing 1 kg and sliding on ice with a velocity of 2 m/s is stopped by friction in 10 s. The force of friction (assuming it to be constant) will be

    • - 20 N

    • -0.2 N

    • 0.2 N

    • 20 N

    Solution

    B.

    -0.2 N

    u = 2 m/s

    v = 0

    f = 0s

    The kinematic eqution is

        v = u - αt      

    Where u is the initial velocity 

              v - final velocity

               α - acceleration

               t - time

    ⇒      α = v - ut

                   = 0 - 210

             α = - 15

    ⇒      α = -0.2 m/s2

    Friction force =  m α

                           = 1 × (-0.2)

                            = -0.2 N

    Question 576
    CBSEENPH11026279

    When a ceiling fan is switched on, it makes 10 revolution in the first 3 s. Assuming a uniform angular acceleration, how many rotation it will make in the next 3 s

    • 10

    • 20

    • 30

    • 40

    Solution

    C.

    30

    Angle turned in three second and θ1 s = 2 n x 10

           = 20 π rad

    According to the kinematic equation of Rotational motion

            θ1 = ωo t + 12 αt2

           20 π = 0 + 12α 32

             α = 40 π9rad/s2

    Now, angle turned in 6 s from the starting

            θ2 = ωo t + 12 αt2

                 = 0 + 12 × 40π9 + 62

           θ2 = 80 π rad

    Since angle turned between t = 3 s to t = 6 s

           θlost =   θ2 -  θ1

               = 80π - 20 π

         θlost =  60 π 

    Number of revolution = 60π2π 

                                 = 30 

    Hence Number of revolution in next 3s is 30

    Question 577
    CBSEENPH11026296

    Two boys are standing at the ends A and B of a ground, where AB= a. The boy at B starts running in a direction perpendicular to AB with velocity The boy at A starts running simultaneously with velocity v and catches the other boy in a time t, where t is

    • av2 + v12

    • a2v2 - v12

    • av - v1  

    • av + v1

    Solution

    B.

    a2v2 - v12

    Distance covered by boy A in time t

                   AC = vt                            ...(i)

    Distance covered by boy B in time t

                BC = v1 t                               ... (ii)

    Using Pythagourus theorem

               AC2 = AB2 + BC2

               ( vt )2 = a2 + (v1 t)2         

                       

    ⇒          v2 t2 - v12 t2 = a2

    ⇒          t2  v2 - v12   = a2

    ⇒          t = a2v2 - v12

    Question 578
    CBSEENPH11026297

    A force F is given F = at + bt2 where, t is time. What are the dimensions of a and b?

    • [ M L T-1 ] and [M L To ]

    • [ M L T-3 ] and [ M L2 T-4 ]

    • [ M L T-4 ] and [ M L T1 ]

    • [ M L T-3 ] and [ M L T-4 ]

    Solution

    B.

    [ M L T-3 ] and [ M L2 T-4 ]

    D.

    [ M L T-3 ] and [ M L T-4 ]

             Force F = at + bt2

       From principle of homogeneity

         Dimension of at  =   F  t 

                                 =  M L T-2  T 

           Dimension of at = [ M L T-3 ]

    Similarly, 

      dimensions of  bt2 F  t2 

                                =  M L T2   T2 

     dimension of bt2   = [ M L T-4 ]

    Question 579
    CBSEENPH11026304

    A particle moves towards east for 2 s with velocity 15 m/s and move towards north for 8 s with velocity 5 m/s. Then, average velocity of the particle is

    • 1 m/s

    • 5 m/s

    • 7 m/s

    • 10 m/s

    Solution

    B.

    5 m/s

    Distance covered towards east 

              = speed × time

              =  2 × 15 = 30 m

    Distance covered towards north

             = speed × time

             = 8  × 5 = 40 m

    Total displacement

              = 402 + 302

               = 50 m

    Average velocity = Total displacementTotal time

                              = 502 + 8

                              = 5 m/s     

    Question 580
    CBSEENPH11026320

    The distance x (in µm) covered by a molecule starting from point A at time t = 0 and stopping at another point B in given by the equation  x = t2 2 - t3 The distance between A and B ( in μm) is closed to

    • 10.7

    • 20.7

    • 40.7

    • 50.7

    Solution

    A.

    10.7

    Given that distance

             x = t2 2 - t3

    On differentiating

            dxdt = 4t - t2

    For x to be maximum,

             dxdt = 0

    ⇒     4t - t2 = 0

    ⇒            t = 4

    ∴ Maximum distance

          xmax42  2 - 43 

                  = 16 × 6 - 43

                  = 16 × 23

        xmax = 10.7

    Question 581
    CBSEENPH11026322

    The density of a rod having length l varies as ρ = c + dx, where x is the distance from the left end. The centre of mass is     

    • 3 cl + 2 D l23  2c + D l 

    • 2 cl + 3D I22  4c + 8I 

    • 2 cl + 3 DI23 2c + I

    • cl + DI22  2 c + DI 

    Solution

    A.

    3 cl + 2 D l23  2c + D l 

    Let the cross-sectional area is α. The mass of an element of length dx located at a distance x away from the left end is (C + Dx)α dx.  The x-coordinate of the centre of mass is given by

             Xcm =  x cm dm

                   = 0lxC + Dx  α dx0l C + Dx  α dx

    Where limit is from 0 to l.

                  = C l22+ D l3Cl + D l22 

                 Xcm3 CI + 2 DI23 2C + DI

    Question 582
    CBSEENPH11026346

    A block is kept on the floor of an elevator at rest. The elevator starts descending with an acceleration of 12 m/s2. Find the displacement of the block during the first 0.2 s after the start. (Take, g = 10 m/s2)

    • 30 cm

    • zero

    • 20 cm

    • 25 cm

    Solution

    C.

    20 cm

    The acceleration (downward) of the elevator is 12 m/s2 which is greater than gravitational acceleration g. So, the block will act as a freely falling body.

    From the motion of the block

          s = ut + 12 gt2

              = 0 × t + 12 × 10 × (0.2)2

               = 12 × 10 × 0.2 × 0.2

            s = 20 cm

    Question 583
    CBSEENPH11026351

    A rod PQ of length l is moving with ends remaining in contact with frictionless wall and floor. If at the instant, shown the velocity of end Q is 2 m/s towards negative direction of x. The speed of end P will be

       

    • 3 ms-1

    • 23 ms-1

    • 2 ms-1

    • 52 ms-1

    Solution

    B.

    23 ms-1

    From the diagram in question

                 x = l cosθ 

                 y = l sinθ

    ∴           vx = dxdt  = -l sinθdt                ..... ddtcosθ =-sinθ 

                 vydydt = l cosθ dt

    ∴            vy vx= - cot θ                         .....cos θsin θ = cot θ

               vy = - vx cot θ

    Since      vx = -2 m/s

              vy = -2 cot 60o                ....( Given:- θ = 60o )

                   = - 2 × 13

             vy- 23ms-1

    The negative sign indicates the motion towards negative direction of Y.

    Question 584
    CBSEENPH11026356

    The displacement x of a particle varies with time t as x = ae-αt + beβt where a, b, α and β are positive constants. The velocity of the particle will

    • decrease with time

    • be independent of α and β

    • drop to zero when α = β

    • increase with time

    Solution

    D.

    increase with time

    Given:- 

            x = a e-αt + beβt

    We know that

       velocity v = dxdt

                  v  =  d dt ae-αt + beβt 

                      = -aα e-αt + bβ eβt

                  v = -a αe-αt +  eβt

                   v = A + B

    where A = - aα e-αt , B = bβ e-βt

    The value ot term A = -aα e-αt increases and of term B = bβ e-βt increases with time. As a result velocity goes on increasing with time. 

    Question 585
    CBSEENPH11026359

    If A + B = C and that C is perpendicular to A. What is the angle between A and B, if A = B

    • π4 rad

    • π2 rad

    • 3 π4 rad

    • π rad

    Solution

    C.

    3 π4 rad

    Given:-

          A + B = C

    Since, B = C - A

    Also

       C  A

        B2 = 2 A2

    ⇒   B = 2 A

    Now

        A2 + B2 + 2A B cosθ = C2 = A2

    ∴          cos θ = - 12

    This gives

           θ = 3π4 rad

    Question 586
    CBSEENPH11026360

    A body is whirled in a horizontal circle of radius 25 cm. It has an angular velocity of 13 rad/s. What is its linear velocity at any point on circular path?

    • 2 m/s

    • 3 m/s

    • 3.25 m/s

    • 4.25 m/s

    Solution

    C.

    3.25 m/s

    Given:- r = 25 cm =  m

      ω = 13 rad/s

       Linear speed  = Radius × angular speed

                V = rω

              = 0.25 × 13

            3.25 m/s

    Question 587
    CBSEENPH11026401

    Assertion: If  α-2t for a particle moving in a straight line starting with initial velocity 4 m/s from the origin, then distance travelled by it in 2 s is same as displacement.

    Reason: Velocity changes direction after 2 s only.

    • If both assertion and reason are true and reason is the correct explanation of assertion.

    • If both assertion and reason are true but reason is not the correct explanation of assertion.

    • If assertion is true but reason is false.

    • If both assertion and reason are false.

    Solution

    B.

    If both assertion and reason are true but reason is not the correct explanation of assertion.

    Acceleration is given by 

        α = - 2t

        dvdt = -2t

    ⇒    4vdv = - 0tt dt

    ⇒     v4v = - t 0t

    ⇒     v - 4 = - t2

    ⇒           v = 4 - t2

    ⇒           v > 0   for   t < 2

                  v = 0   for   t = 2

                  v  < 0   for  t > 2

    The body reverses its direction of motion after 2 s.

    Question 588
    CBSEENPH11026412

    A man is at a distance of  6 m from a bus. The bus begins to move with a constant acceleration of 3ms.In order to catch the bus, the minimum speed with which the man should run towards the bus is

    • 2 m s-1

    • 4 m s-1

    • 6 m s-1

    • 8 m s-1

    Solution

    C.

    6 m s-1

    If the man did not run, the bus would be at a distance s1 at time t given by

       s1 = 6 + 12α t2

           = 6 + 12 × 3 × t2

       s1 = 6 + 32t2

    If v is the speed of man, he would cover a distance s2 = vt in time t.

    To catch the bus,

               s1 = s2

         6 + 32 t2 = v t

    ⇒     t2 - 2 v3t + 4 = 0

    which gives

             t = 2 v6 ±  4 v29 - 16 12  

    Now t will be real if  4 v29 - 16 is positive or zero.

    Minimum v corresponds to 4 v216 - 16 = 0 which gives v = 6 m s-1        

    Question 589
    CBSEENPH11026421

    A ball is dropped from the top of a building 100 m high. At the same instant another ball is thrown upwards with a velocity of 40 m/s from the bottom of the building. The two balls will meet after

    • 3 s

    • 2 s

    • 2.5 s

    • 5 s

    Solution

    C.

    2.5 s

    Let balls meet after t s. The distance travelled by the ball coming down is 

              s112 gt2

    Distance travelled by the other ball

            s2 = 40 t - 12 gt2

             s1 + s2 = 100 m

    ∴        12 gt2 + 40 t - 12gt2 = 100m

                      t = 10040

    ∴                 t = 2.5 s

    Question 590
    CBSEENPH11026442

    The displacement of a body is given to be proportional to the cube of time elapsed. The magnitude of the acceleration of the body is proportional to

    • t

    • t

    • t2

    • t3/2

    Solution

    B.

    t

    As given the displacement of a body is given to be proportional to the cube of time elapsed. 

                  s  ∝ t3

                    s = k t3

    velocity:-  Velocity of an object is the rate of change of its position with respect to time. 

                 v = dsdt

                     = d  k t3 dt

               v = 3 kt2

     Acceleration:- Acceleration is the rate of change of velocity with respect to time.

                 α = dvdt

                     = ddt 3 k t2 

                  α  = 6 kt

    ∴              α  t

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