Physics Part I Chapter 8 Gravitation
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    NCERT Solution For Class 11 Physics Physics Part I

    Gravitation Here is the CBSE Physics Chapter 8 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 Gravitation Chapter 8 NCERT Solutions for Class 11 Physics Gravitation Chapter 8 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 8
    CBSEENPH11020474

    A solid sphere is rotating in free space. If the radius of the sphere is increased keeping mass same which one of the following will not be affected?

    • moment of inertia

    • angular momentum

    • angular velocity

    • rotational kinetic energy

    Solution

    B.

    angular momentum

    In free space, neither acceleration due to gravity for external torque act on the rotating solid sphere. Therefore, taking the same mass of sphere if the radius is increased then a moment of inertia, rotational kinetic energy and angular velocity will change but according to the law of conservation of momentum, angular momentum will not change.

    Question 9
    CBSEENPH11020477

    A satellite of mass m revolves around the earth of radius R at a height x from its surface. If g is the acceleration due to gravity on the surface of the earth, the orbital speed of the satellite is

    • fraction numerator gR over denominator straight R minus straight x end fraction
    • gx

    • fraction numerator gR squared over denominator straight R space plus straight x end fraction
    • open parentheses fraction numerator gR squared over denominator straight R space plus straight x end fraction close parentheses to the power of 1 divided by 2 end exponent

    Solution

    D.

    open parentheses fraction numerator gR squared over denominator straight R space plus straight x end fraction close parentheses to the power of 1 divided by 2 end exponent

    The gravitational force exerted on satellite at a height x is
    straight F subscript straight G space equals space fraction numerator GM subscript straight e straight m over denominator left parenthesis straight R plus straight x right parenthesis squared end fraction
    where Me = mass of earth Since, gravitational force provides the necessary centripetal force, so,
    fraction numerator GM subscript straight e straight m over denominator left parenthesis straight R plus straight x right parenthesis squared end fraction space equals space fraction numerator mv subscript 0 superscript 2 over denominator left parenthesis straight R plus straight x right parenthesis end fraction
rightwards double arrow space fraction numerator GM subscript straight e straight m over denominator left parenthesis straight R plus straight x right parenthesis end fraction space equals space mv subscript 0 superscript 2
rightwards double arrow space fraction numerator begin display style gR squared straight m end style over denominator left parenthesis straight R plus straight x right parenthesis end fraction space equals space mv subscript 0 superscript 2
rightwards double arrow space fraction numerator begin display style gR squared straight m end style over denominator left parenthesis straight R plus straight x right parenthesis end fraction space equals space mv subscript 0 superscript 2 space
open parentheses because space straight g space equals space fraction numerator begin display style GM subscript straight e end style over denominator straight R end fraction close parentheses
straight v subscript straight o space equals space square root of open square brackets fraction numerator gR squared over denominator left parenthesis straight R plus straight x right parenthesis end fraction close square brackets end root
space equals space open square brackets fraction numerator gR squared over denominator left parenthesis straight R plus straight x right parenthesis end fraction close square brackets to the power of 1 divided by 2 end exponent

    Question 10
    CBSEENPH11020478

    The time period of an earth satellite in circular orbit is independent of

    • the mass of the satellite

    • radius of its orbit

    • both the mass and radius of the orbit

    • neither the mass of the satellite nor the radius of its orbit.

    Solution

    A.

    the mass of the satellite

    Time period of satellite
    straight T space equals space 2 straight pi space square root of fraction numerator left parenthesis straight R plus straight h right parenthesis cubed over denominator GM subscript straight e end fraction end root
    where R + h = orbital radius of satellite,
    Me = mass of earth Thus, time period does not depend on the mass of the satellite.

    Question 11
    CBSEENPH11020479

    If g is the acceleration due to gravity on the earth’s surface, the gain in the potential energy of object of mass m raised from the surface of the earth to a height equal to the radius R of the earth is

    • 2 mgR

    • mgR/2

    • mgR/4

    • mgR

    Solution

    B.

    mgR/2

    Gravitational potential energy of body on earth's surface
    straight U space equals space minus space fraction numerator GM subscript straight e straight m over denominator straight R end fraction
    At a height h from earth's surface, its value is
    straight U subscript straight h space equals space minus space fraction numerator GM subscript straight e straight m over denominator left parenthesis straight R plus straight h right parenthesis end fraction
space equals space fraction numerator GM subscript straight e straight m over denominator 2 straight R end fraction
    where Me = mass of earth
    m = mass of body
    R = radius of earth
    therefore comma space Gain space in space potential space energy space equals space straight U subscript straight h minus straight U
equals space fraction numerator negative space GM subscript straight e straight m over denominator 2 straight R end fraction minus open parentheses fraction numerator GM subscript straight e straight m over denominator straight R end fraction close parentheses
space equals space fraction numerator GM subscript straight e straight m over denominator 2 straight R end fraction space plus space fraction numerator GM subscript straight e straight m over denominator straight R end fraction
space equals space fraction numerator GM subscript straight e straight m over denominator 2 straight R end fraction
space equals space fraction numerator gR squared straight m over denominator 2 straight R end fraction space space open parentheses straight g space equals space GM subscript straight e over straight R close parentheses
space equals space 1 half space mgR

    Question 12
    CBSEENPH11020480

    Suppose the gravitational force varies inversely as the nth power of distance. Then the time period planet in circular orbit of radius R around the sun will be proportional to

    • straight R to the power of open parentheses fraction numerator straight n plus 1 over denominator 2 end fraction close parentheses end exponent
    • straight R to the power of open parentheses fraction numerator straight n minus 1 over denominator 2 end fraction close parentheses end exponent
    • Rn

    • straight R to the power of open parentheses fraction numerator straight n minus 2 over denominator 2 end fraction close parentheses end exponent

    Solution

    A.

    straight R to the power of open parentheses fraction numerator straight n plus 1 over denominator 2 end fraction close parentheses end exponent

    The necessary centripetal force required for a planet to move round the sun = gravitational force exerted on it
    straight i. straight e space mv squared over straight R space equals space fraction numerator GM subscript straight e straight m over denominator straight R end fraction
straight v space equals space open parentheses GM over straight R to the power of straight n minus 1 end exponent close parentheses to the power of 1 divided by 2 end exponent
Now comma space straight T space equals space fraction numerator 2 πR over denominator straight v end fraction space equals space 2 πRx space open parentheses straight R to the power of straight n minus 1 end exponent over GM subscript straight e close parentheses to the power of 1 divided by 2 end exponent
space equals space 2 straight pi space open parentheses fraction numerator straight R squared space straight x space straight R to the power of straight n minus 1 end exponent over denominator GM subscript straight e end fraction close parentheses to the power of 1 divided by 2 end exponent
space equals space 2 straight pi space open parentheses fraction numerator straight R to the power of left parenthesis straight n plus 1 right parenthesis divided by 2 end exponent over denominator left parenthesis GM subscript straight e right parenthesis to the power of 1 divided by 2 end exponent end fraction close parentheses
straight T space proportional to space straight R to the power of left parenthesis straight n plus 1 right parenthesis divided by 2 end exponent

    Question 16
    CBSEENPH11020826

    Two astronauts are floating in gravitational free space after having lost contact with their spaceship. The two will:

    • Keep floating at the same distance between them

    • Move towards each other

    • Move away from each other

    • Will become stationary

    Solution

    B.

    Move towards each other

    Both the astronauts are in the condition of weightness. Gravitational force between them pulls towards each other

    Question 17
    CBSEENPH11020839

    Sponsor Area

    Question 18
    CBSEENPH11020857

    Two 20 g flatworms climb over a very thin wall, 10 cm high. One of the worms is 20 cm long, the other is wider and only 10 cm long. which of the following statement is correct regarding them?

    • 20 cm worm has done more work against gravity

    • 10 cm worm has done more work against gravity

    • Both worms have done equal work against gravity

    • Ratio of work done by both the worms is 4:5

    Solution

    B.

    10 cm worm has done more work against gravity

    The work done against gravity can be calculated from the increased in height of the centre of mass. The centre of mass of a worm folded in two is located at the middle of either half, i.e., at a point one-quarter of the worm's total length from one end.

    Thus, the centre of mass of the narrow flat worm travels 5 cm up the wall; while that of the broad worm moves 7.5 cm. Hence, work done by 10 cm worm is more against gravity. Also, the ratio of the work done is 2:3.

    Question 19
    CBSEENPH11020858

    A rocket is intended to leave the Earth's gravitational field. The fuel in its main engine is a little less than the amount that is necessary and an auxiliary engine, (only capable of operating for a short time) has to be used as well. When is it best to switch on the auxiliary engine?

    • at take-off

    • When the rocket has nearly stopped with respect to the Earth

    • It doesn't matter

    • Can't say

    Solution

    A.

    at take-off

    The rocket has to reach the highest possible total energy. If the zero level of gravitational potential energy is infinitely far away, then the energy of the rocket standing on the surface on the earth is negative. The energy released during the operation of the engines increases the total energy of the rocket, and the rocket can leave the earth's gravitational field if the sum of its potential and kinetic energies becomes positive.

    The energy released in the course of the operation of the principal and auxiliary engines increase the total energy of the rocket and its ejected combustion products by a fixed value; this increase is independent of the moment when the engines are scratched on. However, the speed at which the combustion products fall back to the earth does depend on the timing of the rocket's operations. Indeed, if the auxiliary engine starts working when the rocket is at a greater height, the combustion fall further and their speed and total energy are higher when they hit the ground. This means that the sooner the auxiliary engine is switched on. the higher, the energy ultimately acquired by the rocket. The same concept is valid for the principal engine, and if the only energy consideration applies, it is best to operate the engines for the shortest time and at the highest thrust.

    Question 20
    CBSEENPH11020866

    The planets with radii R1 and R2 have densities ρ1, ρ2 respectively. Their atmospheric pressure is p1 and p2 respectively. Therefore, the ratio of masses of their atmospheres, neglecting variation of g within the limits of the atmosphere is

    • ρ1R2ρ1ρ2R1ρ2

    • ρ1R2ρ2pP2R1ρ1

    • p1R1ρ1p2R2ρ2

    • p1R1ρ2p2R2ρ1

    Solution

    D.

    p1R1ρ2p2R2ρ1

    Acceleration due to gravity

    g = GMR2 = 43πρGRg1g2 = ρ1R1ρ2R2

    Atmospheric pressure can be given by

    P = W/S

    Where w = weight of the atmosphere,

    S = surface area of the planet

    m1m2 = ρ1S1g2ρ2S2g1 = p1.(4πR12)p2 (4πR22).ρ2R1ρ1R1 = p1R1ρ2p2R2ρ1 

    Question 21
    CBSEENPH11020867

    A thin symmertical double convex lens of refractive index μ2 = 1.5 is placed between a medium of refractive index μ1 = 1.4 to the left and another medium of refractive index μ3 = 1.6 to the right. Then, the system behaves as

    • a convex lens

    • a concave lens

    • a glass plate

    • a convexo-concave lens

    Solution

    C.

    a glass plate

    For the first refracting surface,

    For a parallel beam of light; 

    -μu + μ2V = μ2-1R = 1.4- + 1.5V = 1.5 - 1.4RV = 15 R

    For second refracting surface,

    μ21.5R + μ3V' = μ3- μ2R - 1.515 R + 1.6V' = 1.6-1.5R V' = 

    Hence, the combination behaves as a glass plate.

    Question 22
    CBSEENPH11024904

    Write the SI unit of Gravitational constant

    Solution

    F =GMmr2

    Therefore, we can obtain G as,

    G=Fr2Mm

    Since SI unit of

    Force = Newton

    Mass = Kilogram

    Distance, r = metre

     

    G = Nm²kg⁻²

         OR

    G = kg⁻¹ m³s⁻²

    Question 23
    CBSEENPH11026109

    Radius of orbit of satellite of earth is R. Its kinetic energy is proportional to

    • 1R

    • 1R

    • R

    • 1R32

    Solution

    A.

    1R

    Kinetic energy of the satellite

    KE = 12mvo2       ...... 1Where vo =GMRNow putting the value of vo is eqn1KE =12mGMR22       =12mGMR KE 1R

    So the kinetic energy of the satellite is inversely proportional 

    to R. Therefore R increases then KE decreases.

    Question 24
    CBSEENPH11026138
    Question 25
    CBSEENPH11026141

    Earth is revolving around the sun. If the distance of the earth from the sun is reduced to 1/4th of the present distance then the length of present day will be reduced by :

    • 14

    • 12

    • 18

    • 16

    Solution

    C.

    18

    Law of periods:- The square of the time period of revolution of a planet is proportional to the cube of the semi-major axis of the ellipse traced out by the planet.

    From Kepeler's law

    T2 ∝ R3

    T1T22 =R1R23T1T2=R1R232T1T2  =RR432        =  432         = 23    T1T2 = 8 T2 = T18

    Question 27
    CBSEENPH11026155

    Imagine a new planet having the same density as that of earth but it is 3 times bigger than the earth in size. If the acceleration due to gravity on the surface of earth is g and that on the surface of the new planet is g', then

    • g' = 3g

    • g' = g9

    • g' = 9g

    • g' = 27g

    Solution

    A.

    g' = 3g

    The acceleration due to gravity on the planet can be found using the relation

    g = GMR2but M = 43πR3ρ ρ- densityg = G × 43πR3ρR2g = G × 43πRρ

    ⇒ g ∝ R

     g'R  =  R'R g'g =  3RR g'g =  3 g'  =  3g

    Question 28
    CBSEENPH11026177

    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 statements 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

    When two satellites of earth are moving in same orbit, then time period of both are equal. 
    From Kepler's third law
    T2 ∝ r3
    Time period is independent of mass, hence their time 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 29
    CBSEENPH11026196

    For a satellite moving in an orbit around the earth, the ratio of kinetic energy to potential energy is

    • 2

    • 12

    • 12

    • 2

    Solution

    B.

    12

    Given:- Kinetic energy of satellite is half of its potential energy. 
    Potential energy of satelite

               U = GMemRe

    Kinetic energy of satellite

              K = 12GMe mRe

    Thus

        KU = 12GMe mRe × ReGMe m KU = 12

    Question 30
    CBSEENPH11026216

    A satellite orbiting the earth in a circular orbit of radius R completes one revolution in 3h. If orbital radius of geostationary satellite is 36000 km, orbital radius of earth is

    • 6000 km

    • 9000 km

    • 12000 km

    • 15000 km

    Solution

    B.

    9000 km

    Time period of satellite

         T24 π2GM r3

         T24π2g R2 r3                      [ since GM = gR2 ]

    ⇒    T = 2πR r3g

    ⇒     T ∝ r3/2 

    ∴    T1T2 = r1r232       

    Given   T1 = 3h , T2 = 24 h 

     (geostationary satellite) r1 = R,  r2 = 36000 km

    ⇒        324 = R3600032

    ⇒         R = 1823 × 36000

    ⇒         R = 14 × 36000 

    ⇒         R = 9000 km

    Question 31
    CBSEENPH11026261

    A launching vehicle carrying an artificial satellite of mass m is set for launch on the surface of the earth of mass M and radius R. If the satellite is intended to move in a circular orbit of radius 7R, the minimum energy required to be spent by the launching vehicle on the satellite is (Gravitational constant= G)

    • GMmR

    • - 13 G M m14 R

    • GMm7R

    • -G M m14 R

    Solution

    B.

    - 13 G M m14 R

    The energy of artificial satellite at the surface of the earth

        E1 =  - G M mR

    When the satellite is intended to move in a circular orbit of radius 7R, then energy of artificial satellite

       E2- 12G M m7R

    The minimum energy required

        E = E1 - E2   

       = - G M mR  + 12 G M m7R

         = - 14 G M m + G  M m14 R

    E   = -13 G M m14R

    Question 32
    CBSEENPH11026282

    The escape velocity for the earth is 11.2 km/s. The mass of another planet 100  times mass of earth and its radius is 4 times radius of the earth. The escape velocity for the planet is

    • 280 km/s

    • 56.0 km/s

    • 112 km/s

    • 56 km/s

    Solution

    B.

    56.0 km/s

    Escape velocity is the minimum speed needed to a free object to escape speed from the gravitational influence of a massive body.

    The escape velocity of the planet

            Vescape 2G mR  MR

             Vescape 2MPRP2 MeRe

    ∴        VP Ve = MPRP× ReRP

    where Vp - velocity of planet, Ve - velocity of earth

     ⇒      VPVe = 100 × 14 

      ⇒      VPVe = 5

     ⇒        VP = 5 × 11.2 

     ⇒        VP = 56 km/s

    Question 33
    CBSEENPH11026307

    Change in acceleration due to gravity is same upto a height h from each other the earth surface and below depth x then relation between x and his ( h and x<<<Re )

    • x = h

    • x = 2h

    • x = h2

    • x = h2

    Solution

    B.

    x = 2h

    Variation in g at the height h from the earth surface 

            g' =  g 1 - 2hRe 

     

    Variation in g at a depth x below the earth surface 

            g'' = g 1 - xRe

    Here d = x i.e depth

    But variations are same

    ∴     g 1 - 2hRe = g 1 - xRe

    ∴            x = 2h 

    Question 34
    CBSEENPH11026326

     A uniform ring of mass m and radius a is placed directly above a uniform sphere of mass m and of equal to radius. The centre of the ring is at a distance 3a from the centre of the sphere. The gravitational force (Fl exerted by the sphere on the ring is

    • 3 G Mm8 a2

    • 2 G M m3 a2

    • 7 G Mm2 a2

    • 3 G Mma2

    Solution

    A.

    3 G Mm8 a2

    Now, the gravitational field due to the ring at a distance d = 3a on its axis is

                 E = G mda2 + d232

                  E = 3 G m aa2 + 3a2 32

                  E = 3 Gm8 a2

          

    The force in a particle of mass M placed here is

    Force on the sphere = Mass of sphere × gravitational field

                         F = M E 

                           F  = 3 GMm8 a2

    This is also the force due to the sphere in the ring. 

    Question 35
    CBSEENPH11026377

    A body of mass 2m is placed on earth's surface. Calculate the change in gravitational potential energy, if this body is taken from earth's surface to a height of h, where h = 4R.

    • 2 mghR

    • 23 mgR

    • 85 mgR

    • mgR2

    Solution

    C.

    85 mgR

     Potential energy, 

                   U = -G Mmr

                      = - 2G Mmr                  [ m = 2m]

    At earth's surface, r = R

    ∴           Ue-2GMmR

    Now, if a body is taken to height, h = 4R, then the potential energy is given by

                   Uh-2GMmR + h                   [ r = h + R ]

                        = -2GMm5R

    Thus, change in gravitational potential energy,

          ΔU = Uh - Ue

               = -2GMm5R + 2GMmR

               = 85 GMmR

           ΔU = 85 gR2 mR                       [  Gm = gR2 ]

            ΔU = 85 mgR                                        

    Question 36
    CBSEENPH11026385

    A communication satellite of 500 kg revolves around the earth in a circular orbit of radius 4.0 x 107 m in the equatorial plane of the earth from west to east. The magnitude of angular momentum of the satellite is

    • ∼ 0.13  × 1014 kg m2 s-1

    • ∼ 1.3  × 1014 kg m2 s-1

    • ∼ 0.58 × 1014 kg m2 s-1

    • ∼ 2.58  × 1014 kg m2 s-1

    Solution

    C.

    ∼ 0.58 × 1014 kg m2 s-1

    As the satellite is moving in equatorial plane with orbital radius 4 x 107 m.

    ∴        Satellite is geostationary satellite.

    Hence, the time taken by satellite to complete its one revolution

           T = 24 h

                = 24 × 3600 s

           T = 86400 s

    Velocity of satellite 

           v = 2πrT

    Angular momentum, L = mvr

           L = m 2 π rT r

               = 2π mT r2

    ∴     L = 2 × 3.14 × 50086400× ( 4 × 107 )2

            L = 0.58 × 1014 kg m2 s-1

    Question 37
    CBSEENPH11026386

    The change in the gravitational potential energy when a body of mass m is raised to a height nR above the surface of the earth is ( here R is the radius of the earth )

    • nn + 1 mgR

    • nn - 1 mgR

    • nmgR

    • mgRn

    Solution

    A.

    nn + 1 mgR

    Gravitational potential energy of mass m at any point at a distance r from the centre of earth is

                  U = - G Mmr

    At the surface of earth r = R,

    ∴          Us- G MmR

                     = - mgR                               g = GMR2

    At the height h = nR from the surface of earth

                     r = R + h                  

                       = R + nR

                   r = R ( 1 + n )

    ∴       Uh- G MmR 1 + n

                  = -mg R1 + n

    Change in gravitational potential energy is

              ΔU = Uh - Us

                    = - m gR1+n - - m gR 

                     = - m gR1 + n 1 - 11 + n

                ΔU = m gR n1 + n

    Question 38
    CBSEENPH11026407

    Assertion:  Two bodies of masses M and m (M > m) are allowed to fall from the same height if the air resistance for each be the same then both the bodies will reach the earth simultaneously. 

    Reason:  For same air resistance, acceleration of both the bodies will be same.

    • 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

    D.

    If both assertion and reason are false.

    The force acting on the body of mass M are its weight Mg acting vertically downwards and air resistance F acting vertically upward.

    ∴          Acceleration of the body

                     α = Mg - FM

                      α = g - FM

    Now, M > m, therefore, the body with larger mass will have greater acceleration and it will reach the ground first.

    Sponsor Area

    Question 40
    CBSEENPH11026417

    Time period of pendulum, on a satellite orbiting the earth, is

    • 1π

    • zero

    • π

    • infinity

    Solution

    D.

    infinity

    On an artificial satellite orbiting the earth the acceleration is given by GMR2 towards the centre of the earth.

    Now for a body of mass m on the satellite the graviational force due to earth is  G MmR2 towards the centre of the earth.

    Let the reaction force on the surface of the satellite be N, then

              G MmR2 - N = m G MR2

    ⇒           N = 0

    That is on the satellite there is a state of weightlessness or  g = 0

               T = 2 π lg = ∞

    Hence     T = ∞

    Question 41
    CBSEENPH11026423

    The additional kinetic energy to be provided to a satellite of mass m revolving around a planet of mass M to transfer it from a circular orbit of radius R, to another of radius R2 (R2 > R1 ) is

    • GmM 1R12 - 1R22

    • GmM 1R1 - 1R2

    • 2 GmM 1R1 - 1R2

    • 12 GmM 1R1- 1R2

    Solution

    D.

    12 GmM 1R1- 1R2

    Additional kinetic energy provided to satelite   

        - GmM2R1 + KE =-GMm2R2

                KE = - G Mm2R2 + G Mm2R1

    ∴           KE = 12 GMm 1R1 - 1R2

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