Q 1 :

Two planets A and B having masses m1 and m2 move around the sun in circular orbits of r1 and r2 radii respectively. If angular momentum of A is L and that of B is 3 L, the ratio of time period (TATB) is                       [2024]

  • 127(m2m1)3

     

  • (r1r2)3

     

  • (r2r1)32

     

  • 27(m1m2)3

     

(1) 

        We know, T2r3(TATB)2=(r1r2)3

        TATB=(r1r2)3/2                                                             ...(1)

       Given LB=3LAm2v2r2=3(m1v1r1)

        v2v1=3m1r1m2r2                                                               ...(2)

       Orbital velocity, v=GMsrv2v1=r1r2                  ...(3)

       From eq. (2) and eq. (3)

       r1r2=3m1m2r1r2r1r2=(m23m1)                               ...(4)

       Using eq. (4) in eq. (1)

       TATB=(m23m1)3=127(m2m1)3



Q 2 :

Given below are two statements: one is labelled as Assertion (A) and the other is labelled as Reason (R).  

 

Assertion (A): The angular speed of the moon in its orbit about the earth is more than the angular speed of the earth in its orbit about the sun.

 

Reason (R): The moon takes less time to move around the earth than the time taken by the earth to move around the sun.

 

In the light of the above statements, choose the most appropriate answer from the given below:                                            [2024]

  • (A) is correct but (R) is not correct

     

  • Both (A) and (R) are correct and (R) is the correct explanation of (A)

     

  • Both (A) and (R) are correct but (R) is not the correct explanation of (A)

     

  • (A) is not correct but (R) is correct

     

(2)  

        ω=2πTω1T

        Tmoon= 27 days

        Tearth= 365 days 4 hour

        ωmoon>ωearth

 



Q 3 :

A planet takes 200 days to complete one revolution around the Sun. If the distance of the planet from Sun is reduced to one fourth of the original distance, how many days will it take to complete one revolution            [2024]

  • 20

     

  • 25

     

  • 50

     

  • 100

     

(2) 

         T2r3T12r13=T22r23

         (200)2r3=T22(r4)3

           200×2004×4×4=T22T2=2004×2

             T2=25 days



Q 4 :

A light planet is revolving around a massive star in a circular orbit of radius R with a period of revolution T. If the force of attraction between planet and star is proportional to R-3/2 then choose the correct option            [2024]

  • T2R5/2

     

  • T2R7/2

     

  • T2R3/2

     

  • T2R3

     

(1) 

        F1R3/2 or F=KR3/2

         Gravitational force provides centripetal force.

         KR3/2=mω2R

           ω2=KmR5/2

           (2πT)2=KmR5/2T2

            =4π2mR5/2KT2R5/2

 



Q 5 :

Four identical particles of mass m are kept at the four corners of a square. If the gravitational force exerted on one of the masses by the other masses is (22+132)Gm2L2, the length of the sides of the square is    [2024]

  • 2 L

     

  • 4 L

     

  • 3 L

     

  • L/2

     

(2)

Fnet=2F+F'

F=Gm2a2 and F'=Gm2(2a)2

Fnet=2Gm2a2+Gm22a2

(22+132)Gm2L2=Gm2a2(22+12)a=4L



Q 6 :

A small point of mass m is placed at a distance 2R from the centre 'O' of a big uniform solid sphere of mass M and radius R. The gravitational force on 'm' due to M is F1. A spherical part of radius R/3 is removed from the big sphere as shown in the figure and the gravitational force on m due to remaining part of M is found to be F2. The value of ratio F1 : F2 is          [2025]

  • 16 : 9

     

  • 11 : 10

     

  • 12 : 11

     

  • 12 : 9

     

(3)

F1=GMm4R2

F2=GMm4R2G(M27)m(4R3)2

 GMmR2(14148)=1148GMmR2

 F1F2=1211



Q 7 :

If a satellite orbiting the earth is 9 times closer to the earth than the moon, what is the time period of rotation of the satellite? Given rotational time period of Moon = 27 days and gravitational attraction between the satellite and the moon is neglected.          [2025]

  • 1 day

     

  • 81 days

     

  • 27 days

     

  • 3 days

     

(1)

From Kepler's law, T2R3

(TmTs)2=(RR/9)3

TmTs=(3)3  Ts=(2727)= 1 day



Q 8 :

A satellite is launched into a circular orbit of radius 'R' around the earth. A second satellite is launched into an orbit of radius 1.03 R. The time period of revolution of the second satellite is larger than the first one approximately by:         [2025]

  • 3%

     

  • 4.5%

     

  • 9%

     

  • 2.5%

     

(2)

T2=KR3

2TT=3RR

2TT=3×0.03RR

TT=3×0.032×100=4.5%



Q 9 :

Given below are two statements: one is labelled as Assertion (A) and the other is labelled as Reason (R).

Assertion (A) : The radius vector from the Sun to a planet sweeps out equal areas in equal intervals of time and thus areal velocity of planet is constant.

Reason (R) : For a central force field the angular momentum is a constant.

In the light of the above statements, choose the most appropriate answer from the options given below:          [2025]

  • Both (A) and (R) are correct and (R) is the correct explanation of (A)

     

  • Both (A) and (R) are correct but (R) is not the correct explanation of (A)

     

  • (A) is correct but (R) is not correct

     

  • (A) is not correct but (R) is correct

     

(1)

From Kepler's 2nd law

                      dAdt=L2m=constant

Due to central force torque is zero and angular momentum is constant.



Q 10 :

Two planets, A and B are orbiting a common star in circular orbits of radii RA and RB, respectively, with RB=2RA. The planet B is 42 times more massive than planet A. The ratio (LBLA) of angular momentum (LB) of planet B to that of planet A(LA) is closest to integer __________.          [2025]



(8)

L=mv0R=mGMRR=mGMR

Here M is mass of star

LBLA=mBmARBRA=4221

 LBLA=8



Q 11 :

Three identical spheres of mass m, are placed at the vertices of an equilateral triangle of length a. When released, they interact only through gravitational force and collide after a time T = 4 seconds. If the sides of the triangle are increased to length 2a and also the masses of the spheres are made 2m, then they will collide after ________ seconds.         [2025]



(8)

TmxGyaz

TMx[M1L3T2]y[L]z

TMxyL3y+zT2y

xy=0  x=y

2y=1  y=12, x=12

3y+z=0  z=3y=32

Hence, Tm1/2G1/2a3/2

T(a3m)1/2  T=4×(232)1/2=8 s



Q 12 :

If the distance of the Earth from the Sun is 1.5×106 km, then the distance of an imaginary planet from the Sun, if its period of revolution is 2.83 years is:      [2023]

  • 6×106 km

     

  • 3×106 km

     

  • 6×107 km

     

  • 3×107 km

     

(2)

T2R3(T1T2)2=(R1R2)3

(12.83)2=(1.5×106R2)3

R2=[(2.83)2×(1.5×106)3]1/3

            =81/3×1.5×106=3×106 km



Q 13 :

Every planet revolves around the Sun in an elliptical orbit:                                           [2023]

A. The force acting on a planet is inversely proportional to the square of the distance from the Sun.

B. Force acting on a planet is inversely proportional to the product of the masses of the planet and the Sun.

C. The centripetal force acting on the planet is directed away from the Sun.

D. The square of the time period of revolution of a planet around the Sun is directly proportional to the cube of the semi-major axis of the elliptical orbit.

Choose the correct answer from the options given below:

  • A and D only

     

  • B and C only

     

  • A and C only

     

  • C and C only

     

(1)

F=Gm1m2r2

F1r2

Fm1m2

This force provides centripetal force and acts towards the Sun.

T2a3  (Kepler’s third law)



Q 14 :

A planet has double the mass of the Earth. Its average density is equal to that of the Earth. An object weighing W on Earth will weigh on that planet       [2023]

  • 21/3W

     

  • W

     

  • 22/3W

     

  • 2W

     

(1)

m=ρ×43πR3

Rm1/3  (ρ=constant)

Weight=WgGmR2

        Wmm2/3m1/3

So, W'=(2)1/3W



Q 15 :

Choose the incorrect statement from the following:       [2023]

  • The speed of a satellite in a given circular orbit remains constant.

     

  • For a planet revolving around the Sun in an elliptical orbit, the total energy of the planet remains constant.

     

  • When a body falls towards the Earth, the displacement of the Earth towards the body is negligible.

     

  • The linear speed of a planet revolving around the Sun remains constant.

     

(4)

Planets revolve in elliptical paths around sun. Thus their linear speed is not constant.



Q 16 :

The radii of two planets 'A' and 'B' are 'R' and '4R' and their densities are ρ and ρ/3 respectively. The ratio of acceleration due to gravity at their surfaces (gA:gB) will be           [2023]

  • 1 : 16

     

  • 3 : 16

     

  • 3 : 4

     

  • 4 : 3

     

(3)

g=GMR2=GR2×ρ×4π3R3=(4π3G)ρR

gAgB=R×ρ4R×ρ3=34



Q 17 :

Two identical particles each of mass 'm' go round a circle of radius a under the action of their mutual gravitational attraction. The angular speed of each particle will be     [2023]

  • Gm2a3

     

  • Gm8a3

     

  • Gm4a3

     

  • Gma3

     

(3)

F=mω2r

Gmm(2a)2=mω2a

ω=Gm4a3



Q 18 :

If the Earth suddenly shrinks to 164th of its original volume with its mass remaining the same, the period of rotation of the Earth becomes 24xh. The value of x is ________ .        [2023]



(16)

From conservation of angular momentum

25MR2ω1=25M(R4)2ω2

MR2ω1=MR216ω2

ω1ω2=116

T2T1=116T2T1=161=t1x

  t1=24161=24t2x=16



Q 19 :

Net gravitational force at the center of a square is found to be F1 when four particles having mass M, 2M, 3M and 4M are placed at the four corners of the square as shown in figure and is F2 when the positions of 3M and 4M are interchanged. The ratio F1F2 is α5. The value of α is ________.           [2026]

  • 1

     

  • 3

     

  • 25

     

  • 2

     

(4)