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Gravitation — JEE Previous Year Questions

Every Gravitation question asked in JEE Main and JEE Advanced across the last 186 papers — 184 questions, each with its correct answer. Free to read, no account needed.

Questions

184

Papers it appeared in

152/186

Appearance rate

82%

All 184 Gravitation questions

Most recent papers first.

Q1·PhysicsSingle correctJEE Advanced 2026
A particle of mass mmm, and angular momentum ℓ\ellℓ is moving in a circular orbit of radius r0r_{0}r0​ under the influence of an attractive force F⃗(r)=−kr2r^\vec{F}(r) = -\frac{k}{r^{2}}\hat{r}F(r)=−r2k​r^. Keeping its angular momentum unchanged, the particle is displaced radially by a small distance δr≪r0\delta r \ll r_{0}δr≪r0​, due to which its radial distance varies periodically. The corresponding time period is:
  1. (A)2πℓ3mk2\frac{2\pi\ell^{3}}{mk^{2}}mk22πℓ3​
  2. (B)2πmk2\pi\sqrt{\frac{m}{k}}2πkm​​
  3. (C)2πℓ33mk2\frac{2\pi\ell^{3}}{3mk^{2}}3mk22πℓ3​
  4. (D)2πℓ35mk2\frac{2\pi\ell^{3}}{5mk^{2}}5mk22πℓ3​

Correct answer: (A)

Step-by-step solution →
Q2·PhysicsSingle correctJEE Main 2026
A body of mass m is taken from the surface of earth to a height equal to twice the radius of earth (Re_ee​). The increase in potential energy will be ________. (g is acceleration due to gravity at the surface of earth)
  1. (A)12mgRe\dfrac{1}{2}mgR_{e}21​mgRe​
  2. (B)34mgRe\dfrac{3}{4}mgR_{e}43​mgRe​
  3. (C)14mgRe\dfrac{1}{4}mgR_{e}41​mgRe​
  4. (D)23mgRe\dfrac{2}{3}mgR_{e}32​mgRe​

Correct answer: (D)

Step-by-step solution →
Q3·PhysicsSingle correctJEE Main 2026
When one moves from a point 16 km below the earth's surface to a point 16 km above the earth's surface. The change in ggg is approximately α\alphaα %. The value of α\alphaα is ______. (Take radius of the earth === 6400 km.)
  1. (A)0.12
  2. (B)0.25
  3. (C)0.50
  4. (D)0.75

Correct answer: (B)

Step-by-step solution →
Q4·PhysicsSingle correctJEE Main 2026
The height in terms of radius of the earth (R)(R)(R), at which the acceleration due to gravity becomes g9\frac{g}{9}9g​, where ggg is acceleration due to gravity on earth's surface, is ______.
  1. (A)3R\sqrt{3}R3​R
  2. (B)22R2\sqrt{2}R22​R
  3. (C)2R2R2R
  4. (D)49R\frac{4}{9}R94​R

Correct answer: (C)

Step-by-step solution →
Q5·PhysicsSingle correctJEE Main 2026
A planet (P1)(P_{1})(P1​) is moving around the star of mass 2M2M2M in the orbit of radius RRR. Another planet (P2)(P_{2})(P2​) is moving around another star of mass 4M4M4M in a orbit of radius 2R2R2R. Ratio of time periods of revolution of P2P_{2}P2​ and P1P_{1}P1​ is ______.
  1. (A)12\frac{1}{2}21​
  2. (B)2
  3. (C)4
  4. (D)14\frac{1}{4}41​

Correct answer: (B)

Step-by-step solution →
Q6·PhysicsSingle correctJEE Main 2026
If a body of mass 1 kg falls on the earth from infinity, it attains velocity (v)(v)(v) and kinetic energy (k)(k)(k) on reaching the surface of earth. The values of vvv and kkk respectively are ______. (Take radius of earth to be 6400 km and g=9.8g = 9.8g=9.8 m/s2^{2}2)
  1. (A)11.2 km/s; 6.27×1076.27 \times 10^{7}6.27×107 J
  2. (B)11.2 km/s; 12.54×10712.54 \times 10^{7}12.54×107 J
  3. (C)8.8 km/s; 6.27×1076.27 \times 10^{7}6.27×107 J
  4. (D)8.8 km/s; 12.54×10712.54 \times 10^{7}12.54×107 J

Correct answer: (A)

Step-by-step solution →
Q7·PhysicsSingle correctJEE Main 2026
Three masses 200 kg, 300 kg and 400 kg are placed at the vertices of an equilateral triangle with sides 20 m. They are rearranged on the vertices of a bigger triangle of side 25 m and with the same centre. The work done in this process _________ J. (Gravitational constant G=6.7×10−11G = 6.7 \times 10^{-11}G=6.7×10−11 N m2^{2}2 / kg2^{2}2)
  1. (A)9.86×10−69.86 \times 10^{-6}9.86×10−6
  2. (B)2.85×10−72.85 \times 10^{-7}2.85×10−7
  3. (C)1.74×10−71.74 \times 10^{-7}1.74×10−7
  4. (D)4.77×10−74.77 \times 10^{-7}4.77×10−7

Correct answer: (C)

Step-by-step solution →
Q8·PhysicsSingle correctJEE Main 2026
Net gravitational force at the centre of a square is found to be F1F_{1}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 it is F2F_{2}F2​ when the positions of 3M and 4M are interchanged. The ratio F1F2\frac{F_{1}}{F_{2}}F2​F1​​ is α5\frac{\alpha}{\sqrt{5}}5​α​. The value of α is _____.
  1. (A)2
  2. (B)3
  3. (C)1
  4. (D)252\sqrt{5}25​

Correct answer: (A)

Step-by-step solution →
Q9·PhysicsSingle correctJEE Main 2026
Given below are two statements : Statement I : A satellite is moving around earth in the orbit very close to the earth surface. The time period of revolution of satellite depends upon the density of earth. Statement II : The time period of revolution of the satellite is T=2πRegT = 2\pi\sqrt{\dfrac{R_e}{g}}T=2πgRe​​​ (for satellite very close to the earth surface), where ReR_eRe​ radius of earth and ggg acceleration due to gravity. In the light of the above statements, choose the correct answer from the options given below :
  1. (A)Both Statement I and Statement II are false
  2. (B)Both Statement I and Statement II are true
  3. (C)Statement I is true but Statement II is false
  4. (D)Statement I is false but Statement II is true

Correct answer: (B)

Step-by-step solution →
Q10·PhysicsSingle correctJEE Main 2026
The escape velocity from a spherical planet A is 10 km/s. The escape velocity from another planet B whose density and radius are 10% of those of planet A, is_______m/s.
  1. (A)1000
  2. (B)2005200\sqrt{5}2005​
  3. (C)10010100\sqrt{10}10010​
  4. (D)100021000\sqrt{2}10002​

Correct answer: (C)

Step-by-step solution →
Q11·PhysicsSingle correctJEE Main 2026
Initially a satellite of 100 kg is in a circular orbit of radius 1.5RE1.5R_E1.5RE​. This satellite can be moved to a circular orbit of radius 3RE3R_E3RE​ by supplying α×106\alpha \times 10^{6}α×106J of energy. The value of α is ____________. (Take Radius of Earth RE=6×106R_E = 6 \times 10^{6}RE​=6×106 m and g = 10 m/s2^{2}2)
  1. (A)150
  2. (B)500
  3. (C)100
  4. (D)1000

Correct answer: (D)

Step-by-step solution →
Q12·PhysicsNumericalJEE Advanced 2025
A geostationary satellite above the equator is orbiting around the earth at a fixed distance r1r_1r1​ from the center of the earth. A second satellite is orbiting in the equatorial plane in the opposite direction to the earth's rotation, at a distance r2r_2r2​ from the center of the earth, such that r1=1.21 r2r_1 = 1.21\ r_2r1​=1.21 r2​. The time period of the second satellite as measured from the geostationary satellite is 24p\frac{24}{p}p24​ hours. The value of p is ______

Correct answer: 2.33

Step-by-step solution →
Q13·PhysicsSingle correctJEE Main 2025
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 areal velocity of planet is constant. Reason (R): For a central force field the angular momentum is constant. In the light of the above statements, choose the most appropriate answer from the options given below:
  1. (A)Both (A) and (R) are correct and (R) is the correct explanation of (A)
  2. (B)Both (A) and (R) are correct but (R) is not the correct explanation of (A)
  3. (C)(A) is correct but (R) is not correct
  4. (D)(A) is not correct but (R) is correct

Correct answer: (A)

Step-by-step solution →
Q14·PhysicsSingle correctJEE Main 2025
Given below are two statements: one is labelled as Assertion A and the other is labelled as Reason R. Assertion A: The kinetic energy needed to project a body of mass mmm from earth surface to infinity is 12mgR\dfrac{1}{2}mgR21​mgR, where RRR is the radius of earth. Reason R: The maximum potential energy of a body is zero when it is projected to infinity from earth surface. In the light of the above statements, choose the correct answer from the option given below:
  1. (A)A False but R is true
  2. (B)Both A and R are true and R is the correct explanation of A
  3. (C)A is true but R is false
  4. (D)Both A and R are true but R is NOT the correct explanation of A

Correct answer: (A)

Step-by-step solution →
Q15·PhysicsSingle correctJEE Main 2025
An object is kept at rest at a distance of 3R3R3R above the earth's surface where R is earth's radius. The minimum speed with which it must be projected so that it does not return to earth is: (Assume M = mass of earth, G = Universal gravitational constant)
  1. (A)GM2R\sqrt{\dfrac{GM}{2R}}2RGM​​
  2. (B)GMR\sqrt{\dfrac{GM}{R}}RGM​​
  3. (C)3GMR\sqrt{\dfrac{3GM}{R}}R3GM​​
  4. (D)2GMR\sqrt{\dfrac{2GM}{R}}R2GM​​

Correct answer: (A)

Step-by-step solution →
Q16·PhysicsIntegerJEE Main 2025
Three identical spheres of mass mmm are placed at the vertices of an equilateral triangle of side length aaa. When released, they interact only through gravitational force and collide after a time T=4T=4T=4 seconds. If the sides of the triangle are increased to length 2a2a2a and also the masses of the spheres are made 2m2m2m, then they will collide after __________ seconds.

Correct answer: 8

Step-by-step solution →
Q17·PhysicsIntegerJEE Main 2025
A satellite of mass 1000 kg is launched to revolve around the earth in an orbit at a height of 270 km from the earth’s surface. Kinetic energy of the satellite in this orbit is ______ ×1010 J\times10^{10}\,J×1010J. (Mass of earth =6×1024=6\times10^{24}=6×1024 kg, Radius of earth =6.4×106=6.4\times10^6=6.4×106 m, Gravitational constant =6.67×10−11 Nm2kg−2=6.67\times10^{-11}\,Nm^2kg^{-2}=6.67×10−11Nm2kg−2)

Correct answer: 3

Step-by-step solution →
Q18·PhysicsIntegerJEE Main 2025
Two planets, A and B are orbiting a common star in circular orbits of radii RAR_ARA​ and RBR_BRB​, respectively, with RB=2RAR_B=2R_ARB​=2RA​. The planet B is 424\sqrt{2}42​ times more massive than planet A. The ratio (LBLA)\left(\dfrac{L_B}{L_A}\right)(LA​LB​​) of angular momentum (LBL_BLB​) of planet B to that of planet A(LA)A(L_A)A(LA​) is closest to integer ______.

Correct answer: 8

Step-by-step solution →
Q19·PhysicsSingle correctJEE Main 2025
Given below are two statements: one is labelled as Assertion (A) and the other is labelled as Reason (R). Assertion (A): Time period of a simple pendulum is longer at the top of a mountain than that at the base of the mountain. Reason (R): Time period of a simple pendulum decreases with increasing value of acceleration due to gravity and vice-versa. In the light of the above statements, choose the most appropriate answer from the options given below:
  1. (A)Both (A) and (R) are true but (R) is not the correct explanation of (A).
  2. (B)Both (A) and (R) are true and (R) is the correct explanation of (A).
  3. (C)(A) is true but (R) is false.
  4. (D)(A) is false but (R) is true.

Correct answer: (B)

Step-by-step solution →
Q20·PhysicsSingle correctJEE Main 2025
Earth has mass 8 times and radius 2 times that of a planet. If the escape velocity from the earth is 11.2 km/s, the escape velocity in km/s from the planet will be:
  1. (A)11.2
  2. (B)5.6
  3. (C)2.8
  4. (D)8.4

Correct answer: (B)

Step-by-step solution →
Q21·PhysicsSingle correctJEE Main 2025
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 :-
  1. (A)3%
  2. (B)4.5%
  3. (C)9%
  4. (D)2.5%

Correct answer: (B)

Step-by-step solution →
Q22·PhysicsIntegerJEE Main 2025
Acceleration due to gravity on the surface of earth is "g". If the diameter of earth is reduced to one third of its original value and mass remains unchanged, then the acceleration due to gravity on the surface of the earth is __________ g.

Correct answer: 9

Step-by-step solution →
Q23·PhysicsIntegerJEE Main 2025
A satellite of mass M2\dfrac{M}{2}2M​ is revolving around earth in a circular orbit at a height of R3\dfrac{R}{3}3R​ from earth surface. The angular momentum of the satellite is MGMRxM\sqrt{\dfrac{GMR}{x}}MxGMR​​. The value of xxx is __________, where MMM and RRR are the mass and radius of earth, respectively. (GGG is the gravitational constant)

Correct answer: 3

Step-by-step solution →
Q24·PhysicsSingle correctJEE Main 2025
If a satellite orbiting the Earth is 999 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=27=27 days and gravitational attraction between the satellite and the moon is neglected.
  1. (A)111 day
  2. (B)818181 days
  3. (C)272727 days
  4. (D)333 days

Correct answer: (A)

Step-by-step solution →
Q25·PhysicsSingle correctJEE Main 2025
A small point 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 (on the side nearer to m, as shown) and the force on m due to the remaining part of M is F₂. The ratio F₁ : F₂ is:
  1. (A)16 : 9
  2. (B)11 : 10
  3. (C)12 : 11
  4. (D)12 : 9

Correct answer: (C)

Step-by-step solution →
Q26·PhysicsSingle correctJEE Main 2025
Given below are two statements: one is labelled as Assertion (A) and the other is labelled as Reason (R). Assertion (A): A simple pendulum is taken to a planet of mass and radius, 4 times and 2 times, respectively, than the earth. The time period of the pendulum remains same on earth and the planet. Reason (R): The mass of the pendulum remains unchanged on earth and the other planet. In the light of the above statements, choose the correct answer from the options given below:
  1. (A)Both (A) and (R) are true but (R) is NOT the correct explanation of (A)
  2. (B)(A) is true but (R) is false
  3. (C)(A) is false but (R) is true
  4. (D)Both (A) and (R) are true and (R) is the correct explanation of (A)

Correct answer: (A)

Step-by-step solution →
Q27·PhysicsSingle correctJEE Advanced 2024
A particle of mass m is under the influence of the gravitational field of a body of mass M(≫m)M(\gg m)M(≫m) . The particle is moving in a circular orbit of radius r0r_0r0​ with time period T0T_0T0​ around the mass M . Then, the particle is subjected to an additional central force, corresponding to the potential energy Vc(r)=mα/r3V_c(r) = m\alpha/r^3Vc​(r)=mα/r3 , where α is a positive constant of suitable dimensions and r is the distance from the center of the orbit. If the particle moves in the same circular orbit of radius r0r_0r0​ in the combined gravitational potential due to M and Vc(r)V_c(r)Vc​(r) , but with a new time period T1T_1T1​ , then ( T12−T02T_1^2 - T_0^2T12​−T02​ / T12T_1^2T12​ is given by [G is the gravitational constant.]
  1. (A)3αGMr02\frac{3\alpha}{GMr_0^2}GMr02​3α​
  2. (B)α2GMr02\frac{\alpha}{2GMr_0^2}2GMr02​α​
  3. (C)αGMr02\frac{\alpha}{GMr_0^2}GMr02​α​
  4. (D)2αGMr02\frac{2\alpha}{GMr_0^2}GMr02​2α​

Correct answer: (A)

Step-by-step solution →
Q28·PhysicsSingle correctJEE Main 2024
An astronaut takes a ball of mass m from earth to space. He throws the ball into a circular orbit about earth at an altitude of 318.5 km. From earth's surface to the orbit, the change in total mechanical energy of the ball is xGMem21Rex\dfrac{GM_e m}{21 R_e}x21Re​GMe​m​. The value of x is (take Re=6370R_e = 6370Re​=6370 km):
  1. (A)11
  2. (B)9
  3. (C)12
  4. (D)10

Correct answer: (A)

Step-by-step solution →
Q29·PhysicsSingle correctJEE Main 2024
A satellite of 10310^{3}103 kg mass is revolving in a circular orbit of radius 2R2R2R. If 104R6\dfrac{10^{4}R}{6}6104R​ J energy is supplied to the satellite, it would revolve in a new circular orbit of radius: (use g=10g=10g=10 m/s2^{2}2, R=R=R= radius of earth)
  1. (A)2.5 R
  2. (B)3 R
  3. (C)4 R
  4. (D)6 R

Correct answer: (D)

Step-by-step solution →
Q30·PhysicsSingle correctJEE Main 2024
Two satellites A and B go round a planet in circular orbits having radii 4R and R respectively. If the speed of A is 3v, the speed of B will be:
  1. (A)43v\dfrac{4}{3}v34​v
  2. (B)3v3v3v
  3. (C)6v6v6v
  4. (D)12v12v12v

Correct answer: (C)

Step-by-step solution →
Q31·PhysicsSingle correctJEE Main 2024
Two planets AAA and BBB having masses m1m_1m1​ and m2m_2m2​ move around the sun in circular orbits of r1r_1r1​ and r2r_2r2​ radii respectively. If angular momentum of AAA is LLL and that of BBB is 3L3L3L, the ratio of time period (TATB)\left(\dfrac{T_A}{T_B}\right)(TB​TA​​) is:
  1. (A)(r1r2)3/2\left(\dfrac{r_1}{r_2}\right)^{3/2}(r2​r1​​)3/2
  2. (B)(r1r2)2\left(\dfrac{r_1}{r_2}\right)^2(r2​r1​​)2
  3. (C)127(m2m1)3\dfrac{1}{27}\left(\dfrac{m_2}{m_1}\right)^3271​(m1​m2​​)3
  4. (D)27(m2m1)327\left(\dfrac{m_2}{m_1}\right)^327(m1​m2​​)3

Correct answer: (C)

Step-by-step solution →
Q32·PhysicsSingle correctJEE Main 2024
Assuming the earth to be a sphere of uniform mass density, a body weighed 300 N on the surface of earth. How much it would weigh at R/4R/4R/4 depth under surface of earth ?
  1. (A)75 N
  2. (B)375 N
  3. (C)300 N
  4. (D)225 N

Correct answer: (D)

Step-by-step solution →
Q33·PhysicsSingle correctJEE Main 2024
To project a body of mass mmm from earth's surface to infinity, the required kinetic energy is (assume the radius of earth is RER_{E}RE​, g=g=g= acceleration due to gravity on the surface of earth):
  1. (A)2mgRE2mgR_{E}2mgRE​
  2. (B)mgREmgR_{E}mgRE​
  3. (C)12mgRE\tfrac{1}{2}mgR_{E}21​mgRE​
  4. (D)4mgRE4mgR_{E}4mgRE​

Correct answer: (B)

Step-by-step solution →
Q34·PhysicsSingle correctJEE Main 2024
Match List-I (energies of a planet of mass mmm in a circular orbit of radius aaa around the Sun of mass MMM; rrr = radius of the planet) with List-II (their expressions). Choose the correct answer from the options given below.
List-IList-II
A.Kinetic energy of planetI.GMma\dfrac{GMm}{a}aGMm​
B.Gravitational potential energy of Sun–planet systemII.GMm2a\dfrac{GMm}{2a}2aGMm​
C.Total mechanical energy of planetIII.−GMma-\dfrac{GMm}{a}−aGMm​
D.Escape energy at the surface of planet for a unit-mass objectIV.Gmr\dfrac{Gm}{r}rGm​
  1. (A)(A)–II, (B)–I, (C)–IV, (D)–III
  2. (B)(A)–III, (B)–IV, (C)–I, (D)–II
  3. (C)(A)–I, (B)–IV, (C)–II, (D)–III
  4. (D)(A)–I, (B)–II, (C)–III, (D)–IV

Correct answer: (A)

Step-by-step solution →
Q35·PhysicsSingle correctJEE Main 2024
A simple pendulum doing small oscillations at a place at height R above the earth’s surface has time period T1=4T_1 = 4T1​=4 s. T2T_2T2​ would be its time period if it is brought to a point at height 2R from the earth’s surface. Choose the correct relation [R = radius of Earth]:
  1. (A)T1=T2T_1 = T_2T1​=T2​
  2. (B)2T1=3T22T_1 = 3T_22T1​=3T2​
  3. (C)3T1=2T23T_1 = 2T_23T1​=2T2​
  4. (D)2T1=T22T_1 = T_22T1​=T2​

Correct answer: (C)

Step-by-step solution →
Q36·PhysicsSingle correctJEE Main 2024
A satellite revolving around a planet in a stationary orbit has a time period of 6 hours. The mass of the planet is one-fourth the mass of the earth. The radius of the orbit of the planet is: (Given: radius of the geo-stationary orbit for earth is 4.2×1044.2\times10^44.2×104 km)
  1. (A)1.4×1041.4\times10^41.4×104 km
  2. (B)8.4×1048.4\times10^48.4×104 km
  3. (C)1.68×1051.68\times10^51.68×105 km
  4. (D)1.05×1041.05\times10^41.05×104 km

Correct answer: (D)

Step-by-step solution →
Q37·PhysicsSingle correctJEE Main 2024
A metal wire of uniform mass density having length LLL and mass MMM is bent to form a semicircular arc and a particle of mass mmm is placed at the centre of the arc. The gravitational force on the particle by the wire is:
  1. (A)GMmπ2L2\dfrac{GMm\pi}{2L^2}2L2GMmπ​
  2. (B)000
  3. (C)GMmπL2\dfrac{GMm\pi}{L^2}L2GMmπ​
  4. (D)2GMmπL2\dfrac{2GMm\pi}{L^2}L22GMmπ​

Correct answer: (D)

Step-by-step solution →
Q38·PhysicsSingle correctJEE Main 2024
Correct formula for height of a satellite from earth's surface is
  1. (A)(T2R2g4π)1/2−R\left(\tfrac{T^2R^2g}{4\pi}\right)^{1/2}-R(4πT2R2g​)1/2−R
  2. (B)(T2R2g4π2)1/3−R\left(\tfrac{T^2R^2g}{4\pi^2}\right)^{1/3}-R(4π2T2R2g​)1/3−R
  3. (C)(T2R24π2g)1/3−R\left(\tfrac{T^2R^2}{4\pi^2 g}\right)^{1/3}-R(4π2gT2R2​)1/3−R
  4. (D)(T2R24π2)−1/3+R\left(\tfrac{T^2R^2}{4\pi^2}\right)^{-1/3}+R(4π2T2R2​)−1/3+R

Correct answer: (B)

Step-by-step solution →
Q39·PhysicsSingle correctJEE Main 2024
A 90 kg body placed at 2R2R2R distance from surface of earth experiences gravitational pull of (R = Radius of earth, g=10 ms−2g=10\ \text{ms}^{-2}g=10 ms−2)
  1. (A)300 N
  2. (B)225 N
  3. (C)450 N
  4. (D)100 N

Correct answer: (D)

Step-by-step solution →
Q40·PhysicsSingle correctJEE Main 2024
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−3/2R^{-3/2}R−3/2 then choose the correct option:
  1. (A)T2∝R5/2T^2\propto R^{5/2}T2∝R5/2
  2. (B)T2∝R7/2T^2\propto R^{7/2}T2∝R7/2
  3. (C)T2∝R3/2T^2\propto R^{3/2}T2∝R3/2
  4. (D)T2∝R3T^2\propto R^3T2∝R3

Correct answer: (A)

Step-by-step solution →
Q41·PhysicsSingle correctJEE Main 2024
If RRR is the radius of the earth and the acceleration due to gravity on the surface of earth is g=π2 m/s2g=\pi^2\,m/s^2g=π2m/s2, then the length of the second’s pendulum at a height h=2Rh=2Rh=2R from the surface of earth will be:
  1. (A)29 m\dfrac29\,m92​m
  2. (B)19 m\dfrac19\,m91​m
  3. (C)49 m\dfrac49\,m94​m
  4. (D)89 m\dfrac89\,m98​m

Correct answer: (B)

Step-by-step solution →
Q42·PhysicsSingle correctJEE Main 2024
The mass of the moon is 1144\dfrac{1}{144}1441​ times the mass of a planet and its diameter is 116\dfrac{1}{16}161​ times the diameter of a planet. If the escape velocity on the planet is vvv, the escape velocity on the moon will be:
  1. (A)v4\dfrac{v}{4}4v​
  2. (B)v3\dfrac{v}{3}3v​
  3. (C)v12\dfrac{v}{12}12v​
  4. (D)v6\dfrac{v}{6}6v​

Correct answer: (B)

Step-by-step solution →
Q43·PhysicsSingle correctJEE Main 2024
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\left(\dfrac{2\sqrt{2}+1}{32}\right)\dfrac{Gm^2}{L^2}(3222​+1​)L2Gm2​, the length of the sides of the square is :
  1. (A)L2\dfrac{L}{2}2L​
  2. (B)4L4L4L
  3. (C)3L3L3L
  4. (D)2L2L2L

Correct answer: (B)

Step-by-step solution →
Q44·PhysicsSingle correctJEE Main 2024
Escape velocity of a body from earth is 11.2 km/s. If the radius of a planet be one-third the radius of earth and mass be one-sixth that of earth, the escape velocity from the planet is:
  1. (A)11.2 km/s
  2. (B)8.4 km/s
  3. (C)4.2 km/s
  4. (D)7.9 km/s

Correct answer: (D)

Step-by-step solution →
Q45·PhysicsSingle correctJEE Main 2024
The gravitational potential at a point above the surface of earth is −5.12×107 J/kg-5.12\times10^7\,J/kg−5.12×107J/kg and the acceleration due to gravity at that point is 6.4 m/s26.4\,m/s^26.4m/s2. Assume that the mean radius of earth to be 6400 km6400\,km6400km. The height of this point above the earth’s surface is:
  1. (A)1600 km1600\,km1600km
  2. (B)540 km540\,km540km
  3. (C)1200 km1200\,km1200km
  4. (D)1000 km1000\,km1000km

Correct answer: (A)

Step-by-step solution →
Q46·PhysicsNumericalJEE Main 2024
A simple pendulum is placed at a place where its distance from the earth's surface is equal to the radius of the earth. If the length of the string is 4m, then the time period of small oscillations will be ______ s. [take g=π2g=\pi^2g=π2 m/s2^22]

Correct answer: 8

Step-by-step solution →
Q47·PhysicsSingle correctJEE Main 2024
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?
  1. (A)25
  2. (B)50
  3. (C)100
  4. (D)20

Correct answer: (A)

Step-by-step solution →
Q48·PhysicsSingle correctJEE Main 2024
At what distance above and below the surface of the earth a body will have same weight, (take radius of earth as R.)
  1. (A)5R−R\sqrt{5}R-R5​R−R
  2. (B)3R−R2\dfrac{\sqrt{3}R-R}{2}23​R−R​
  3. (C)R2\dfrac{R}{2}2R​
  4. (D)5R−R2\dfrac{\sqrt{5}R-R}{2}25​R−R​

Correct answer: (D)

Step-by-step solution →
Q49·PhysicsSingle correctJEE Main 2024
The acceleration due to gravity on the surface of earth is ggg. If the diameter of earth reduces to half of its original value and mass remains constant, then acceleration due to gravity on the surface of earth would be:
  1. (A)g/4g/4g/4
  2. (B)2g2g2g
  3. (C)g/2g/2g/2
  4. (D)4g4g4g

Correct answer: (D)

Step-by-step solution →
Q50·PhysicsSingle correctJEE Main 2024
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 options given below : (1) (A) is correct but (R) is not correct (2) Both (A) and (R) are correct and (R) is the correct explanation of (A) (3) (A) is not correct but (R) is correct (4) (A) is correct but (R) is not the correct explanation of (A)
  1. (A)(A) is correct but (R) is not correct
  2. (B)Both (A) and (R) are correct and (R) is the correct explanation of (A)
  3. (C)(A) is not correct but (R) is correct
  4. (D)(A) is correct but (R) is not the correct explanation of (A)

Correct answer: (B)

Step-by-step solution →
Q51·PhysicsSingle correctJEE Advanced 2023
Two satellites P and Q are moving in different circular orbits around the Earth (radius R). The heights of P and Q from the Earth surface are hPh_PhP​ and hQh_QhQ​, respectively, where hP=R/3h_P = R/3hP​=R/3. The accelerations of P and Q due to Earth's gravity are gPg_PgP​ and gQg_QgQ​, respectively. If gP/gQ=36/25g_P/g_Q = 36/25gP​/gQ​=36/25, what is the value of hQh_QhQ​?
  1. (A)3R/5
  2. (B)R/6
  3. (C)6R/5
  4. (D)5R/6

Correct answer: (A)

Step-by-step solution →
Q52·PhysicsSingle correctJEE Main 2023
A body is released from a height equal to the radius (R)(R)(R) of the earth. The velocity of the body when it strikes the surface of the earth will be : (Given ggg = acceleration due to gravity on the earth.)
  1. (A)gR\sqrt{gR}gR​
  2. (B)4gR\sqrt{4gR}4gR​
  3. (C)2gR\sqrt{2gR}2gR​
  4. (D)gR2\sqrt{\frac{gR}{2}}2gR​​

Correct answer: (A)

Step-by-step solution →
Q53·PhysicsSingle correctJEE Main 2023
Two identical particles each of mass 'm' go round a circle of radius aaa under the action of their mutual gravitational attraction. The angular speed of each particle will be :
  1. (A)Gm2a3\sqrt{\frac{Gm}{2a^{3}}}2a3Gm​​
  2. (B)Gm8a3\sqrt{\frac{Gm}{8a^{3}}}8a3Gm​​
  3. (C)Gm4a3\sqrt{\frac{Gm}{4a^{3}}}4a3Gm​​
  4. (D)Gma3\sqrt{\frac{Gm}{a^{3}}}a3Gm​​

Correct answer: (C)

Step-by-step solution →
Q54·PhysicsSingle correctJEE Main 2023
Given below are two statements: Statement I: For a planet, if the ratio of mass of the planet to its radius increases, the escape velocity from the planet also increases. Statement II: Escape velocity is independent of the radius of the planet. In the light of above statements, choose the most appropriate answer from the options given below
  1. (A)Both Statement I and Statement II are incorrect
  2. (B)Statement I is correct but Statement II is incorrect
  3. (C)Statement I is incorrect but Statement II is correct
  4. (D)Both Statement I and Statement II are correct

Correct answer: (B)

Step-by-step solution →
Q55·PhysicsSingle correctJEE Main 2023
Two planets A and B of radii R and 1.5 R have densities ρ\rhoρ and ρ/2\rho/2ρ/2 respectively. The ratio of acceleration due to gravity at the surface of B to A is:
  1. (A)2 : 3
  2. (B)2 : 1
  3. (C)3 : 4
  4. (D)4 : 3

Correct answer: (C)

Step-by-step solution →
Q56·PhysicsSingle correctJEE Main 2023
A planet having mass 9 Me9\,M_{e}9Me​ and radius 4 Re4\,R_{e}4Re​, where MeM_{e}Me​ and ReR_{e}Re​ are mass and radius of earth respectively, has escape velocity in km/skm/skm/s given by: (Given escape velocity on earth Ve=11.2×103 m/sV_{e}=11.2\times10^{3}\,m/sVe​=11.2×103m/s)
  1. (A)67.267.267.2
  2. (B)16.816.816.8
  3. (C)33.633.633.6
  4. (D)11.211.211.2

Correct answer: (B)

Step-by-step solution →
Q57·PhysicsSingle correctJEE Main 2023
Given below are two statements: one is labelled as Assertion A and the other is labelled as Reason R. Assertion A: If we move from poles to equator, the direction of acceleration due to gravity of earth always points downward. Reason R: Electric dipole consists of two equal and opposite charges. In the light of above statements, choose the correct answer from the options given below:
  1. (A)Both A and R are true and R is the correct explanation of A
  2. (B)A is true but R is false
  3. (C)Both A and R true but R is NOT the correct explanation of A
  4. (D)A is false but R is true

Correct answer: (A)

Step-by-step solution →
Q58·PhysicsSingle correctJEE Main 2023
Two satellites A and B move round the earth in the same orbit. The mass of A is twice the mass of B. The quantity which is same for the two satellites will be:
  1. (A)Potential energy
  2. (B)Total energy
  3. (C)Kinetic energy
  4. (D)Speed

Correct answer: (D)

Step-by-step solution →
Q59·PhysicsSingle correctJEE Main 2023
The ratio of escape velocity of a planet to the escape velocity of earth will be: Given: Mass of the planet is 16 times mass of the earth and 4 times the radius of earth.
  1. (A)4 : 1
  2. (B)2 : 1
  3. (C)1 : 2\sqrt22​
  4. (D)1 : 4

Correct answer: (B)

Step-by-step solution →
Q60·PhysicsSingle correctJEE Main 2023
If V is the gravitational potential due to a sphere of uniform density on its surface, then its value at the centre of the sphere will be:
  1. (A)3V2\dfrac{3V}{2}23V​
  2. (B)VVV
  3. (C)43V\dfrac{4}{3}V34​V
  4. (D)V2\dfrac{V}{2}2V​

Correct answer: (A)

Step-by-step solution →
Q61·PhysicsSingle correctJEE Main 2023
The radii of two planets 'A' and 'B' are 'R' and '4R' and their densities are ρ\rhoρ and ρ/3\rho/3ρ/3 respectively. The ratio of acceleration due to gravity at their surfaces (gA:gB)(g_A : g_B)(gA​:gB​) will be :
  1. (A)1:161 : 161:16
  2. (B)3:163 : 163:16
  3. (C)3:43 : 43:4
  4. (D)4:34 : 34:3

Correct answer: (C)

Step-by-step solution →
Q62·PhysicsSingle correctJEE Main 2023
A spaceship of mass 2×1042\times10^{4}2×104 kg is launched into a circular orbit close to the earth's surface. The additional velocity to be imparted to the spaceship in the orbit to overcome the gravitational pull will be (if g=10g=10g=10 m/s2^22 and radius of earth =6400=6400=6400 km)
  1. (A)11.2(2−1)11.2\left(\sqrt{2}-1\right)11.2(2​−1) km/s
  2. (B)7.9(2−1)7.9\left(\sqrt{2}-1\right)7.9(2​−1) km/s
  3. (C)8(2−1)8\left(\sqrt{2}-1\right)8(2​−1) km/s
  4. (D)4(2−1)4\left(\sqrt{2}-1\right)4(2​−1) km/s

Correct answer: (C)

Step-by-step solution →
Q63·PhysicsSingle correctJEE Main 2023
The time period of a satellite, revolving above the earth's surface at a height equal to R, will be (given g=π2g=\pi^2g=π2 m/s2^22, R = radius of earth):
  1. (A)4R\sqrt{4R}4R​
  2. (B)8R\sqrt{8R}8R​
  3. (C)32R\sqrt{32R}32R​
  4. (D)2R\sqrt{2R}2R​

Correct answer: (C)

Step-by-step solution →
Q64·PhysicsSingle correctJEE Main 2023
Given below are two statements: Statement I: Rotation of the earth shows effect on the value of acceleration due to gravity (g). Statement II: The effect of rotation of the earth on the value of 'g' at the equator is minimum and that at the pole is maximum. In the light of the above statements, choose the correct answer from the options given below:
  1. (A)Statement I is false but Statement II is true
  2. (B)Statement I is true but Statement II is false
  3. (C)Both Statement I and Statement II are true
  4. (D)Both Statement I and Statement II are false

Correct answer: (B)

Step-by-step solution →
Q65·PhysicsSingle correctJEE Main 2023
Assuming the earth to be a sphere of uniform mass density, the weight of a body at a depth R2\dfrac{R}{2}2R​ from the surface of earth, if its weight on the surface of earth is 200 N, will be: (Given R = Radius of earth)
  1. (A)400 N
  2. (B)500 N
  3. (C)300 N
  4. (D)100 N

Correct answer: (D)

Step-by-step solution →
Q66·PhysicsSingle correctJEE Main 2023
Two satellites of masses m and 3m revolve around the earth in circular orbits of radii r and 3r respectively. The ratio of orbital speeds of the satellites respectively is:
  1. (A)1 : 1
  2. (B)2 : 3
  3. (C)3:1\sqrt3:13​:1
  4. (D)9 : 1

Correct answer: (C)

Step-by-step solution →
Q67·PhysicsSingle correctJEE Main 2023
The orbital angular momentum of a satellite revolving in a circular orbit at a distance rrr from the centre of the earth is LLL. If the distance of the satellite from the centre of the earth is increased to nine times its initial value, then the new angular momentum will be
  1. (A)8L8L8L
  2. (B)4L4L4L
  3. (C)9L9L9L
  4. (D)3L3L3L

Correct answer: (D)

Step-by-step solution →
Q68·PhysicsSingle correctJEE Main 2023
Given below are two statements: Statement I: If E be the total energy of a satellite moving around the earth, then its potential energy will be E2\dfrac{E}{2}2E​. Statement II: The kinetic energy of a satellite revolving in an orbit is equal to the half the magnitude of total energy E. In the light of the above statements, choose the most appropriate answer from the options given below
  1. (A)Both Statement I and Statement II are correct
  2. (B)Both Statement I and Statement II are incorrect
  3. (C)Statement I is incorrect but Statement II is correct
  4. (D)Statement I is correct but Statement II is incorrect

Correct answer: (B)

Step-by-step solution →
Q69·PhysicsSingle correctJEE Main 2023
The acceleration due to gravity at height h above the earth, if h≪Rh\ll Rh≪R (radius of the earth), is given by
  1. (A)g′=g(1−2hR)g'=g\left(1-\dfrac{2h}{R}\right)g′=g(1−R2h​)
  2. (B)g′=g(1−h2R2)g'=g\left(1-\dfrac{h^{2}}{R^{2}}\right)g′=g(1−R2h2​)
  3. (C)g′=g(1−h2R)g'=g\left(1-\dfrac{h}{2R}\right)g′=g(1−2Rh​)
  4. (D)g′=g(1−h22R2)g'=g\left(1-\dfrac{h^{2}}{2R^{2}}\right)g′=g(1−2R2h2​)

Correct answer: (A)

Step-by-step solution →
Q70·PhysicsSingle correctJEE Main 2023
The weight of a body on the earth is 400400400 N. Then weight of the body when taken to a depth half of the radius of the earth will be:
  1. (A)Zero
  2. (B)300300300 N
  3. (C)100100100 N
  4. (D)200200200 N

Correct answer: (D)

Step-by-step solution →
Q71·PhysicsSingle correctJEE Main 2023
A planet has double the mass of the earth. Its average density is equal to that of the earth. An object weighing WWW on earth will weigh on that planet:
  1. (A)22/3 W2^{2/3}\,W22/3W
  2. (B)2 W2\,W2W
  3. (C)21/3 W2^{1/3}\,W21/3W
  4. (D)WWW

Correct answer: (C)

Step-by-step solution →
Q72·PhysicsSingle correctJEE Main 2023
Given below are two statements: one is labelled as Assertion A and the other is labelled as Reason R. Assertion A: Earth has atmosphere whereas moon doesn't have any atmosphere. Reason R: The escape velocity on moon is very small as compared to that on earth. In the light of the above statement, choose the correct answer from the options given below:
  1. (A)A is true but R is false
  2. (B)A is false but R is true
  3. (C)Both A and R are correct but R is NOT the correct explanation of A
  4. (D)Both A and R are correct and R is correct explanation of A

Correct answer: (D)

Step-by-step solution →
Q73·PhysicsSingle correctJEE Main 2023
Choose the incorrect statement from the following:
  1. (A)The speed of satellite in a given circular orbit remains constant.
  2. (B)For a planet revolving around the sun in an elliptical orbit, the total energy of the planet remains constant.
  3. (C)When a body falls towards earth, the displacement of earth towards the body is negligible.
  4. (D)The linear speed of a planet revolving around the sun remains constant.

Correct answer: (D)

Step-by-step solution →
Q74·PhysicsSingle correctJEE Main 2023
The weight of a body on the surface of the earth is 100 N. The gravitational force on it when taken at a height, from the surface of earth, equal to one-fourth the radius of the earth is:
  1. (A)100 N100\ N100 N
  2. (B)64 N64\ N64 N
  3. (C)50 N50\ N50 N
  4. (D)25 N25\ N25 N

Correct answer: (B)

Step-by-step solution →
Q75·PhysicsSingle correctJEE Main 2023
If earth has a mass nine times and radius twice to that of a planet PPP, then ve3x\tfrac{v_e}{3}\sqrt{x}3ve​​x​ ms−1^{-1}−1 will be the minimum velocity required by a rocket to pull out of gravitational force of PPP, where vev_eve​ is escape velocity on earth. The value of xxx is:
  1. (A)111
  2. (B)333
  3. (C)181818
  4. (D)222

Correct answer: (D)

Step-by-step solution →
Q76·PhysicsSingle correctJEE Main 2023
Given below are two statements: Statement I: Acceleration due to gravity is different at different places on the surface of earth. Statement II: Acceleration due to gravity increases as we go down below the earth's surface. In the light of the above statements, choose the correct answer from the options given below:
  1. (A)Statement I is false but Statement II is true
  2. (B)Statement I is true but Statement II is false
  3. (C)Both Statement I and Statement II are false
  4. (D)Both Statement I and Statement II are true

Correct answer: (B)

Step-by-step solution →
Q77·PhysicsSingle correctJEE Main 2023
The escape velocities of two planets A and B are in the ratio 1:21:21:2. If the ratio of their radii respectively is 1:31:31:3, then the ratio of acceleration due to gravity of planet A to the acceleration of gravity of planet B will be:
  1. (A)32\dfrac3223​
  2. (B)23\dfrac2332​
  3. (C)34\dfrac3443​
  4. (D)43\dfrac4334​

Correct answer: (C)

Step-by-step solution →
Q78·PhysicsSingle correctJEE Main 2023
A body weight W, is projected vertically upwards from earth's surface to reach a height above the earth which is equal to nine times the radius of earth. The weight of the body at that height will be:
  1. (A)w100\dfrac{w}{100}100w​
  2. (B)w91\dfrac{w}{91}91w​
  3. (C)w3\dfrac{w}{3}3w​
  4. (D)w9\dfrac{w}{9}9w​

Correct answer: (A)

Step-by-step solution →
Q79·PhysicsSingle correctJEE Main 2023
At a certain depth "ddd" below surface of earth, value of acceleration due to gravity becomes four times that of its value at a height 3R3R3R above earth surface. Where RRR is Radius of earth (Take R=6400R = 6400R=6400 km). The depth ddd is equal to
  1. (A)480048004800 km
  2. (B)256025602560 km
  3. (C)640640640 km
  4. (D)526052605260 km

Correct answer: (A)

Step-by-step solution →
Q80·PhysicsSingle correctJEE Main 2023
An object is allowed to fall from a height RRR above the earth, where RRR is the radius of the earth. Its velocity when it strikes the earth's surface, ignoring air resistance, will be:
  1. (A)2gR\sqrt{2gR}2gR​
  2. (B)gR2\sqrt{\dfrac{gR}{2}}2gR​​
  3. (C)2gR2\sqrt{gR}2gR​
  4. (D)gR\sqrt{gR}gR​

Correct answer: (D)

Step-by-step solution →
Q81·PhysicsSingle correctJEE Main 2023
If the gravitational field in the space is given as (−Kr2)\left(-\dfrac{K}{r^2}\right)(−r2K​). Taking the reference point to be at r=2r=2r=2 cm with gravitational potential V=10V=10V=10 J/kg. Find the gravitational potential at r=3r=3r=3 cm in SI unit (Given that K=6K=6K=6 J cm/kg):
  1. (A)999
  2. (B)101010
  3. (C)111111
  4. (D)121212

Correct answer: (C)

Step-by-step solution →
Q82·PhysicsSingle correctJEE Main 2023
Two particles of equal mass 'mmm' move in a circle of radius 'rrr' under the action of their mutual gravitational attraction. The speed of each particle will be:
  1. (A)4Gmr\sqrt{\dfrac{4Gm}{r}}r4Gm​​
  2. (B)Gm4r\sqrt{\dfrac{Gm}{4r}}4rGm​​
  3. (C)Gmr\sqrt{\dfrac{Gm}{r}}rGm​​
  4. (D)Gm2r\sqrt{\dfrac{Gm}{2r}}2rGm​​

Correct answer: (B)

Step-by-step solution →
Q83·PhysicsSingle correctJEE Main 2023
The time period of a satellite of earth is 242424 hours. If the separation between the earth and the satellite is decreased to one fourth of the previous value, then its new time period will become.
  1. (A)444 hours
  2. (B)666 hours
  3. (C)333 hours
  4. (D)121212 hours

Correct answer: (C)

Step-by-step solution →
Q84·PhysicsSingle correctJEE Main 2023
Assume that the earth is a solid sphere of uniform density and a tunnel is dug along its diameter throughout the earth. It is found that when a particle is released in this tunnel, it executes a simple harmonic motion. The mass of the particle is 100 g. The time period of the motion of the particle will be (approximately): (Take g=10g=10g=10 m s−2^{-2}−2, radius of earth = 6400 km)
  1. (A)121212 hours
  2. (B)111 hour 404040 minutes
  3. (C)242424 hours
  4. (D)111 hour 242424 minutes

Correct answer: (D)

Step-by-step solution →
Q85·PhysicsSingle correctJEE Main 2023
Every planet revolves around the sun in an elliptical orbit: A. The force acting on a planet is inversely proportional to square of distance from sun. B. Force acting on planet is inversely proportional to 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 time period of revolution of planet around sun is directly proportional to cube of semi-major axis of elliptical orbit. Choose the correct answer from the options given below:
  1. (A)B and C only
  2. (B)A and C Only
  3. (C)A and D only
  4. (D)C and D only

Correct answer: (C)

Step-by-step solution →
Q86·PhysicsSingle correctJEE Main 2023
A body of mass is taken from earth surface to the height hhh equal to twice the radius of earth (ReR_eRe​), the increase in potential energy will be: ( ggg = acceleration due to gravity on the surface of Earth)
  1. (A)3 mgRe3\,mgR_e3mgRe​
  2. (B)13mgRe\frac{1}{3}mgR_e31​mgRe​
  3. (C)23mgRe\frac{2}{3}mgR_e32​mgRe​
  4. (D)12mgRe\frac{1}{2}mgR_e21​mgRe​

Correct answer: (C)

Step-by-step solution →
Q87·PhysicsSingle correctJEE Main 2023
Given below are two statements: Statement I: Acceleration due to earth's gravity decreases as you go 'up' or 'down' from earth's surface. Statement II: Acceleration due to earth's gravity is same at a height hhh and depth ddd from earth's surface, if h=dh=dh=d. In the light of above statements, choose the most appropriate answer from the options given below
  1. (A)Both Statement I and Statement II are incorrect
  2. (B)Statement I is incorrect but statement II is correct
  3. (C)Both Statement I and Statement II are correct
  4. (D)Statement I is correct but statement II is incorrect

Correct answer: (D)

Step-by-step solution →
Q88·PhysicsSingle correctJEE Main 2023
If the distance of the earth from Sun is 1.5×1061.5\times10^61.5×106 km, then the distance of an imaginary planet from Sun, if its period of revolution is 2.832.832.83 years is:
  1. (A)6×1066\times10^66×106 km
  2. (B)3×1063\times10^63×106 km
  3. (C)3×1073\times10^73×107 km
  4. (D)6×1076\times10^76×107 km

Correct answer: (B)

Step-by-step solution →
Q89·PhysicsSingle correctJEE Main 2023
Given below are two statements: one is labelled as Assertion A and the other is labelled as Reason R. Assertion A: A pendulum clock when taken to Mount Everest becomes fast. Reason R: The value of ggg (acceleration due to gravity) is less at Mount Everest than on the surface of earth. In the light of the above statements, choose the most appropriate answer from the options given below
  1. (A)Both A and R are correct and R is the correct explanation of A
  2. (B)A is correct but R is not correct
  3. (C)Both A and R are correct but R is NOT the correct explanation of A
  4. (D)A is not correct but R is correct

Correct answer: (D)

Step-by-step solution →
Q90·PhysicsSingle correctJEE Main 2023
The weight of a body at the surface of earth is 18 N. The weight of the body at an altitude of 3200 km above the earth's surface is (given, radius of earth Re=6400R_e=6400Re​=6400 km):
  1. (A)8 N
  2. (B)4.9 N
  3. (C)9.8 N
  4. (D)19.6 N

Correct answer: (A)

Step-by-step solution →
Q91·PhysicsNumericalJEE Advanced 2022
Two spherical stars A and B have densities ρA\rho_AρA​ and ρB\rho_BρB​, respectively. A and B have the same radius, and their masses MAM_AMA​ and MBM_BMB​ are related by MB=2MAM_B = 2 M_AMB​=2MA​. Due to an interaction process, star A loses some of its mass, so that its radius is halved, while its spherical shape is retained, and its density remains ρA\rho_AρA​. The entire mass lost by A is deposited as a thick spherical shell on B with the density of the shell being ρA\rho_AρA​. If vAv_AvA​ and vBv_BvB​ are the escape velocities from A and B after the interaction process, the ratio vBvA=10n151/3\frac{v_B}{v_A} = \sqrt{\frac{10n}{15^{1/3}}}vA​vB​​=151/310n​​ . The value of n is

Correct answer: 2.30

Step-by-step solution →
Q92·PhysicsNumericalJEE Main 2022
If the acceleration due to gravity experienced by a point mass at a height h above the surface of earth is same as that of the acceleration due to gravity at a depth α\alphaαh (h << ReR_eRe​) from the earth surface. The value of α\alphaα will be __________. (use ReR_eRe​ = 6400 km)

Correct answer: 2

Step-by-step solution →
Q93·PhysicsSingle correctJEE Main 2022
An object of mass 1 kg is taken to a height from the surface of earth which is equal to three times the radius of earth. The gain in potential energy of the object will be [If, g=10ms−2^{-2}−2 and radius of earth = 6400 km]
  1. (A)48 MJ
  2. (B)24 MJ
  3. (C)36 MJ
  4. (D)12 MJ

Correct answer: (A)

Step-by-step solution →
Q94·PhysicsSingle correctJEE Main 2022
If the radius of earth shrinks by 2% while its mass remains same. The acceleration due to gravity on the earth's surface will approximately :
  1. (A)decrease by 2%
  2. (B)decrease by 4%
  3. (C)increase by 2%
  4. (D)increase by 4%

Correct answer: (D)

Step-by-step solution →
Q95·PhysicsSingle correctJEE Main 2022
Assume there are two identical simple pendulum Clocks-1 is placed on the earth and Clock-2 is placed on a space station located at a height h above the earth surface. Clock-1 and Clock-2 operate at time periods 4s and 6s respectively. Then the value of h is −-− (consider radius of earth RER_ERE​ = 6400 km and g on earth 10 m/s2m/s^2m/s2)
  1. (A)1200 km
  2. (B)1600 km
  3. (C)3200 km
  4. (D)4800 km

Correct answer: (C)

Step-by-step solution →
Q96·PhysicsSingle correctJEE Main 2022
A body of mass m is projected with velocity λve\lambda v_eλve​ in vertically upward direction from the surface of the earth into space. It is given that vev_eve​ is escape velocity and λ < 1. If air resistance is considered to the negligible, then the maximum height from the centre of earth, to which the body can go, will be (R : radius of earth)
  1. (A)R1+λ2\frac{R}{1 + \lambda^{2}}1+λ2R​
  2. (B)R1−λ2\frac{R}{1 - \lambda^{2}}1−λ2R​
  3. (C)R1−λ\frac{R}{1 - \lambda}1−λR​
  4. (D)λ2R1−λ2\frac{\lambda^{2}R}{1 - \lambda^{2}}1−λ2λ2R​

Correct answer: (B)

Step-by-step solution →
Q97·PhysicsSingle correctJEE Main 2022
Two satellites A and B having masses in the ratio 4:3 are revolving in circular orbits of radii 3r and 4 r respectively around the earth. The ratio of total mechanical energy of A to B is :
  1. (A)9 : 16
  2. (B)16 : 9
  3. (C)1 : 1
  4. (D)4 : 3

Correct answer: (B)

Step-by-step solution →
Q98·PhysicsSingle correctJEE Main 2022
The percentage decrease in the weight of a rocket, when taken to a height of 32 km above the surface of earth will, be : (Radius of earth = 6400km)
  1. (A)1 %
  2. (B)3%
  3. (C)4%
  4. (D)0.5%

Correct answer: (A)

Step-by-step solution →
Q99·PhysicsSingle correctJEE Main 2022
A body is projected vertically upwards from the surface of earth with a velocity equal to one third of escape velocity. The maximum height attained by the body will be: (Take radius of earth = 6400 km and g=10 ms−2\text{ms}^{-2}ms−2 )
  1. (A)800 km
  2. (B)1600 km
  3. (C)2133 km
  4. (D)4800 km

Correct answer: (A)

Step-by-step solution →
Q100·PhysicsSingle correctJEE Main 2022
Three identical particle A, B and C of mass 100 kg each are placed in a straight line with AB = BC = 13 m. The gravitational force on a fourth particle P of the same mass is F, when placed at a distance 13 m from the particle B on the perpendicular bisector of the line AC. The value of F will be approximately :
  1. (A)21 G
  2. (B)100 G
  3. (C)59 G
  4. (D)42 G

Correct answer: (B)

Step-by-step solution →
Q101·PhysicsSingle correctJEE Main 2022
The time period of a satellite revolving around earth in a given orbit is 7 hours. If the radius of orbit is increased to three times its previous value, then approximate new time period of the satellite will be :
  1. (A)40 hours
  2. (B)36 hours
  3. (C)30 hours
  4. (D)25 hours

Correct answer: (B)

Step-by-step solution →
Q102·PhysicsSingle correctJEE Main 2022
The escape velocity of a body on a planet 'A' is 12 kms−1kms^{-1}kms−1. The escape velocity of the body on another planet 'B', whose density is four times and radius is half of the planet 'A', is :
  1. (A)12 kms−1kms^{-1}kms−1
  2. (B)24 kms−1kms^{-1}kms−1
  3. (C)36 kms−1kms^{-1}kms−1
  4. (D)6 kms−1kms^{-1}kms−1

Correct answer: (A)

Step-by-step solution →
Q103·PhysicsSingle correctJEE Main 2022
Two objects of equal masses placed at certain distance from each other attracts each other with a force of F. If one-third mass of one object is transferred to the other object, then the new force will be :
  1. (A)29\frac{2}{9}92​F
  2. (B)169\frac{16}{9}916​F
  3. (C)89\frac{8}{9}98​F
  4. (D)F

Correct answer: (C)

Step-by-step solution →
Q104·PhysicsSingle correctJEE Main 2022
Four spheres each of mass m form a square of side d (as shown in figure). A fifth sphere of mass M is situated at the centre of square. The total gravitational potential energy of the system is :
  1. (A)−Gmd[(4+2)m+42M]-\dfrac{Gm}{d}\left[(4+\sqrt{2})m+4\sqrt{2}M\right]−dGm​[(4+2​)m+42​M]
  2. (B)−Gmd[(4+2)M+42m]-\dfrac{Gm}{d}\left[(4+\sqrt{2})M+4\sqrt{2}m\right]−dGm​[(4+2​)M+42​m]
  3. (C)−Gmd[3m2+42M]-\dfrac{Gm}{d}\left[3m^{2}+4\sqrt{2}M\right]−dGm​[3m2+42​M]
  4. (D)−Gmd[6m2+42M]-\dfrac{Gm}{d}\left[6m^{2}+4\sqrt{2}M\right]−dGm​[6m2+42​M]

Correct answer: (A)

Step-by-step solution →
Q105·PhysicsSingle correctJEE Main 2022
Given below are two statements : Statement I : The law of gravitation holds good for any pair of bodies in the universe. Statement II : The weight of any person becomes zero when the person is at the centre of the earth. In the light of the above statements, choose the correct answer from the options given below.
  1. (A)Both statement I and Statement II are true
  2. (B)Both statement I and Statement II are false
  3. (C)Statement I is true but Statement II are false
  4. (D)Statement I is false but Statement II is true

Correct answer: (A)

Step-by-step solution →
Q106·PhysicsSingle correctJEE Main 2022
The variation of acceleration due to gravity (g) with distance (r) from the center of the earth is correctly represented by : (Given R = radius of earth)
  1. (A)(A)
  2. (B)(B)
  3. (C)(C)
  4. (D)(D)

Correct answer: (A)

Step-by-step solution →
Q107·PhysicsSingle correctJEE Main 2022
Given below are two statements : One is labelled as Assertion A and the other is labelled as Reason R. Assertion A : If we move from poles to equator, the direction of acceleration due to gravity of earth always points towards the center of earth without any variation in its magnitude. Reason R : At equator, the direction of acceleration due to the gravity is towards the center of earth. In the light of above statements, choose the correct answer from the options given below :
  1. (A)Both A and R are true and R is the correct explanation of A.
  2. (B)Both A and R are true but R is NOT the correct explanation of A.
  3. (C)A is true but R is false
  4. (D)A is false but R is true

Correct answer: (D)

Step-by-step solution →
Q108·PhysicsSingle correctJEE Main 2022
The height of any point P above the surface of earth is equal to diameter of earth. The value of acceleration due to gravity at point P will be : (Given g = acceleration due to gravity at the surface of earth)
  1. (A)g/2
  2. (B)g/4
  3. (C)g/3
  4. (D)g/9

Correct answer: (D)

Step-by-step solution →
Q109·PhysicsNumericalJEE Main 2022
Two satellites S1S_1S1​ and S2S_2S2​ are revolving in circular orbits around a planet with radius R1R_1R1​ = 3200 km and R2R_2R2​ = 800 km respectively. The ratio of speed of satellite S1S_1S1​ to the speed of satellite S2S_2S2​ in their respective orbits would be 1x\frac{1}{x}x1​ where x =

Correct answer: 2

Step-by-step solution →
Q110·PhysicsSingle correctJEE Main 2022
The distance between Sun and Earth is R. The duration of year if the distance between Sun and Earth becomes 3R will be :
  1. (A)3\sqrt{3}3​ years
  2. (B)3 years
  3. (C)9 years
  4. (D)333\sqrt{3}33​ years

Correct answer: (D)

Step-by-step solution →
Q111·PhysicsSingle correctJEE Main 2022
The approximate height from the surface of earth at which the weight of the body becomes 13\dfrac{1}{3}31​ of its weight on the surface of earth is : [Radius of earth R = 6400 km and 3=1.732\sqrt{3} = 1.7323​=1.732]
  1. (A)3840 km
  2. (B)4685 km
  3. (C)2133 km
  4. (D)4267 km

Correct answer: (B)

Step-by-step solution →
Q112·PhysicsIntegerJEE Advanced 2021
The distance between two stars of masses 3MS3M_S3MS​ and 6MS6M_S6MS​ is 9R9R9R. Here RRR is the mean distance between the centers of the Earth and the Sun, and MSM_SMS​ is the mass of the Sun. The two stars orbit around their common center of mass in circular orbits with period nTnTnT, where TTT is the period of Earth's revolution around the Sun. The value of nnn is ___.

Correct answer: 9

Step-by-step solution →
Q113·PhysicsNumericalJEE Main 2021
Two satellites revolve around a planet in coplanar circular orbits in anticlockwise direction. Their period of revolutions are 1 hour and 8 hours respectively. The radius of the orbit of nearer satellite is 2 × 103^{3}3 km. The angular speed of the farther satellite as observed from the nearer satellite at the instant when both the satellites are closest is πx\frac{\pi}{x}xπ​ rad h−1^{-1}−1 where x is ..........

Correct answer: 3

Step-by-step solution →
Q114·PhysicsSingle correctJEE Main 2021
Four particles each of mass M, move along a circle of radius R under the action of their mutual gravitational attraction as shown in figure. The speed of each particle is :
  1. (A)12GMR(22+1)\frac{1}{2}\sqrt{\frac{GM}{R(2\sqrt{2}+1)}}21​R(22​+1)GM​​
  2. (B)12GMR(22+1)\frac{1}{2}\sqrt{\frac{GM}{R}(2\sqrt{2}+1)}21​RGM​(22​+1)​
  3. (C)12GMR(22−1)\frac{1}{2}\sqrt{\frac{GM}{R}(2\sqrt{2}-1)}21​RGM​(22​−1)​
  4. (D)GMR\sqrt{\frac{GM}{R}}RGM​​

Correct answer: (B)

Step-by-step solution →
Q115·PhysicsSingle correctJEE Main 2021
The masses and radii of the earth and moon are (M1M_{1}M1​, R1R_{1}R1​) and (M2M_{2}M2​, R2R_{2}R2​) respectively. Their centres are at a distance 'r' apart. Find the minimum escape velocity for a particle of mass 'm' to be projected from the middle of these two masses:
  1. (A)V=124G(M1+M2)rV=\frac{1}{2}\sqrt{\frac{4G(M_{1}+M_{2})}{r}}V=21​r4G(M1​+M2​)​​
  2. (B)V=4G(M1+M2)rV=\sqrt{\frac{4G(M_{1}+M_{2})}{r}}V=r4G(M1​+M2​)​​
  3. (C)V=122G(M1+M2)rV=\frac{1}{2}\sqrt{\frac{2G(M_{1}+M_{2})}{r}}V=21​r2G(M1​+M2​)​​
  4. (D)V=2G (M1+M2)rV=\frac{\sqrt{2G}\,(M_{1}+M_{2})}{r}V=r2G​(M1​+M2​)​

Correct answer: (B)

Step-by-step solution →
Q116·PhysicsSingle correctJEE Main 2021
If RER_ERE​ be the radius of Earth, then the ratio between the acceleration due to gravity at a depth 'r' below and a height 'r' above the earth surface is : (Given : r<REr < R_Er<RE​)
  1. (A)1−rRE−r2RE2−r3RE31 - \frac{r}{R_E} - \frac{r^2}{R_E^2} - \frac{r^3}{R_E^3}1−RE​r​−RE2​r2​−RE3​r3​
  2. (B)1+rRE+r2RE2+r3RE31 + \frac{r}{R_E} + \frac{r^2}{R_E^2} + \frac{r^3}{R_E^3}1+RE​r​+RE2​r2​+RE3​r3​
  3. (C)1+rRE−r2RE2+r3RE31 + \frac{r}{R_E} - \frac{r^2}{R_E^2} + \frac{r^3}{R_E^3}1+RE​r​−RE2​r2​+RE3​r3​
  4. (D)1+rRE−r2RE2−r3RE31 + \frac{r}{R_E} - \frac{r^2}{R_E^2} - \frac{r^3}{R_E^3}1+RE​r​−RE2​r2​−RE3​r3​

Correct answer: (D)

Step-by-step solution →
Q117·PhysicsNumericalJEE Main 2021
A body of mass (2M) splits into four masses {m,M−m,m,M−m}\{m, M - m, m, M - m\}{m,M−m,m,M−m}, which are rearranged to form a square as shown in the figure. The ratio of Mm\frac{M}{m}mM​ for which, the gravitational potential energy of the system becomes maximum is x : 1. The value of x is ....... .

Correct answer: 2

Step-by-step solution →
Q118·PhysicsSingle correctJEE Main 2021
A mass of 50 kg is placed at the centre of a uniform spherical shell of mass 100 kg and radius 50 m. If the gravitational potential at a point, 25 m from the centre is V kg/m. The value of V is :
  1. (A)− 60 G
  2. (B)+ 2 G
  3. (C)− 20 G
  4. (D)− 4 G

Correct answer: (D)

Step-by-step solution →
Q119·PhysicsSingle correctJEE Main 2021
Inside a uniform spherical shell : (a) the gravitational field is zero (b) the gravitational potential is zero (c) the gravitational field is same everywhere (d) the gravitation potential is same everywhere (e) all of the above Choose the most appropriate answer from the options given below :
  1. (A)(a), (c) and (d) only
  2. (B)(e) only
  3. (C)(a), (b) and (c) only
  4. (D)(b), (c) and (d) only

Correct answer: (A)

Step-by-step solution →
Q120·PhysicsSingle correctJEE Main 2021
Two identical particles of mass 1 kg each go round a circle of radius R, under the action of their mutual gravitational attraction. The angular speed of each particle is :
  1. (A)12GR3\frac{1}{2}\sqrt{\frac{G}{R^3}}21​R3G​​
  2. (B)G2R3\sqrt{\frac{G}{2R^3}}2R3G​​
  3. (C)12R1G\frac{1}{2R}\sqrt{\frac{1}{G}}2R1​G1​​
  4. (D)2GR3\sqrt{\frac{2G}{R^3}}R32G​​

Correct answer: (A)

Step-by-step solution →
Q121·PhysicsSingle correctJEE Main 2021
The planet Mars has two moons, if one of them has a period 7 hours, 30 minutes and an orbital radius of 9.0 × 103^33 km. Find the mass of Mars. {Given 4π2G\frac{4\pi^2}{G}G4π2​ = 6 × 1011^{11}11 N−2^{-2}−2 kg2^22}
  1. (A)3.25 × 1021^{21}21 kg
  2. (B)6.00 × 1023^{23}23 kg
  3. (C)5.96 × 1019^{19}19 kg
  4. (D)7.02 × 1025^{25}25 kg

Correct answer: (B)

Step-by-step solution →
Q122·PhysicsNumericalJEE Main 2021
Suppose two planets (spherical in shape) of radii R and 2R, but mass M and 9 M respectively have a centre to centre separation 8 R as shown in the figure. A satellite of mass 'm' is projected from the surface of the planet of mass 'M' directly towards the centre of the second planet. The minimum speed 'υ' required for the satellite to reach the surface of the second plant is a7GMR\sqrt{\frac{a}{7}\frac{GM}{R}}7a​RGM​​ then the value of 'a' is ____. [ Given : The two planets are fixed in their position ]

Correct answer: 4

Step-by-step solution →
Q123·PhysicsSingle correctJEE Main 2021
The minimum and maximum distances of a planet revolving around the Sun are x1x_1x1​ and x2x_2x2​. If the minimum speed of the planet on its trajectory is v0v_0v0​ then its maximum speed will be :
  1. (A)v0x12x22\frac{v_0 x_1^2}{x_2^2}x22​v0​x12​​
  2. (B)v0x22x12\frac{v_0 x_2^2}{x_1^2}x12​v0​x22​​
  3. (C)v0x2x1\frac{v_0 x_2}{x_1}x1​v0​x2​​
  4. (D)v0x1x2\frac{v_0 x_1}{x_2}x2​v0​x1​​

Correct answer: (C)

Step-by-step solution →
Q124·PhysicsSingle correctJEE Main 2021
Consider a planet in some solar system which has a mass double the mass of earth and density equal to the average density of earth. If the weight of an object on earth is W, the weight of the same object on that planet will be :
  1. (A)2 W
  2. (B)W
  3. (C)2\sqrt{2}2​ W
  4. (D)2132^{\frac{1}{3}}231​ W

Correct answer: (D)

Step-by-step solution →
Q125·PhysicsSingle correctJEE Main 2021
A body is projected vertically upwards from the surface of earth with a velocity sufficient enough to carry it to infinity. The time taken by it to earth reach height h is ________ s.
  1. (A)2Reg[(1+hRe)3/2−1]\sqrt{\frac{2R_e}{g}}\left[\left(1+\frac{h}{R_e}\right)^{3/2}-1\right]g2Re​​​[(1+Re​h​)3/2−1]
  2. (B)132Reg[(1+hRe)3/2−1]\frac{1}{3}\sqrt{\frac{2R_e}{g}}\left[\left(1+\frac{h}{R_e}\right)^{3/2}-1\right]31​g2Re​​​[(1+Re​h​)3/2−1]
  3. (C)13Re2g[(1+hRe)3/2−1]\frac{1}{3}\sqrt{\frac{R_e}{2g}}\left[\left(1+\frac{h}{R_e}\right)^{3/2}-1\right]31​2gRe​​​[(1+Re​h​)3/2−1]
  4. (D)Re2g[(1+hRe)3/2−1]\sqrt{\frac{R_e}{2g}}\left[\left(1+\frac{h}{R_e}\right)^{3/2}-1\right]2gRe​​​[(1+Re​h​)3/2−1]

Correct answer: (B)

Step-by-step solution →
Q126·PhysicsSingle correctJEE Main 2021
Consider a binary star system of star A and star B with masses mAm_AmA​ and mBm_BmB​ revolving in a circular orbit of radii rAr_ArA​ and rBr_BrB​, respectively. If TAT_ATA​ and TBT_BTB​ are the time period of star A and star B, respectively, then :
  1. (A)TA=TBT_A = T_BTA​=TB​
  2. (B)TA>TB(if mA>mB)T_A > T_B \left( \text{if } m_A > m_B \right)TA​>TB​(if mA​>mB​)
  3. (C)TATB=(rArB)32\dfrac{T_A}{T_B} = \left( \dfrac{r_A}{r_B} \right)^{\frac{3}{2}}TB​TA​​=(rB​rA​​)23​
  4. (D)TA>TB(if rA>rB)T_A > T_B \left( \text{if } r_A > r_B \right)TA​>TB​(if rA​>rB​)

Correct answer: (A)

Step-by-step solution →
Q127·PhysicsSingle correctJEE Main 2021
A person whose mass is 100 kg travels from Earth to Mars in a spaceship. Neglect all other object in sky and take acceleration due to gravity on the surface of the Earth and mars as 10m / s2^22 and 4m / s2^22 respectively. Identify from the below figures, the curve that fits best for the weight of the passenger as a function of time.
  1. (A)(c)
  2. (B)(a)
  3. (C)(b)
  4. (D)(d)

Correct answer: (A)

Step-by-step solution →
Q128·PhysicsSingle correctJEE Main 2021
A satellite is launched into a circular orbit of radius R around earth, while a second satellite is launched into a circular orbit of radius 1.02R. The percentage difference in the time periods of the two satellites is :
  1. (A)0.7
  2. (B)3.0
  3. (C)2.0
  4. (D)1.5

Correct answer: (B)

Step-by-step solution →
Q129·PhysicsSingle correctJEE Main 2021
The angular momentum of a planet of mass M moving around the sun in an elliptical orbit is L⃗\vec{L}L. The magnitude of the areal velocity of the planet is :
  1. (A)4LM\frac{4L}{M}M4L​
  2. (B)LM\frac{L}{M}ML​
  3. (C)2LM\frac{2L}{M}M2L​
  4. (D)L2M\frac{L}{2M}2ML​

Correct answer: (D)

Step-by-step solution →
Q130·PhysicsSingle correctJEE Main 2021
If the angular velocity of earth's spin is increased such that the bodies at the equator start floating, the duration of the day would be approximately : (Take : g = 10 ms−2^{-2}−2, the radius of earth, R = 6400 × 103^{3}3 m, Take π = 3.14)
  1. (A)60 minutes
  2. (B)does not change
  3. (C)1200 minutes
  4. (D)84 minutes

Correct answer: (D)

Step-by-step solution →
Q131·PhysicsSingle correctJEE Main 2021
A particle of mass m moves in a circular orbit under the central potential field, U(r)=−CrU(r)=\frac{-C}{r}U(r)=r−C​, where C is a positive constant. The correct radius – velocity graph of the particle's motion is :
  1. (A)(1)
  2. (B)(2)
  3. (C)(3)
  4. (D)(4)

Correct answer: (A)

Step-by-step solution →
Q132·PhysicsSingle correctJEE Main 2021
The time period of a satellite in a circular orbit of radius R is T. The period of another satellite in a circular orbit of radius 9R is :
  1. (A)9 T
  2. (B)27 T
  3. (C)12 T
  4. (D)3 T

Correct answer: (B)

Step-by-step solution →
Q133·PhysicsNumericalJEE Main 2021
The radius in kilometer to which the present radius of earth (R = 6400 km) to be compressed so that the escape velocity is increased 10 time is ________.

Correct answer: 64

Step-by-step solution →
Q134·PhysicsSingle correctJEE Main 2021
A geostationary satellite is orbiting around an arbitary planet 'P' at a height of 11R above the surface of 'P' , R being the radius of 'P'. The time period of another satellite in hours at a height of 2R from the surface of 'P' is__________.'P' has the time period of 24 hours.
  1. (A)626\sqrt{2}62​
  2. (B)62\frac{6}{\sqrt{2}}2​6​
  3. (C)3
  4. (D)5

Correct answer: (C)

Step-by-step solution →
Q135·PhysicsSingle correctJEE Main 2021
The maximum and minimum distances of a comet from the Sun are 1.6 × 1012^{12}12 m and 8.0 × 1010^{10}10 m respectively. If the speed of the comet at the nearest point is 6 × 104^{4}4 ms−1^{-1}−1, the speed at the farthest point is :
  1. (A)1.5 × 103^{3}3 m/s
  2. (B)6.0 × 103^{3}3 m/s
  3. (C)3.0 × 103^{3}3 m/s
  4. (D)4.5 × 103^{3}3 m/s

Correct answer: (C)

Step-by-step solution →
Q136·PhysicsNumericalJEE Main 2021
If one wants to remove all the mass of the earth to infinity in order to break it up completely. The amount of energy that needs to be supplied will be x5GM2R\frac{x}{5}\frac{GM^{2}}{R}5x​RGM2​ where x is ____ (Round off to the Nearest Integer) (M is the mass of earth, R is the radius of earth, G is the gravitational constant)

Correct answer: 3

Step-by-step solution →
Q137·PhysicsNumericalJEE Main 2021
In the reported figure of earth, the value of acceleration due to gravity is same at point A and C but it is smaller than that of its value at point B (surface of the earth). The value of OA : AB will be x : y. The value of x is ________________.

Correct answer: 4

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Q138·PhysicsSingle correctJEE Main 2021
A planet revolving in elliptical orbit has : A. a constant velocity of revolution. B. has the least velocity when it is nearest to the sun. C. its areal velocity is directly proportional to its velocity. D. areal velocity is inversely proportional to its velocity. E. to follow a trajectory such that the areal velocity is constant. Choose the correct answer from the options given below :
  1. (A)A only
  2. (B)E only
  3. (C)D only
  4. (D)C only

Correct answer: (B)

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Q139·PhysicsSingle correctJEE Main 2021
Assume that a tunnel is dug along a chord of the earth, at a perpendicular distance (R/2) from the earth's centre, where 'R' is the radius of the Earth. The wall of the tunnel is frictionless. If a particle is released in this tunnel, it will execute a simple harmonic motion with a time period :
  1. (A)2πRg\sqrt{\frac{R}{g}}gR​​
  2. (B)12πgR\frac{1}{2\pi}\sqrt{\frac{g}{R}}2π1​Rg​​
  3. (C)2πRg\frac{2\pi R}{g}g2πR​
  4. (D)g2πR\frac{g}{2\pi R}2πRg​

Correct answer: (A)

Step-by-step solution →
Q140·PhysicsSingle correctJEE Main 2021
Find the gravitational force of attraction between the ring and sphere as shown in the diagram, where the plane of the ring is perpendicular to the line joining the centres. If 8\sqrt{8}8​R is the distance between the centres of a ring (of mass 'm') and a sphere (mass 'M') where both have equal radius 'R'.
  1. (A)89⋅GmMR\frac{\sqrt{8}}{9}\cdot\frac{GmM}{R}98​​⋅RGmM​
  2. (B)827⋅GmMR2\frac{\sqrt{8}}{27}\cdot\frac{GmM}{R^{2}}278​​⋅R2GmM​
  3. (C)223⋅GMmR2\frac{2\sqrt{2}}{3}\cdot\frac{GMm}{R^{2}}322​​⋅R2GMm​
  4. (D)138⋅GMmR2\frac{1}{3\sqrt{8}}\cdot\frac{GMm}{R^{2}}38​1​⋅R2GMm​

Correct answer: (B)

Step-by-step solution →
Q141·PhysicsSingle correctJEE Main 2021
Given below are two statements : one is labelled as Assertion A and the other is labelled as reason R. Assertion A : The escape velocities of planet A and B are same. But A and B are of unequal mass. Reason R : The product of their mass and radius must be same. M1R1=M2R2M_1R_1 = M_2R_2M1​R1​=M2​R2​ In the light of the above statements, choose the most appropriate answer from the options given below:
  1. (A)Both A and R are correct but R is NOT the correct explanation of A
  2. (B)A is correct but R is not correct
  3. (C)Both A and R are correct and R is the correct explanation of A
  4. (D)A is not correct but R is correct

Correct answer: (B)

Step-by-step solution →
Q142·PhysicsSingle correctJEE Main 2021
A solid sphere of radius R gravitationally attracts a particle placed at 3R from its centre with a force F1F_1F1​. Now a spherical cavity of radius (R2)\left(\dfrac{R}{2}\right)(2R​) is made in the sphere (as shown in figure) and the force becomes F2F_2F2​. The value of F1:F2F_1 : F_2F1​:F2​ is :
  1. (A)41 : 50
  2. (B)36 : 25
  3. (C)50 : 41
  4. (D)25 : 36

Correct answer: (A)

Step-by-step solution →
Q143·PhysicsNumericalJEE Main 2021
The initial velocity υi required to project a body vertically upward from the surface of the earth to reach a height of 10R, where R is the radius of the earth, may be described in terms of x escape velocity υe such that υi = y ×υe_{e}e​. The value of x will be_______.

Correct answer: 10

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Q144·PhysicsSingle correctJEE Main 2021
Two satellites A and B of masses 200 kg and 400 kg are revolving round the earth at height of 600 km and 1600 km respectively. If TAT_ATA​ and TBT_BTB​ are the time periods of A and B respectively then the value of TB−TAT_B - T_ATB​−TA​ : [Given : radius of earth = 6400 km, mass of earth = 6×10246 \times 10^{24}6×1024 kg]
  1. (A)4.24×1024.24 \times 10^24.24×102 s
  2. (B)3.33×1023.33 \times 10^23.33×102 s
  3. (C)1.33×1031.33 \times 10^31.33×103 s
  4. (D)4.24×1034.24 \times 10^34.24×103 s

Correct answer: (C)

Step-by-step solution →
Q145·PhysicsSingle correctJEE Main 2021
Two stars of masses m and 2m at a distance d rotate about their common centre of mass in free space. The period of revolution is -
  1. (A)2πd33Gm2\pi\sqrt{\frac{d^3}{3Gm}}2π3Gmd3​​
  2. (B)12π3Gmd3\frac{1}{2\pi}\sqrt{\frac{3Gm}{d^3}}2π1​d33Gm​​
  3. (C)12πd33Gm\frac{1}{2\pi}\sqrt{\frac{d^3}{3Gm}}2π1​3Gmd3​​
  4. (D)2π3Gmd32\pi\sqrt{\frac{3Gm}{d^3}}2πd33Gm​​

Correct answer: (A)

Step-by-step solution →
Q146·PhysicsSingle correctJEE Main 2021
Four identical particles of equal masses 1 kg made to move along the circumference of a circle of radius 1 m under the action of their own mutual gravitational attraction. The speed of each particle will be -
  1. (A)(1+22)G2\frac{\sqrt{(1+2\sqrt{2})G}}{2}2(1+22​)G​​
  2. (B)G(1+22)\sqrt{G(1+2\sqrt{2})}G(1+22​)​
  3. (C)G2(22−1)\sqrt{\frac{G}{2}\left(2\sqrt{2}-1\right)}2G​(22​−1)​
  4. (D)G2(1+22)\sqrt{\frac{G}{2}\left(1+2\sqrt{2}\right)}2G​(1+22​)​

Correct answer: (A)

Step-by-step solution →
Q147·PhysicsSingle correctJEE Main 2021
Consider two satellites S1S_1S1​ and S2S_2S2​ with periods of revolution 1 hr. and 8 hr. respectively revolving around a planet in circular orbits.The ratio of angular velocity of satellite S1S_1S1​ to the angular velocity of satellite S2S_2S2​ is -
  1. (A)8 : 1
  2. (B)1 : 8
  3. (C)2 : 1
  4. (D)1 : 4

Correct answer: (A)

Step-by-step solution →
Q148·PhysicsSingle correctJEE Main 2021
A body weighs 49 N on a spring balance at the north pole. What will be its weight recorded on the same weighing machine, if it is shifted to the equator? [Use g=GMR2=\frac{GM}{R^2}=R2GM​ = 9.8 ms−2ms^{-2}ms−2 and radius of earth, R = 6400 km.]
  1. (A)49 N
  2. (B)49.83 N
  3. (C)49.17 N
  4. (D)48.83 N

Correct answer: (D)

Step-by-step solution →
Q149·PhysicsSingle correctJEE Main 2020
Two planets have masses M and 16 M and their radii are a and 2a, respectively. The separation between the centres of the planets is 10a. A body of mass m is fired from the surface of the larger planet towards the smaller planet along the line joining their centres. For the body to be able to reach at the surface of smaller planet, the minimum firings speed needed is:
  1. (A)2GMa2\sqrt{\frac{GM}{a}}2aGM​​
  2. (B)4GMa4\sqrt{\frac{GM}{a}}4aGM​​
  3. (C)GM2ma\sqrt{\frac{GM^{2}}{ma}}maGM2​​
  4. (D)325GMa\frac{3}{2}\sqrt{\frac{5GM}{a}}23​a5GM​​

Correct answer: (D)

Step-by-step solution →
Q150·PhysicsSingle correctJEE Main 2020
A satellite is in an elliptical orbit around a planet P. It is observed that the velocity of the satellite when it is farthest from the planet is 6 times less than that when it is closest to the planet. The ratio of distances between the satellite and the planet at closest and farthest points is:
  1. (A)1 : 6
  2. (B)1 : 3
  3. (C)1 : 2
  4. (D)3 : 4

Correct answer: (A)

Step-by-step solution →
Q151·PhysicsSingle correctJEE Main 2020
The value of the acceleration due to gravity is g1g_1g1​ at a height h=R2h=\dfrac{R}{2}h=2R​ (R= radius of the earth) from the surface of the earth. It is again equal to g1g_1g1​ at a depth d below the surface of the earth. The ratio (dR)\left(\dfrac{d}{R}\right)(Rd​) equals:
  1. (A)79\frac{7}{9}97​
  2. (B)13\frac{1}{3}31​
  3. (C)59\frac{5}{9}95​
  4. (D)49\frac{4}{9}94​

Correct answer: (C)

Step-by-step solution →
Q152·PhysicsSingle correctJEE Main 2020
The acceleration due to gravity on the earth's surface at the poles is g an and angular velocity of the earth about the axis passing through the pole is ω\omegaω. An object is weight at the equator and at a height h above the poles by using a spring balance. If the weights are found to be same, then h is : (h<<R, where R is the radius of the earth)
  1. (A)R2ω22g\dfrac{R^{2}\omega^{2}}{2g}2gR2ω2​
  2. (B)R2ω2g\dfrac{R^{2}\omega^{2}}{g}gR2ω2​
  3. (C)R2ω28g\dfrac{R^{2}\omega^{2}}{8g}8gR2ω2​
  4. (D)R2ω24g\dfrac{R^{2}\omega^{2}}{4g}4gR2ω2​

Correct answer: (A)

Step-by-step solution →
Q153·PhysicsSingle correctJEE Main 2020
A body is moving in a low circular orbit about a planet of mass M and radius R. The radius of the orbit can be taken to be R itself. Then the ratio of the speed of this body in the orbit to the escape velocity from the planet is :
  1. (A)2
  2. (B)1
  3. (C)2\sqrt{2}2​
  4. (D)12\frac{1}{\sqrt{2}}2​1​

Correct answer: (D)

Step-by-step solution →
Q154·PhysicsSingle correctJEE Main 2020
On the x-axis and at a distance x from the origin, the gravitational field due to a mass distribution is given by Ax(x2+a2)3/2\frac{Ax}{\left(x^{2}+a^{2}\right)^{3/2}}(x2+a2)3/2Ax​ in the x-direction. The magnitude of gravitational potential on the x-axis at a distance x, taking its value to be zero at infinity is:
  1. (A)A(x2+a2)1/2A\left(x^{2}+a^{2}\right)^{1/2}A(x2+a2)1/2
  2. (B)A(x2+a2)3/2\frac{A}{\left(x^{2}+a^{2}\right)^{3/2}}(x2+a2)3/2A​
  3. (C)A(x2+a2)3/2A\left(x^{2}+a^{2}\right)^{3/2}A(x2+a2)3/2
  4. (D)A(x2+a2)1/2\frac{A}{\left(x^{2}+a^{2}\right)^{1/2}}(x2+a2)1/2A​

Correct answer: (D)

Step-by-step solution →
Q155·PhysicsSingle correctJEE Main 2020
A satellite is moving in a low nearly circular orbit around the earth. Its radius is roughly equal to that of the earth's radius ReR_eRe​. By firing rockets attached to it, its speed is instantaneously increased in the direction of its motion so that it become 32\sqrt{\dfrac{3}{2}}23​​ times larger. Due to this the farthest distance from the centre of the earth that the satellite reaches is R. Value of R is:
  1. (A)2.5Re2.5R_e2.5Re​
  2. (B)3Re3R_e3Re​
  3. (C)2Re2R_e2Re​
  4. (D)4Re4R_e4Re​

Correct answer: (B)

Step-by-step solution →
Q156·PhysicsSingle correctJEE Main 2020
The mass density of a planet of radius R varies with the distance r from its centre as ρ(r)=ρ0(1−r2R2)\rho\left(r\right)=\rho_{0}\left(1-\frac{r^{2}}{R^{2}}\right)ρ(r)=ρ0​(1−R2r2​). Then the gravitational field is maximum at:
  1. (A)r=13Rr=\frac{1}{\sqrt{3}}Rr=3​1​R
  2. (B)r=Rr=Rr=R
  3. (C)r=59Rr=\sqrt{\frac{5}{9}}Rr=95​​R
  4. (D)r=34Rr=\sqrt{\frac{3}{4}}Rr=43​​R

Correct answer: (C)

Step-by-step solution →
Q157·PhysicsSingle correctJEE Main 2020
Planet A has mass M and radius R. Plant B has half the mass and half the radius of Planet A. If the escape velocities from the planets A and B are vAv_{A}vA​ and vBv_{B}vB​, respectively, then vAvB=n4\frac{v_{A}}{v_{B}} = \frac{n}{4}vB​vA​​=4n​. The value of ‘n’ is:
  1. (A)2
  2. (B)1
  3. (C)4
  4. (D)3

Correct answer: (C)

Step-by-step solution →
Q158·PhysicsSingle correctJEE Main 2020
A body A of mass m is moving in a circular orbit of radius R about a planet. Another body B of mass m2\dfrac{m}{2}2m​ collides with A with a velocity which is half (v⃗2)\left(\dfrac{\vec{v}}{2}\right)(2v​) the instantaneous velocity v⃗\vec{v}v of A. The collision is completely inelastic. Then, the combined body:
  1. (A)starts moving in an elliptical orbit around the planet.
  2. (B)continues to move in a circular orbit.
  3. (C)Escapes from the Planet's Gravitational field.
  4. (D)Falls vertically downwards towards the planet.

Correct answer: (A)

Step-by-step solution →
Q159·PhysicsSingle correctJEE Main 2020
Consider two solid spheres of radii R1=1R_{1} = 1R1​=1m, R2=2R_{2} = 2R2​=2m and masses M1M_{1}M1​ and M2M_{2}M2​, respectively. The gravitational field due to sphere (1) and (2) are shown. The value of M1M2\dfrac{M_{1}}{M_{2}}M2​M1​​ is:
  1. (A)12\dfrac{1}{2}21​
  2. (B)13\dfrac{1}{3}31​
  3. (C)16\dfrac{1}{6}61​
  4. (D)23\dfrac{2}{3}32​

Correct answer: (C)

Step-by-step solution →
Q160·PhysicsNumericalJEE Main 2020
An asteroid is moving directly towards the centre of the earth. When at a distance of 10 R (R is the radius of the earth) from he earths centre, it has a speed of 12 km/s. Neglecting the effect of earths atmosphere, what will be the speed of the asteroid when it hits the surface of the earth (escape velocity form the earth is 11.2 km/s)? Given your answer to the nearest integer in kilometer/s ____.

Correct answer: 16

Step-by-step solution →
Q161·PhysicsSingle correctJEE Main 2020
A box weighs 196 N on a spring balance at the north pole. Its weight recorded on the same balance if it is shifted to the equator is close to (Take g = 10 ms−2^{-2}−2 at the north pole and the radius of the earth = 6400 km)
  1. (A)194.66 N
  2. (B)194.32 N
  3. (C)195.32 N
  4. (D)195.66 N

Correct answer: (C)

Step-by-step solution →
Q162·PhysicsSingle correctJEE Main 2020
A satellite of mass m is launched vertically upwards with an initial speed u from the surface of the earth. After it reaches height R (R = radius of the earth), it ejects a rocket of mass m10\frac{m}{10}10m​ so that subsequently the satellite moves in a circular orbit. The kinetic energy of the rocket is (G is the gravitational constant ; M is the mass of the earth)
  1. (A)m20(u−2GM3R)2\frac{m}{20}\left(u-\sqrt{\frac{2GM}{3R}}\right)^{2}20m​(u−3R2GM​​)2
  2. (B)3m8(u+5GM6R)2\frac{3m}{8}\left(u+\sqrt{\frac{5GM}{6R}}\right)^{2}83m​(u+6R5GM​​)2
  3. (C)m20(u2−113200GMR)\frac{m}{20}\left(u^{2}-\frac{113}{200}\frac{GM}{R}\right)20m​(u2−200113​RGM​)
  4. (D)5m(u2−119200GMR)5m\left(u^{2}-\frac{119}{200}\frac{GM}{R}\right)5m(u2−200119​RGM​)

Correct answer: (D)

Step-by-step solution →
Q163·PhysicsSingle correctJEE Advanced 2019
Consider a spherical gaseous cloud of mass density ρ(r) in free space where r is the radial distance from its center. The gaseous cloud is made of particles of equal mass m moving in circular orbits about the common center with the same kinetic energy K. The force acting on the particles is their mutual gravitational force. If ρ(r) is constant in time, the particle number density n(r) = ρ(r)/m is [ G is universal gravitational constant]
  1. (A)3Kπr2m2G\frac{3K}{\pi r^2 m^2 G}πr2m2G3K​
  2. (B)K2πr2m2G\frac{K}{2\pi r^2 m^2 G}2πr2m2GK​
  3. (C)K6πr2m2G\frac{K}{6\pi r^2 m^2 G}6πr2m2GK​
  4. (D)Kπr2m2G\frac{K}{\pi r^2 m^2 G}πr2m2GK​

Correct answer: (B)

Step-by-step solution →
Q164·PhysicsSingle correctJEE Main 2019
The ratio of the weights of a body on the Earth's surface to that on the surface of a planet is 9 : 4. The mass of the planet is 19\dfrac{1}{9}91​th of that of the Earth. If 'R' is the radius of the Earth, what is the radius of the planet ? (Take the planets to have the same mass density)
  1. (A)R3\dfrac{R}{3}3R​
  2. (B)R4\dfrac{R}{4}4R​
  3. (C)R9\dfrac{R}{9}9R​
  4. (D)R2\dfrac{R}{2}2R​

Correct answer: (D)

Step-by-step solution →
Q165·PhysicsSingle correctJEE Main 2019
The value of acceleration due to gravity at Earth's surface is 9.8 ms−2^{-2}−2. The altitude above its surface at which the acceleration due to gravity decreases to 4.9 ms−2^{-2}−2, is close to: (Radius of earth = 6.4 ×\times× 106^{6}6 m)
  1. (A)6.4 ×\times× 106^{6}6 m
  2. (B)9.0 ×\times× 106^{6}6 m
  3. (C)2.6 ×\times× 106^{6}6 m
  4. (D)1.6 ×\times× 106^{6}6 m

Correct answer: (C)

Step-by-step solution →
Q166·PhysicsSingle correctJEE Main 2019
A spaceship orbits around a planet at a height of 20 km from its surface. Assuming that only gravitational field of the plant acts on the spaceship. What will be the number of complete revolutions made by the spaceship in 24 hours around the plane? [Given: Mass of plane = 8×10228 \times 10^{22}8×1022 kg, Radius of planet = 2×1062 \times 10^{6}2×106 m, Gravitational constant G = 6.67×10−116.67 \times 10^{-11}6.67×10−11 Mn2^{2}2/kg2^{2}2]
  1. (A)9
  2. (B)11
  3. (C)13
  4. (D)17

Correct answer: (B)

Step-by-step solution →
Q167·PhysicsSingle correctJEE Main 2019
A solid sphere of mass 'M' and radius 'a' is surrounded by a uniform concentric spherical shell of thickness 2a and mass 2M. The gravitational field at distance '3a' from the centre will be:
  1. (A)2GM3a2\dfrac{2GM}{3a^2}3a22GM​
  2. (B)2GM9a2\dfrac{2GM}{9a^2}9a22GM​
  3. (C)GM9a2\dfrac{GM}{9a^2}9a2GM​
  4. (D)GM3a2\dfrac{GM}{3a^2}3a2GM​

Correct answer: (D)

Step-by-step solution →
Q168·PhysicsSingle correctJEE Main 2019
A test particle is moving in a circular orbit in the gravitational field produced by a mass density ρ(r)=Kr2\rho(r) = \dfrac{K}{r^{2}}ρ(r)=r2K​. Identify the correct relation between the radius R of the particle's orbit and its period T:
  1. (A)T/R2^{2}2 is a constant
  2. (B)TR is constant
  3. (C)T2^{2}2/R3^{3}3 is a constant
  4. (D)T/R is a constant

Correct answer: (D)

Step-by-step solution →
Q169·PhysicsSingle correctJEE Main 2019
A rocket has to be launched from each in such a way that it never returns. If E is the minimum energy delivered by the rocket launcher, what should be the minimum energy that the launcher should have if the same rocket is to be launched from the surface of the moon? Assume that the density of the earth and the moon are equal and that the earth's volume is 64 times the volume of the moon.
  1. (A)E32\dfrac{E}{32}32E​
  2. (B)E16\dfrac{E}{16}16E​
  3. (C)E64\dfrac{E}{64}64E​
  4. (D)E4\dfrac{E}{4}4E​

Correct answer: (B)

Step-by-step solution →
Q170·PhysicsSingle correctJEE Main 2019
Four identical particles of mass M are located at the corners of a square of side 'a'. What should be their speed if each of them revolves under the influence of other's gravitational field in a circular orbit circumscribing the square?
  1. (A)1.35GMa1.35\sqrt{\dfrac{GM}{a}}1.35aGM​​
  2. (B)1.16GMa1.16\sqrt{\dfrac{GM}{a}}1.16aGM​​
  3. (C)1.41GMa1.41\sqrt{\dfrac{GM}{a}}1.41aGM​​
  4. (D)1.21GMa1.21\sqrt{\dfrac{GM}{a}}1.21aGM​​

Correct answer: (B)

Step-by-step solution →
Q171·PhysicsSingle correctJEE Main 2019
Two satellites, A and B, have masses m and 2m respectively. A is in a circular orbit of radius R, and B is in a circular orbit of radius 2R around the earth. The ratio of their kinetic energies, TA_AA​/TB_BB​, is :
  1. (A)12\frac{1}{2}21​
  2. (B)1
  3. (C)2
  4. (D)12\sqrt{\frac{1}{2}}21​​

Correct answer: (B)

Step-by-step solution →
Q172·PhysicsSingle correctJEE Main 2019
A straight rod of length L extends from x = a to x = L + a. The gravitational force it exerts on a point mass 'm' at x = 0, if the mass per unit length of the rod is A + Bx2^{2}2, is given by:
  1. (A)Gm[A(1a+L−1a)−BL]Gm\left[A\left(\frac{1}{a+L} - \frac{1}{a}\right) - BL\right]Gm[A(a+L1​−a1​)−BL]
  2. (B)Gm[A(1a−1a+L)−BL]Gm\left[A\left(\frac{1}{a} - \frac{1}{a+L}\right) - BL\right]Gm[A(a1​−a+L1​)−BL]
  3. (C)Gm[A(1a+L−1a)+BL]Gm\left[A\left(\frac{1}{a+L} - \frac{1}{a}\right) + BL\right]Gm[A(a+L1​−a1​)+BL]
  4. (D)Gm[A(1a−1a+L)+BL]Gm\left[A\left(\frac{1}{a} - \frac{1}{a+L}\right) + BL\right]Gm[A(a1​−a+L1​)+BL]

Correct answer: (D)

Step-by-step solution →
Q173·PhysicsSingle correctJEE Main 2019
A satellite of mass M is in a circular orbit of radius R about the centre of the earth. A meteorite of the same mass, falling towards the earth collides with the satellite completely inelastically. The speeds of the satellite and the meteorite are the same, just before the collision. The subsequent motion of the combined body will be:
  1. (A)such that it escapes to infinity
  2. (B)in an elliptical orbit
  3. (C)in the same circular orbit of radius R
  4. (D)in a circular orbit of a different radius

Correct answer: (B)

Step-by-step solution →
Q174·PhysicsSingle correctJEE Main 2019
The mass and the diameter of a planet are three times the respective values for the Earth. The period of oscillation of a simple pendulum on the Earth is 2s. The period of oscillation of the same pendulum on the planet would be:
  1. (A)32\frac{\sqrt{3}}{2}23​​ s
  2. (B)23\frac{2}{\sqrt{3}}3​2​ s
  3. (C)32\frac{3}{2}23​ s
  4. (D)232\sqrt{3}23​ s

Correct answer: (D)

Step-by-step solution →
Q175·PhysicsSingle correctJEE Main 2019
A satellite is revolving in a circular orbit at a height h from the earth surface, such that h < < R where R is the radius of the earth. Assuming that the effect of earth's atmosphere can be neglected the minimum increase in the speed required so that the satellite could escape from the gravitational field of earth is:
  1. (A)2gR\sqrt{2gR}2gR​
  2. (B)gR\sqrt{gR}gR​
  3. (C)gR2\sqrt{\frac{gR}{2}}2gR​​
  4. (D)gR(2−1)\sqrt{gR}(\sqrt{2}-1)gR​(2​−1)

Correct answer: (D)

Step-by-step solution →
Q176·PhysicsSingle correctJEE Main 2019
A satellite is moving with a constant speed vvv in circular orbit around the earth. An object of mass 'm' is ejected from the satellite such that it just escapes from the gravitational pull of the earth. At the time of ejection, the kinetic energy of the object is:
  1. (A)2mv22mv^{2}2mv2
  2. (B)mv2mv^{2}mv2
  3. (C)12mv2\frac{1}{2}mv^{2}21​mv2
  4. (D)32mv2\frac{3}{2}mv^{2}23​mv2

Correct answer: (B)

Step-by-step solution →
Q177·PhysicsSingle correctJEE Main 2019
If the angular momentum of a planet of mass m, moving a round the Sun in a circular orbit its L, about the center of the Sun, its areal velocity is:
  1. (A)Lm\frac{L}{m}mL​
  2. (B)4Lm\frac{4L}{m}m4L​
  3. (C)L2m\frac{L}{2m}2mL​
  4. (D)2Lm\frac{2L}{m}m2L​

Correct answer: (C)

Step-by-step solution →
Q178·PhysicsSingle correctJEE Main 2019
The energy required to take a satellite to a height 'h' above Earth surface (radius of Earth =6.4×103= 6.4 \times 10^{3}=6.4×103 km) is E1E_1E1​ and kinetic energy required for the satellite to be in a circular orbit at this height is E2E_2E2​. The value of h for which E1E_1E1​ and E2E_2E2​ are equal is
  1. (A)1.6×1031.6 \times 10^{3}1.6×103 km
  2. (B)3.2×1033.2 \times 10^{3}3.2×103 km
  3. (C)6.4×1036.4 \times 10^{3}6.4×103 km
  4. (D)1.28×1041.28 \times 10^{4}1.28×104 km

Correct answer: (B)

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Q179·PhysicsSingle correctJEE Advanced 2018
A planet of mass M, has two natural satellites with masses m1m_{1}m1​ and m2m_{2}m2​. The radii of their circular orbits are R1R_{1}R1​ and R2R_{2}R2​ respectively. Ignore the gravitational force between the satellites. Define v1v_{1}v1​, L1L_{1}L1​, K1K_{1}K1​ and T1T_{1}T1​ to be, respectively, the orbital speed, angular momentum, kinetic energy and time period of revolution of satellite 1; and v2v_{2}v2​, L2L_{2}L2​, K2K_{2}K2​ and T2T_{2}T2​ to be the corresponding quantities of satellite 2. Given m1/m2=2m_{1}/m_{2} = 2m1​/m2​=2 and R1/R2=1/4R_{1}/R_{2} = 1/4R1​/R2​=1/4, match the ratios in List-I to the numbers in List-II.
LIST-ILIST-II
P.v1v2\dfrac{v_{1}}{v_{2}}v2​v1​​1.18\dfrac{1}{8}81​
Q.L1L2\dfrac{L_{1}}{L_{2}}L2​L1​​2.1
R.K1K2\dfrac{K_{1}}{K_{2}}K2​K1​​3.2
S.T1T2\dfrac{T_{1}}{T_{2}}T2​T1​​4.8
  1. (A)P →\rightarrow→ 4; Q →\rightarrow→ 2; R →\rightarrow→ 1; S →\rightarrow→ 3
  2. (B)P →\rightarrow→ 3; Q →\rightarrow→ 2; R →\rightarrow→ 4; S →\rightarrow→ 1
  3. (C)P →\rightarrow→ 2; Q →\rightarrow→ 3; R →\rightarrow→ 1; S →\rightarrow→ 4
  4. (D)P →\rightarrow→ 2; Q →\rightarrow→ 3; R →\rightarrow→ 4; S →\rightarrow→ 1

Correct answer: (B)

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Q180·PhysicsSingle correctJEE Advanced 2017
A rocket is launched normal to the surface of earth, away from the Sun, along the line joining the Sun and the Earth. The Sun is 3×1053\times10^{5}3×105 times heavier than the Earth and is at a distance 2.5×1042.5\times10^{4}2.5×104 times larger than the radius of the Earth. The escape velocity from Earth's gravitational field is ve=11.2 km s−1v_{e} = 11.2\ \text{km s}^{-1}ve​=11.2 km s−1. The minimum initial (vs)(v_{s})(vs​) required for the rocket to be able to leave the Sun-earth system is closest to (Ignore the rotation and revolution of the Earth and the presence of any other planet)
  1. (A)vs=22 km s−1v_{s} = 22\ \text{km s}^{-1}vs​=22 km s−1
  2. (B)vs=42 km s−1v_{s} = 42\ \text{km s}^{-1}vs​=42 km s−1
  3. (C)vs=62 km s−1v_{s} = 62\ \text{km s}^{-1}vs​=62 km s−1
  4. (D)vs=72 km s−1v_{s} = 72\ \text{km s}^{-1}vs​=72 km s−1

Correct answer: (B)

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Q181·PhysicsIntegerJEE Advanced 2015
A large spherical mass M is fixed at one position and two identical point masses m are kept on a line passing through the centre of M (see figure). The point masses are connected by a rigid massless rod of length ℓ\ellℓ and this assembly is free to move along the line connecting them. All three masses interact only through their mutual gravitational interaction. When the point mass nearer to M is at a distance r=3ℓr = 3\ellr=3ℓ from M, the tension in the rod is zero for m=k(M288)m = k\left(\dfrac{M}{288}\right)m=k(288M​). The value of k is

Correct answer: 7

Step-by-step solution →
Q182·PhysicsIntegerJEE Advanced 2015
A bullet is fired vertically upwards with velocity vvv from the surface of a spherical planet. When it reaches its maximum height, its acceleration due to the planet's gravity is 1/4th1/4^{\text{th}}1/4th of its value at the surface of the planet. If the escape velocity from the planet is vesc=vNv_{\text{esc}} = v\sqrt{N}vesc​=vN​, then the value of NNN is (ignore energy loss due to atmosphere)

Correct answer: 2

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Q183·PhysicsSingle correctJEE Advanced 2014
A planet of radius R=110×R = \frac{1}{10} \timesR=101​× (radius of Earth) has the same mass density as Earth. Scientists dig a well of depth R5\frac{R}{5}5R​ on it and lower a wire of the same length and of linear mass density 10−310^{-3}10−3 kgm−1^{-1}−1 into it. If the wire is not touching anywhere, the force applied at the top of the wire by a person holding it in place is (take the radius of Earth =6×106= 6 \times 10^{6}=6×106 m and the acceleration due to gravity of Earth is 10 ms−2^{-2}−2)
  1. (A)96 N
  2. (B)108 N
  3. (C)120 N
  4. (D)150 N

Correct answer: (B)

Step-by-step solution →
Q184·PhysicsMultiple correctJEE Advanced 2013
Two bodies, each of mass MMM, are kept fixed with a separation 2L2L2L. A particle of mass mmm is projected from the midpoint of the line joining their centres, perpendicular to the line. The gravitational constant is GGG. The correct statement(s) is (are)
  1. (A)The minimum initial velocity of the mass mmm to escape the gravitational field of the two bodies is 4GML4\sqrt{\frac{GM}{L}}4LGM​​
  2. (B)The minimum initial velocity of the mass mmm to escape the gravitational field of the two bodies is 2GML2\sqrt{\frac{GM}{L}}2LGM​​.
  3. (C)The minimum initial velocity of the mass mmm to escape the gravitational field of the two bodies is 2GML\sqrt{\frac{2GM}{L}}L2GM​​
  4. (D)The energy of the mass mmm remains constant.

Correct answer: (B)

Step-by-step solution →

Gravitation — frequently asked

How many questions from Gravitation appear in JEE?

Gravitation has appeared in 152 of the last 186 JEE Main and JEE Advanced papers — about 82% of them — contributing 184 questions in total across those papers.

Is Gravitation an important chapter for JEE?

Judged by how often it is actually tested, it appears in roughly 82% of papers. Chapters above about 50% are effectively guaranteed to show up every session, so they repay thorough preparation; lower-frequency chapters are better treated as targeted revision.

Where do these Gravitation questions come from?

Every question is from an official JEE Main or JEE Advanced paper, transcribed from the original paper and tagged to this chapter. Answers follow the official answer key.

Other Physics chapters

  • Properties of Solids and Liquids 344
  • Current Electricity 309
  • Rotational Motion 266
  • Geometrical Optics 259
  • Kinematics 255
  • Magnetic Field of Current 211
  • Electromagnetic Waves 208
  • Thermodynamics 201

All 28 Physics chapters →

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