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JEE • PHYSICS • GRAVITATION

Gravitation practice · Mixed difficulty

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Practice format and chapter quiz

These sets use single-correct-answer MCQs. They do not cover every JEE Main or JEE Advanced question format and are not full-length mock papers.

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Showing 271-300 of 450 questions

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271 Medium
271 of 450
Understanding +4 | -1
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:
View answer and explanation
Correct option: D
A and D only
Option D (A and D only) is correct because Kepler's laws state that the force is inversely proportional to the square of the distance and the square of the period is proportional to the cube of the semi-major axis.
272 Medium
272 of 450
Analyzing +4 | -1
Two identical satellites are at RR and 7R7R away from earth surface, the wrong statement is
View answer and explanation
Correct option: D
Ratio of total energy will be 4 but ratio of potential and kinetic energies will be 2.
Option D is the wrong statement because the ratio of potential and kinetic energies should also be 4, similar to the total energy ratio.
273 Medium
273 of 450
Understanding +4 | -1
Assertion: For circular orbits, the law of periods is T2∝r3{T}^2 \propto {r}^3, where M{M} is the mass of sun and rr is the radius of orbit.
Reason: The square of the period TT of any planet about the sun is proportional to the cube of the semi-major axis aa of the orbit.
View answer and explanation
Correct option: A
If both Assertion & Reason are True & the Reason is a correct explanation of the Assertion
Option A is correct because Kepler's Third Law states that the square of the period is proportional to the cube of the semi-major axis, which applies to circular orbits as well.
274 Medium
274 of 450
Application +4 | -1
A satellite of mass 200 kg is orbiting around the earth at a height of 400 km above its surface. The energy required to send it out of the earth’s gravitational influence is (Mass of earth =6×1024kg= 6 \times {10^{24}}kg, Radius of earth = 6000 km, G=6.67×10−11 Nm2 kg−2G = 6.67 \times {10^{ - 11}}\,N{m^2}\,k{g^{ - 2}})
View answer and explanation
Correct option: C
4×109J4 \times {10^9}J
Option C (4×109J4 \times {10^9}J) is correct because it is the energy required to escape Earth's gravitational influence, calculated using the gravitational potential energy formula.
275 Medium
275 of 450
Application +4 | -1
The masses and radii of the earth and moon are M1,R1{M_{1,}}{R_1} and M2,R2{M_{2,}}{R_2} respectively, their centres are distance dd apart. The minimum velocity with which a particle of mass mm should be projected from a point midway between their centres so that it escapes to infinity is
View answer and explanation
Correct option: A
2Gd(M1+M2)2\sqrt {\frac{G}{d}({M_1} + {M_2})}
Option A is correct because it calculates the escape velocity based on the gravitational potential at the midpoint between the Earth and Moon.
276 Easy
276 of 450
Application +4 | -1
The radius of a planet is twice the radius of earth. Both have almost equal average mass-densities. If VPV_P and VEV_E are escape velocities of the planet and the earth, respectively, then
View answer and explanation
Correct option: B
VP=2VEV_P=2 V_E
Option B (VP=2VEV_P=2 V_E) is correct because the escape velocity is directly proportional to the radius when the average mass-densities are equal.
277 Medium
277 of 450
Application +4 | -1
Two bodies of masses m1 and m2{m_1}\,and\,{m_2} are initially at rest at infinite distance apart. They are then allowed to move towards each other under mutual gravitational attraction. Their relative velocity of approach at a separation distance rr between them is
View answer and explanation
Correct option: B
[2Gr(m1+m2)]1/2{\left[ {\frac{{2G}}{r}\left( {{m_1} + {m_2}} \right)} \right]^{1/2}}
Option B ({\left[ {\frac{{2G}}{r}\left( {{{m_1} + {m_2}} \right)} \right]^{1/2}}{\left[ {\frac{{2G}}{r}\left( {{{m_1} + {m_2}} \right)} \right]^{1/2}}) is correct because it represents the relative velocity derived from conservation of energy and momentum for two masses approaching each other under gravitational attraction.
278 Medium
278 of 450
Analyzing +4 | -1
In the following four periods
(i) Time of revolution of a satellite just above the earth’s surface (Tst)\left( {{T_{st}}} \right) , (ii) Period of oscillation of mass inside the tunnel bored along the diameter of the earth (Tma)\left( {{T_{ma}}} \right) , (iii) Period of simple pendulum having a length equal to the earth’s radius in a uniform field of 9.8N/kg(Tsp)9.8N/kg\left( {{T_{sp}}} \right) , (iv) Period of an infinite length simple pendulum in the earth’s real gravitational filed (Tis)\left( {{T_{is}}} \right)
View answer and explanation
Correct option: C
Tsp>Tis{T_{sp}} > {T_{is}}
Option C (Tsp>Tis{T_{sp}} > {T_{is}}) is correct because the period of a simple pendulum with length equal to Earth's radius in a uniform field is less than that of an infinite length simple pendulum in Earth's real gravitational field.
279 Medium
279 of 450
Understanding +4 | -1
The ratio of orbital velocity of an earth's satellite to the escape velocity of the same earth's satellite is very nearly equal to
View answer and explanation
Correct option: D
12\frac{1}{\sqrt{2}}
Option D (12\frac{1}{\sqrt{2}}) is correct because the ratio of orbital velocity to escape velocity is derived from their respective formulas and simplifies to 12\frac{1}{\sqrt{2}}.
280 Medium
280 of 450
Application +4 | -1
Two isolated point masses m{m} and M{M} are separated by a distance ℓ\ell. The moment of inertia of the system about an axis passing through a point where gravitational field is zero and perpendicular to the line joining the two masses, is
View answer and explanation
Correct option: A
m2+M2(m+M)2ℓ2\frac{m^2+M^2}{(\sqrt{m}+\sqrt{M})^2}\ell^2
Option A (m2+M2(m+M)2ℓ2\frac{m^2+M^2}{(\sqrt{m}+\sqrt{M})^2}\ell^2) is correct because it represents the moment of inertia calculated about the point where the gravitational field is zero, considering the distances of both masses from that point.
281 Medium
281 of 450
Understanding +4 | -1
Two spheres of mass mm and MM are situated in air and the gravitational force between them is 2F.2 F. The space around the masses is now filled with a liquid of specific gravity 4. The gravitational force will now be
View answer and explanation
Correct option: D
2F2 F
Option D (2F2 F) is correct because the gravitational force between two masses is independent of the medium separating them.
282 Medium
282 of 450
Understanding +4 | -1
A satellite of the earth is revolving in a circular orbit with a uniform speed vv. If the gravitational force suddenly disappears, the satellite will
View answer and explanation
Correct option: B
Move with a velocity vv, tangentially to the original orbit
Option B (Move with a velocity vv, tangentially to the original orbit) is correct because, due to inertia, the satellite will continue moving in a straight line tangent to its orbit at the point where gravity disappeared.
283 Easy
283 of 450
Analyzing +4 | -1
Assertion: The magnitude of gravitational potential at the surface of solid sphere is less than that of the centre of sphere.
Reason: Due to solid sphere, gravitational potential is same within the sphere.
View answer and explanation
Correct option: C
Assertion is True, Reason is False
Option C (Assertion is True, Reason is False) is correct because the gravitational potential at the surface of a solid sphere is less than at the center, but the potential inside the sphere is not constant.
284 Medium
284 of 450
Understanding +4 | -1
The gravitational attraction between the two bodies increases when their masses are
View answer and explanation
Correct option: B
Increased and distance is reduced
Option B (Increased and distance is reduced) is correct because gravitational force is directly proportional to the product of masses and inversely proportional to the square of the distance.
285 Medium
285 of 450
Understanding +4 | -1
An artificial satellite of mass ‘m’ is revolving round the earth in a circle of radius RR. Then work done in one revolution is
View answer and explanation
Correct option: D
Zero
Option D (Zero) is correct because the gravitational force is a conservative force, and the work done by a conservative force in a closed path (one revolution) is always zero.
286 Medium
286 of 450
Analyzing +4 | -1
Six stars of equal mass mm are moving about the centre of mass of the system such that they are always on the vertices of a regular hexagon of side length a. Their common time period of revolution around center will be
View answer and explanation
Correct option: B
2π43a3Gm(53+4)2 \pi \sqrt{\frac{4 \sqrt{3} a^3}{G m(5 \sqrt{3}+4)}}
Option B is correct because the time period is derived by equating the centripetal force on each star to the net gravitational force due to the other stars in the hexagonal arrangement.
287 Medium
287 of 450
Analyzing +4 | -1
Assertion: If the earth stops rotating about its axis, the value of the weight of the body at equator will decrease.
Reason: The centripetal force does not act on the body at the equator.
View answer and explanation
Correct option: D
If both Assertion & Reason are false
Option D (If both Assertion & Reason are false) is correct because if the Earth stops rotating, the weight at the equator will increase, and the centripetal force is acting on the body at the equator.
288 Medium
288 of 450
Application +4 | -1
What is the energy required to launch a m kgm \mathrm{~kg} satellite from earth's surface to a circular orbit at an altitude of 7R7 R? (R=R= radius of the earth)
View answer and explanation
Correct option: C
1516mgR\frac{15}{16} m g R
Option C (1516mgR\frac{{15}}{{16}} m g R) is correct because it accurately calculates the energy required to launch the satellite to the specified altitude by considering the change in potential energy.
289 Medium
289 of 450
Application +4 | -1
Satellite A and B are revolving around the orbit of earth. The mass of A is 10 times of mass of B.The ratio of time period (TATB)\left( {\frac{{{T_A}}}{{{T_B}}}} \right) is
View answer and explanation
Correct option: B
1
Option B (1) is correct because the time period of a satellite depends only on the radius of the orbit and the mass of the central body (Earth), not on the mass of the satellite itself.
290 Medium
290 of 450
Understanding +4 | -1
The Earth's gravity provides the centripetal force on a orbiting satellite to keep it moving in circle at constant speed. Which statements best explains why the speed of the satellite does not change even through there is a net force exerted on it?
View answer and explanation
Correct option: D
The centripetal force is always perpendicular to the velocity
Option D (The centripetal force is always perpendicular to the velocity) is correct because a force perpendicular to velocity only changes direction, not speed.
291 Medium
291 of 450
Application +4 | -1
If the angular speed of earth is increased so much that the objects start flying from the equator, then the length of the day will be nearly
View answer and explanation
Correct option: A
1.5 hours
Option A (1.5 hours) is correct because the length of the day is calculated using the formula for the time period of rotation.
292 Medium
292 of 450
Analyzing +4 | -1
A mass MM is broken into two parts of masses m1m_1 and m2m_2. How are m1m_1 and m2m_2 related so that force of gravitational attraction between the two parts is maximum?
View answer and explanation
Correct option: A
M/2M / 2
Option A (M/2M / 2) is correct because the gravitational force is maximized when the two masses are equal, which occurs when each mass is half of the total mass.
293 Medium
293 of 450
Application +4 | -1
Satellite A and B are revolving around the orbit of earth. The mass of A is 10 times of mass of B.The ratio of time period (TATB)\left( {\frac{{{T_A}}}{{{T_B}}}} \right) is
View answer and explanation
Correct option: B
1
Option B (1) is correct because the time period of a satellite depends only on the radius of the orbit and the mass of the central body (Earth), not on the mass of the satellite itself.
294 Medium
294 of 450
Analyzing +4 | -1
Two identical trains PP and QQ move with equal speeds on parallel tracks along the equator. PP moves from east to west and QQ from west to east
View answer and explanation
Correct option: D
Train PP exerts greater force on track
Option B (R, L) is correct because a circuit containing only a resistor and an inductor cannot have a phase difference of exactly π/2 between the alternating current and emf.
295 Medium
295 of 450
Understanding +4 | -1
A planet revolving in elliptical orbit has
I. a constant velocity of revolution
II. has the least velocity when it is nearest to the sun
III. its areal velocity is directly proportional to its velocity.
IV. areal velocity is inversely proportional to its velocity.
V. to follow a trajectory such that the areal velocity is constant.
Choose the correct answer from the options given below.
View answer and explanation
Correct option: D
Only V
Option D (Only V) is correct because Kepler's Second Law states that a planet's areal velocity remains constant throughout its orbit.
296 Medium
296 of 450
Understanding +4 | -1
A cylindrical vessel partially filled with water is rotated about its vertical central axis. It’s surface will
View answer and explanation
Correct option: B
Rise from the sides
Option B is correct because when a cylindrical vessel filled with water is rotated, the water rises from the sides due to centrifugal force.
297 Medium
297 of 450
Application +4 | -1
The length of a seconds pendulum if it is placed at height 2R2 R (RR : radius of earth) is 10xπ2\frac{10}{x \pi^2} metres. Find xx.
View answer and explanation
Correct option: A
9
Option A (9) is correct because the length of the pendulum is calculated using the formula for the time period of a simple pendulum, considering the change in gravity at a height of 2R.
298 Medium
298 of 450
Understanding +4 | -1
Three particles each of mass mm are kept at the vertices of an equilateral triangle of side LL. The gravitational field at, centre due to these particles is
View answer and explanation
Correct option: A
Zero
Option A (Zero) is correct because the gravitational fields due to the three particles at the center of the equilateral triangle cancel each other out due to symmetry.
299 Medium
299 of 450
Understanding +4 | -1
Gas escapes from the surface of a planet because it acquires an escape velocity. The escape velocity will depend on which of the following factors :
I. Mass of the planet
II. Mass of the particle escaping
III. Temperature of the planet
IV. Radius of the planet
Select the correct answer from the codes given below :
View answer and explanation
Correct option: C
I and IV
Option C (I and IV) is correct because escape velocity depends on the mass and radius of the planet, as shown by the formula.
300 Medium
300 of 450
Application +4 | -1
If suppose moon is suddenly stopped and then released (given radius of moon is one-fourth the radius of earth)and the acceleration of moon with respect to earth is 0.0027 ms−20.0027\,\mathrm{m}{\mathrm{s}^{ - 2}}), then the acceleration of the moon just before striking the earth’s surface is (Take g=10ms−2)g = 10m{s^{ - 2}})
View answer and explanation
Correct option: C
6.4  ms−26.4\mathrm{\;m}{\mathrm{s}^{ - 2}}
Option C (6.4  ms−26.4\mathrm{\;m}{\mathrm{s}^{ - 2}}) is correct because it calculates the acceleration of the moon just before striking the Earth's surface using the gravitational force at that distance.
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