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NEET • PHYSICS • GRAVITATION

Gravitation practice · Mixed difficulty

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Showing 1-30 of 255 questions

1 Medium

1 of 255

Application +4 | -1

If a particle of mass m is moving in a horizontal circle of radius r with a centripetal force

\left( { - K/{r^2}} \right)

the total energy is

View answer and explanation

Correct option: A

- \frac{K}{{2r}}

Option A (

- \frac{K}{{2r}}

) is correct because the total energy is the sum of kinetic and potential energy, where potential energy is -K/r and kinetic energy is K/2r.

2 Easy

2 of 255

Analyzing +4 | -1

Suppose the law of gravitational attraction suddenly changes and becomes an inverse cube law i.e.

F \propto \frac{1}{r^3}

, but still remaining a central force. Then

View answer and explanation

Correct option: A

Kepler's law of area still holds

Option A (Kepler's law of area still holds) is correct because Kepler's second law (law of areas) is a consequence of conservation of angular momentum, which holds for any central force.

3 Easy

3 of 255

Application +4 | -1

Two satellites A and B go round a planet P is circular orbits having radii 16 R and 2 R respectively. If the speed of satellite A is

6\,m{s^{ - 1}}

the speed of the satellite B will be [in m/s] nearly

View answer and explanation

Correct option: A

17\,m{s^{ - 1}}

Option A (

17\,m{s^{ - 1}}

) is correct because the orbital speed is inversely proportional to the square root of the radius.

4 Medium

4 of 255

Understanding +4 | -1

The total energy of satellite moving with an orbital velocity

v

around the earth is

View answer and explanation

Correct option: B

- \frac{1}{2}m{v^2}

Option B is correct because the total energy of a satellite in orbit is half of its kinetic energy and is negative, indicating a bound system.

5 Medium

5 of 255

Analyzing +4 | -1

Two satellites of same mass are launched in the same orbit of radius

r

around the earth so as to rotate opposite to each other. If they collide inelastically and stick together as wreckage, the total energy of the system just after collision is

View answer and explanation

Correct option: A

-\frac{2 G M m}{r}

Option B (

-\frac{G M m}{r}

) is correct because the total energy after the inelastic collision is the sum of the kinetic and potential energies of the wreckage.

6 Easy

6 of 255

Understanding +4 | -1

Statement-1: There is almost no effect of rotation of earth at poles.
Statement-2: Because rotation of earth is about polar axis.

View answer and explanation

Correct option: A

Statement-1 is true, Statement-2 is true

Option A (Statement-1 is true, Statement-2 is true) is correct because the rotation of the Earth has minimal effect at the poles, and this is because the Earth rotates about its polar axis.

7 Easy

7 of 255

Application +4 | -1

The escape velocity of a body on the surface of an imaginary planet which has thrice the radius as of earth and double the mass of earth, is (

{v_e}

is the escape velocity at the surface of earth)

View answer and explanation

Correct option: C

\sqrt {\frac{2}{3}} {v_e}

Option C is correct because the escape velocity is proportional to the square root of (M/R), so with the given ratios, the escape velocity becomes sqrt(2/3) times the Earth's escape velocity.

8 Medium

8 of 255

Remembering +4 | -1

According to Kelper’s law of planetary motion if T represent time period and

r

is orbital radius, then for two planets these are related as

View answer and explanation

Correct option: D

{\left( {\frac{{{T_1}}}{{{T_2}}}} \right)^2} = {\left( {\frac{{{r_1}}}{{{r_2}}}} \right)^3}

Option D (

{\left( {\frac{{{T_1}}}{{{T_2}}}} \right)^2} = {\left( {\frac{{{r_1}}}{{{r_2}}}} \right)^3}

) is correct because it represents Kepler's Third Law, relating the period and orbital radius of two planets.

9 Hard

9 of 255

Application +4 | -1

If potential energy is given by

\mathrm{U}=\frac{\mathrm{a}}{\mathrm{r}^2}-\frac{\mathrm{b}}{\mathrm{r}}

. Then find out maximum force. (given

\mathrm{a=2, b}=4

)

View answer and explanation

Correct option: A

-\frac{16}{27} \mathrm{~N}

Option A (

-\frac{16}{27} \mathrm{~N}

) is correct because it represents the maximum force calculated by finding the point where the derivative of the force function is zero.

10 Easy

10 of 255

Remembering +4 | -1

If the mass of a body is

M

on the surface of the earth, then mass of the same body on the surface of the moon is

View answer and explanation

Correct option: D

M

Option D (

M

) is correct because mass is an intrinsic property of an object and does not change with location.

11 Easy

11 of 255

Application +4 | -1

A planet in a distant solar system is 5 times more massive than the earth and its radius is 5 times smaller. Given that the escape velocity from the earth is

11\,k\,m{s^{ - 1}},

the escape velocity from the surface of the planet would be [in kmps]

View answer and explanation

Correct option: B

Option B (55) is correct because the escape velocity is proportional to

\sqrt{M/R}

, and with the given changes, the new escape velocity is 5 times the original.

12 Medium

12 of 255

Application +4 | -1

If M = mass of the earth, R = radius of the earth, then what is the gravitational potential at a distance

r = \frac{R}{2}

from its centre? (Consider gravitational potential at infinite is zero)

View answer and explanation

Correct option: D

- \frac{{11GM}}{{8R}}

Option D (

- \frac{{11GM}}{{8R}}

) is correct because it accounts for the gravitational potential both inside and outside the Earth, considering the given distance from the center.

13 Medium

13 of 255

Understanding +4 | -1

Kepler’s second law states that the straight line joining the planet to the sun sweeps out equal times. This statement is equivalent to saying that

View answer and explanation

Correct option: B

Tangential acceleration is zero

Option B (Tangential acceleration is zero) is correct because Kepler's second law implies that the planet's speed changes as it orbits the sun, but the area swept out is constant, meaning there's no tangential acceleration.

14 Hard

14 of 255

Application +4 | -1

An artificial satellite is moving in a circular orbit around the earth with a speed equal to half the magnitude of escape velocity from the earth. If the satellite is stopped in its orbit and allowed to fall freely onto the earth, the speed with which it hits the surface _____

\mathrm{km} / \mathrm{s}.

[g=

9.8 \mathrm{~ms}^{-2}

and

\left.{R}_e=6400 \mathrm{~km}\right]

View answer and explanation

Correct option: A

7.92

Option A (T) is correct because the period depends on the density of the planet, and since the density is the same, the period remains the same.

15 Medium

15 of 255

Remembering +4 | -1

According to Kepler's law of planetary motion, if

T

represents time period and

r

is orbital radius, then for two planets these are related as

View answer and explanation

Correct option: C

\left[\frac{T_1}{T_2}\right]^2=\left[\frac{r_1}{r_2}\right]^3

Option C is correct because Kepler's third law states that the square of the period is proportional to the cube of the orbital radius.

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16 Medium

16 of 255

Application +4 | -1

A man weighs 80 kg on earth surface. The height above ground where he will weigh 40kg, is (radius of earth is 6400 km)

View answer and explanation

Correct option: B

0.41 times r

Option B (0.41 times r) is correct because it is the height calculated using the variation of g with height formula.

17 Medium

17 of 255

Understanding +4 | -1

When a satellite moves around the earth in a certain orbit, the quantity which remains constant is:

View answer and explanation

Correct option: C

Areal velocity

Option C (Areal velocity) is correct because according to Kepler's second law, the areal velocity of a planet (or satellite) remains constant.

18 Easy

18 of 255

Understanding +4 | -1

Assertion: Smaller the orbit of a planet around the sun, shorter is the time it takes to complete.
Reason: According to Kepler's third law of planetary motion, square of time period is proportional to cube of mean distance from the sun.

View answer and explanation

Correct option: A

Assertion is True, Reason is True; Reason is a correct explanation for Assertion

Option A is correct because Kepler's third law directly links orbital size to orbital period.

19 Medium

19 of 255

Remembering +4 | -1

The escape velocity of a sphere of mass

m

from earth having mass

M

and radius

R

is given by

View answer and explanation

Correct option: A

\sqrt {\frac{{2GM}}{R}}

Option A (

\sqrt {\frac{{2GM}}{R}}

) is correct because the escape velocity formula is derived from equating kinetic energy and gravitational potential energy.

20 Medium

20 of 255

Analyzing +4 | -1

Six stars of equal mass

m

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

2 \pi \sqrt{\frac{4 \sqrt{3} a^3}{G m(5 \sqrt{3}+4)}}

Option B (

2 \pi \sqrt{\frac{4 \sqrt{3} a^3}{G m(5 \sqrt{3}+4)}}

) is correct because it accounts for the gravitational forces from all the stars in the hexagon.

21 Hard

21 of 255

Application +4 | -1

A body of mass

m

is projected with a velocity of

\frac{1}{\sqrt{2}}

times the escape velocity from the surface of earth in the vertical upward direction. If the air resistance is negligible, then the maximum distance of the body from is surface of earth is

x R

where

R

is the radius of earth. Find

x

View answer and explanation

Correct option: A

Option A (1) is correct because the maximum distance from the Earth's center is 2R, making the distance from the surface R, hence x=1.

22 Medium

22 of 255

Analyzing +4 | -1

A smooth wedge

\mathrm{A}

is fitted in a chamber hanging from a fixed ceiling near the earth's surface. A block

\mathrm{B}

placed at the top of the wedge takes time

\mathrm{T}

to slide down the length of the wedge and the cable supporting the chamber is broken at the same instant, the block will be

View answer and explanation

Correct option: C

Remain at the top of the wedge

Option C is correct because when the cable breaks, both the wedge and the block become free-falling, resulting in no relative motion between them, and the block remains at the top of the wedge.

23 Medium

23 of 255

Application +4 | -1

The height at which the weight of an object becomes

\left( {\frac{1}{{16}}} \right)th

of its weight on the surface of earth is (R is radius of earth)

View answer and explanation

Correct option: A

Option A (3R) is correct because the weight is inversely proportional to the square of the distance from the center of the earth.

24 Hard

24 of 255

Application +4 | -1

If an object is having same weight at same distance above and below the surface of earth, find its distance from surface of earth.

View answer and explanation

Correct option: B

(\sqrt{5}-1) \frac{R}{2}

Option B (

(\sqrt{5}-1) \frac{R}{2}

) is correct because it involves equating the gravitational acceleration at a height h and depth d and solving for h.

25 Medium

25 of 255

Understanding +4 | -1

M

is the mass of the earth and

R

its radius, the ratio of the gravitational acceleration and the gravitational constant is

View answer and explanation

Correct option: B

\frac{M}{{{R^2}}}

Option B (

\frac{M}{{{R^2}}}

) is correct because the gravitational acceleration (g) is given by

g = \frac{GM}{R^2}

, so the ratio of g to G is

\frac{g}{G} = \frac{M}{R^2}

26 Medium

26 of 255

Remembering +4 | -1

A satellite is placed in a circular orbit around earth at such a height that it always remains stationary with respect to earth surface. In such case, its height from the earth surface is

View answer and explanation

Correct option: B

36000 km

Option B (36000 km) is correct because a geostationary satellite must be at this height to have a period of 24 hours, matching Earth's rotation.

27 Easy

27 of 255

Understanding +4 | -1

Statement-1: It is possible to put an artificial satellite into orbit in such a way that it will always remain directly over New Delhi.
Statement-2: To put an artificial satellite into orbital such that it remains directly over head at a fixed location, its time period should be related to that of earth in the equatorial plane.

View answer and explanation

Correct option: D

Statement-1 is false, Statement-2 is true

Statement-1 is false, and Statement-2 is true. A geostationary satellite must be above the equator, so it can't be over New Delhi.

28 Easy

28 of 255

Understanding +4 | -1

Statement-1: Gravitational potential is zero inside a shell.
Statement-2: Gravitational potential is equal to double the work done in bringing a unit mass from infinity to a point inside the gravitational field.

View answer and explanation

Correct option: D

Statement-1 is false, Statement-2 is false

Option D is correct because both statements are false: the gravitational potential inside a shell is constant and non-zero, and gravitational potential is equal to the work done, not double the work done.

29 Medium

29 of 255

Analyzing +4 | -1

Observers on the 10th, 5th and ground floor of a tall building measure the velocity of a certain raindrop by some accurate method. Surprisingly the velocity of the raindrop measured by the three observers was found to be the same. This is because

View answer and explanation

Correct option: A

there is no gravitational force acting on the drop

Option B (gravitational force on the raindrop is balanced by the force produced by the surrounding air) is correct because the raindrop reaches terminal velocity where the gravitational force equals air resistance.

30 Easy

30 of 255

Understanding +4 | -1

A planet is moving in an elliptical orbit around the sun. If

T

\mathrm{V}, \mathrm{E}

and

\mathrm{L}

stand respectively for its kinetic energy, gravitational potential energy, total energy and magnitude of angular momentum about the centre of force, then which of the following is correct?

View answer and explanation

Correct option: C

E is always negative

Option C (E is always negative) is correct because the total energy of a planet in an elliptical orbit is always negative.

Revision notes and key takeaways

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Key Takeaways

✓ This set contains 255 single-correct-answer MCQs on GRAVITATION.
✓ Try each question before opening its answer. Read the explanation where one is available.
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