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NEET • PHYSICS • GRAVITATION

Gravitation practice · Medium

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Showing 1-30 of 135 questions

1 Medium

1 of 135

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 Medium

2 of 135

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.

3 Medium

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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.

4 Medium

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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.

5 Medium

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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.

6 Medium

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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.

7 Medium

7 of 135

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.

8 Medium

8 of 135

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.

9 Medium

9 of 135

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.

10 Medium

10 of 135

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.

11 Medium

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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.

12 Medium

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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.

13 Medium

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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.

14 Medium

14 of 135

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}

15 Medium

15 of 135

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.

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16 Medium

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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.

17 Medium

17 of 135

Application +4 | -1

The distance of a geo-stationary satellite from the centre of the earth (Radius R=6400 km) is nearest to

View answer and explanation

Correct option: B

Option B (7R) is correct because a geostationary satellite orbits at approximately 36,000 km above the Earth's surface, which is about 6 times the Earth's radius, plus one Earth radius to the center.

18 Medium

18 of 135

Application +4 | -1

The earth revolves around the sun in one year. If the distance between them becomes double, the new period of revolution will be

View answer and explanation

Correct option: B

2\sqrt 2 \,\,years

Option B (

2\sqrt 2 \,years

) is correct because, according to Kepler's third law, the period is proportional to the distance raised to the power of 3/2.

19 Medium

19 of 135

Application +4 | -1

Two planets have same radii and different uniform densities. The acceleration due to gravity on surface would be

View answer and explanation

Correct option: B

Greater on planet with heavy density

Option B (Greater on planet with heavy density) is correct because acceleration due to gravity is directly proportional to density when radii are the same.

20 Medium

20 of 135

Analyzing +4 | -1

Statement I: At height

h

from ground and at depth

h

below ground, where

h

is approximately equal to

0.62 \mathrm{R}

, the value of

\mathrm{g}

acceleration due to gravity is same.

Statement II: Value of

\mathrm{g}

decreases both sides, in going up and down.

View answer and explanation

Correct option: B

If both Statement - I and Statement - II are true but Statement - II is not the correct explanation of Statement - I.

Option B is correct because both statements are true, but Statement II doesn't fully explain the specific height mentioned in Statement I. The decrease in g with height and depth are due to different reasons.

21 Medium

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Understanding +4 | -1

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

View answer and explanation

Correct option: C

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

22 Medium

22 of 135

Analyzing +4 | -1

If a planet of given density were made larger its force of attraction for an object on its surface would increase because of planet’s greater mass but would decrease because of the greater distance from the object to the centre of the planet. Which effect predominate?

View answer and explanation

Correct option: A

Increases in mass

Option A (Increases in mass) is correct because the gravitational force is directly proportional to the mass of the planet and inversely proportional to the square of the distance to the center. The increase in mass has a stronger effect than the increase in radius.

23 Medium

23 of 135

Application +4 | -1

A body is released from height equal to radius R of the earth. The velocity of body with which it will strike the earth surface is

View answer and explanation

Correct option: C

\sqrt {gR}

Option C (

\sqrt {gR}

) is correct because the velocity can be found using conservation of energy, equating the initial potential energy to the final kinetic energy.

24 Medium

24 of 135

Analyzing +4 | -1

If a planet of given density were made larger (keeping its density unchanged) its force of attraction for an object on its surface would increase because of increased mass of the planet but would decrease because of larger separation between the centre of the planet and its surface. Which effect would dominate?

View answer and explanation

Correct option: B

Increase in radius

Option A (Increase in mass) is correct because the force of attraction is directly proportional to the mass and inversely proportional to the square of the radius. The increase in mass dominates.

25 Medium

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Understanding +4 | -1

S-I: The moon moves around the earth in a circular orbit and the earth moves around the sun in elliptical orbit then.
S-II: The sun's attraction on the earth and moon are external force acting on the centre of mass of the system

View answer and explanation

Correct option: A

Both S-I and S-II are correct

Option A is correct because both statements accurately describe the motion of the moon and earth, and the forces acting on the system.

26 Medium

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Application +4 | -1

An astronaut of mass m is working in a satellite orbiting the earth at a distance h from the earth’s surface. The radius of the earth is R, while its mass is M. The gravitational pull

{F_G}

on the astronaut is

View answer and explanation

Correct option: C

{F_G} = \frac{{GMm}}{{{{\left( {R + h} \right)}^2}}}

Option C (

{F_G} = \frac{{GMm}}{{{{\left( {R + h} \right)}^2}}})

is correct because it directly applies Newton's law of universal gravitation at a distance (R+h) from Earth's center.

27 Medium

27 of 135

Understanding +4 | -1

Consider a binary star system of star

A

and star

B

with masses

{m_A}

and

{m_B}

revolving in a circular orbit of radii

{r_A}

and

{r_B}

respectively. If

{T_A}

and

{T_B}

are the time period of star

A

and star

B

respectively, then

View answer and explanation

Correct option: B

{T_A} = {T_B}

Option B is correct because in a binary star system, both stars orbit their common center of mass with the same period.

28 Medium

28 of 135

Understanding +4 | -1

The mass of the moon is about

1.2\%

of the mass of the earth. Compared to the gravitational force the earth exerts on the moon, the gravitational force the moon exerts on earth

View answer and explanation

Correct option: A

Is the same

Option A (Is the same) is correct because, according to Newton's third law, the gravitational force exerted by the Earth on the Moon is equal in magnitude to the gravitational force exerted by the Moon on the Earth.

29 Medium

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Analyzing +4 | -1

The ratio of acceleration due to gravity at a height

h

above the surface of the earth and at a depth

h

below the surface of the earth for

h < \; <

radius of earth

View answer and explanation

Correct option: C

Decreases linearly with

h

Option C (Decreases linearly with

h

) is correct because the ratio of the acceleration due to gravity at height h and depth h decreases linearly with h when h is much smaller than the radius of the Earth.

30 Medium

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Understanding +4 | -1

Two spheres of mass

m

and M are situated in air and the gravitational force between them is

F

. The space around the masses is now filled with a liquid of specific gravity 3. The gravitational force will now be

View answer and explanation

Correct option: A

F

Option A (

F

) is correct because the gravitational force between two masses is independent of the surrounding medium.

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Key Takeaways

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