Skip to main content
Back to GRAVITATION

Your Progress

0 / 450 Answered
0
Correct
0
Incorrect
0%
Accuracy
JEE • PHYSICS • GRAVITATION

Gravitation practice · Mixed difficulty

Choose an answer to check it, or open “View answer” to study the explanation. Sign in to save progress to your account.

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.

Take a chapter quiz →
Showing 361-390 of 450 questions

Preparing practice and saved progress…

361 Medium
361 of 450
Application +4 | -1
A geostationary satellite is orbiting around an arbitrary planet PP at a height of 11R11R above the surface of PP, RR being the radius of PP. The time period of another satellite in hours at a height of 2R2R from the surface of PP is……
PP has the time period of 24 h.
View answer and explanation
Correct option: C
3
Option C (3) is correct because it accurately calculates the time period of the second satellite using Kepler's third law.
362 Medium
362 of 450
Application +4 | -1
A body is moving in a low circular orbit about a planet of mass MM and radiusRR. The radius of the orbit can be taken to be RR itself. Then, the ratio of the speed of this body in the orbit to the escape velocity from the planet is
View answer and explanation
Correct option: B
12\frac{1}{{\sqrt 2 }}
Option B (12\frac{1}{{\sqrt 2 }}) is correct because the ratio of orbital speed to escape velocity is always 12\frac{1}{{\sqrt 2 }} for a circular orbit at the planet's surface.
363 Medium
363 of 450
Understanding +4 | -1
When a satellite going round the earth in a circular orbit of radius rr and speed vv loses some of its energy, then rr and vv change as
View answer and explanation
Correct option: C
rr will decrease and vv will increase
Option C (rr will decrease and vv will increase) is correct because as a satellite loses energy, its orbital radius decreases, and its orbital speed increases to maintain equilibrium.
364 Easy
364 of 450
Understanding +4 | -1
A planet is moving in an elliptical orbit around the sun. If TT, V,E\mathrm{V}, \mathrm{E} and L\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, indicating a bound system.
365 Medium
365 of 450
Understanding +4 | -1
What effect occurs on the frequency of a pendulum of it is taken from the earth surface to deep into a mine?
View answer and explanation
Correct option: B
decreases
The correct answer is B (decreases) because the effective value of g decreases as you go deeper into a mine, and frequency is proportional to the square root of g.
366 Medium
366 of 450
Application +4 | -1
The radius of earth is R. Height of a point vertically above the earth's surface at which acceleration due to gravity becomes 1% of its value of the surface is
View answer and explanation
Correct option: B
9R
Option B (9R) is correct because it is the height at which the acceleration due to gravity is 1% of its surface value, calculated using the inverse square law.
367 Medium
367 of 450
Application +4 | -1
Four particles of masses m,2m,3mm, 2 m, 3 m and 4m4 m are kept in sequence at the corners of a square of side aa. The magnitude of gravitational force acting on a particle of mass m\mathrm{m} placed at the centre of the square will be
View answer and explanation
Correct option: C
42Gm2a2\frac{4 \sqrt{2} G m^2}{a^2}
Option C (42Gm2a2\frac{{4 \sqrt{{2}} G m^2}}{{a^2}}) is correct because it represents the net gravitational force calculated by vectorially summing the forces from each mass at the corners of the square.
368 Medium
368 of 450
Application +4 | -1
A satellite of mass m is moving around the earth in a circular orbit of radius r. If the mass of the earth is M, the energy required to take the satellite to an orbit of radius 3r is
View answer and explanation
Correct option: C
GMm3r\frac{{GMm}}{{3r}}
Option C (GMm3r\frac{{GMm}}{{3r}}) is correct because the energy required is the difference in the total mechanical energy of the satellite in the two orbits.
369 Medium
369 of 450
Understanding +4 | -1
A man weighs 80kg80 \mathrm{kg} on the surface of the earth. His weight on a geostationary satellite is
View answer and explanation
Correct option: C
Zero
Option C (Zero) is correct because objects in geostationary orbit experience weightlessness due to the absence of a supporting normal force.
370 Medium
370 of 450
Understanding +4 | -1
Gravitational field due to a solid sphere
View answer and explanation
Correct option: B
Increases inside the sphere and decreases outside the
sphere
Option D (Increases inside the sphere and decreases outside the sphere) is correct because the gravitational field increases linearly inside and decreases inversely with the square of the distance outside.
371 Medium
371 of 450
Analyzing +4 | -1
A particle of mass mm is dropped from a great height hh above the hole in the earth dug along its diameter. Then, which of the following statement is correct?
View answer and explanation
Correct option: D
The speed of the particle at the centre of earth equals GM(R+3h)R(R+h)\sqrt{\frac{G M(R+3 h)}{R(R+h)}}, where RR and MM are the radius and mass of the earth, respectively.
Option D is correct because it accurately calculates the particle's speed at the Earth's center considering its initial height and the varying gravitational force inside the Earth.
372 Medium
372 of 450
Understanding +4 | -1
The orbit of geostationary satellite is circular, the time period of satellite depends on
(i) mass of the satellite
(ii) mass of the earth
(iii) radius of the orbit
(iv) height of the satellite from the surface of earth
Which of the following correct?
View answer and explanation
Correct option: D
(ii), (iii) and (iv)
Option D ((ii), (iii) and (iv)) is correct because the time period of a geostationary satellite depends on the mass of the Earth, the radius of the orbit, and the height of the satellite.
373 Medium
373 of 450
Understanding +4 | -1
When a satellite is shifted from a nearer orbit to a farther orbit, its

(a) Gravitational potential energy increases

(b) Kinetic energy increases

(c) Gravitational potential energy decreases

(d) Kinetic energy decreases

Choose the correct option.
View answer and explanation
Correct option: B
(a) and (d) are correct
Option B ((a) and (d) are correct) is correct because as a satellite moves to a farther orbit, its potential energy increases while its kinetic energy decreases due to a reduction in orbital speed.
374 Medium
374 of 450
Analyzing +4 | -1
Statement I: If ice on the pole melts, length of the day will shorten.

Statement II: Ice will flow towards equator and will result in the decrease in moment of inertia of earth resulting in decrease in frequency of rotation of earth.
View answer and explanation
Correct option: C
If Statement - I is true but Statement - II is false.
Option C (If Statement - I is true but Statement - II is false) is correct because melting ice at the poles increases the moment of inertia, slowing down Earth's rotation and lengthening the day.
375 Medium
375 of 450
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
The correct answer is B (Increase in radius) because the acceleration due to gravity is directly proportional to the radius of the planet when density is constant.
376 Medium
376 of 450
Analyzing +4 | -1
Assertion: The time period of revolution of a satellite around a planet is directly proportional to the radius of the orbit of the satellite.

Reason: Artificial satellite do not follow Kepler's laws of planetary motion.
View answer and explanation
Correct option: C
If assertion is true, but reason is false.
Option C (If assertion is true, but reason is false) is correct because the time period is proportional to r(3/2) and artificial satellites do follow Kepler's laws.
377 Medium
377 of 450
Understanding +4 | -1
Consider a binary star system of star AA and star BB with masses mA{m_A} and mB{m_B} revolving in a circular orbit of radii rA{r_A} and rB{r_B} respectively. If TA{T_A} and TB{T_B} are the time period of star AA and star BB respectively, then
View answer and explanation
Correct option: B
TA=TB{T_A} = {T_B}
Option B (TA=TB{T_A} = {T_B}) is correct because both stars in a binary system orbit their common center of mass with the same angular velocity, hence the same period.
378 Medium
378 of 450
Application +4 | -1
Find the distance of a point from the earth's centre where the resultant gravitational field due to the earth and the moon is zero. The mass of the earth is 6.0×1024 kg6.0 \times 10^{24} \mathrm{~kg} and that of the moon is 7.4×1022 kg7.4 \times 10^{22} \mathrm{~kg}. The distance between the earth and the moon is 4.0×105 km4.0 \times 10^5 \mathrm{~km}.
View answer and explanation
Correct option: A
3.6×105 km3.6 \times 10^5 \mathrm{~km}
Option A (3.6×105 km3.6 \times 10^5 \mathrm{~km}) is correct because it is the distance where the gravitational fields of the Earth and Moon cancel each other out.
379 Medium
379 of 450
Application +4 | -1
If a body of mass mm has to be taken from the surface of the earth to a height h=4Rh=4 R, then the amount of energy required is (R=R= radius of the earth)
View answer and explanation
Correct option: C
4mgR5\frac{4 m g R}{5}
Option C (4mgR5\frac{{4 m g R}}{{5}}) is correct because it represents the difference in gravitational potential energy between the surface and the height h=4R.
380 Medium
380 of 450
Analyzing +4 | -1
A spherical uniform planet is rotating about its axis. The velocity of a point on its equator is V.V. Due to the rotation of the planet about its axis the acceleration due to gravity g{g} at equator is 1/2 of gg at poles. The escape velocity of a particle on the planet in terms of V{V} from the pole of the planet is
View answer and explanation
Correct option: A
Ve=2V{V_e}=2{V}
Option A (Ve=2V{V_e}=2{V}) is correct because by relating the acceleration due to gravity at the equator and poles with the planet's rotation, we can derive the escape velocity in terms of the equatorial velocity V.
381 Medium
381 of 450
Understanding +4 | -1
The weight of an astronaut, in an artificial satellite revolving around the earth, is
View answer and explanation
Correct option: A
Zero
Option A is correct because in an artificial satellite, the astronaut experiences weightlessness due to freefall, making their apparent weight zero.
382 Medium
382 of 450
Application +4 | -1
An earth's satellite makes a circle around the earth in 90 minutes. Calculate the height of the satellite above the earth's surface. Given radius of the earth is 6400 km6400 \mathrm{~km} and g=980 cms−2g=980 \mathrm{~cms}^{-2}.
View answer and explanation
Correct option: A
268 km268 \mathrm{~km}
Option A (268 km268 \mathrm{~km}) is correct because it is the height calculated using the time period of the satellite's orbit and the given values of g and R.
383 Easy
383 of 450
Understanding +4 | -1
On which quantity the escape velocity for earth does not depend on
View answer and explanation
Correct option: B
mass of the projectile
Option B is correct because the escape velocity depends on the mass of the Earth, the gravitational constant, and the point of projection relative to the Earth (radius), but not the mass of the projectile.
384 Medium
384 of 450
Application +4 | -1
A diametrical tunnel is dug across the earth. A ball is dropped into the tunnel from one earth. A ball is dropped into the tunnel from one side. The velocity of the ball when it reached the centre of the earth is (Given: gravitational potential at the centre of earth =−32 (GMR)= - \frac{3}{2}\,\left( {\frac{{GM}}{R}} \right))
View answer and explanation
Correct option: B
gR\sqrt {gR}
Option B (gR\sqrt {gR}) is correct because it is derived from equating the loss in potential energy to the gain in kinetic energy as the ball moves to the center of the Earth.
385 Easy
385 of 450
Understanding +4 | -1
When we move from equator to pole, the value of acceleration due to gravity(g)
View answer and explanation
Correct option: A
increases
Option A (increases) is correct because the acceleration due to gravity (g) increases as you move from the equator to the poles due to the Earth's shape and rotation.
386 Medium
386 of 450
Analyzing +4 | -1
A satellite is revolving around the earth at a height hh. The satellite explodes into two equal parts, one part has zero velocity after the explosion, and the other moves in the same direction with double the orbital speed of the original satellite. The maximum height reached by the part with double the velocity is (Neglect the gravitational effects between the fragments)
View answer and explanation
Correct option: D
None of these
Option D is correct. The velocity after the collision becomes greater than the escape velocity, so the fragment will escape to infinity.
387 Medium
387 of 450
Application +4 | -1
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/TB{T_A}/{T_B} is
View answer and explanation
Correct option: D
1
Option D (1) is correct because the ratio of kinetic energies is 1, as both satellites have the same kinetic energy calculated using their orbital speeds and masses.
388 Medium
388 of 450
Application +4 | -1
A satellite is orbiting just above the surface of the earth with period TT. If dd is the density of the earth and GG is the universal constant of gravitation, the quantity 3πGd\frac{3 \pi}{G d} represents
View answer and explanation
Correct option: D
T2T^2
Option D (T2T^2) is correct because the quantity 3πGd\frac{{3 \pi}}{{G d}} is equal to the square of the orbital period TT of a satellite orbiting just above the Earth's surface.
389 Easy
389 of 450
Application +4 | -1
When the radius and mass of earth are increased by 0.5%0.5 \%. Which of the following statement is correct at the surface of the earth?
View answer and explanation
Correct option: A
gg will decrease.
Option A (gg will decrease) is correct because gg is inversely proportional to the square of the radius, so increasing the radius decreases gg.
390 Easy
390 of 450
Remembering +4 | -1
The value of ' g\mathrm{g} ' on poles is
View answer and explanation
Correct option: A
Maximum
Option A (Maximum) is correct because the value of 'g' is maximum at the poles due to the Earth's shape.
Revision notes and key takeaways

Review GRAVITATION in your textbook or class notes, then use this set to check your understanding.

Showing 361-390 of 450 questions

Your practice

0 / 450 answered · 450 remaining · 0% accuracy

Progress is stored on this device. Sign in to save to your account.

✓ Correct · ✗ Incorrect · ○ Unanswered

Practice settings

Type at least 2 characters or jump directly:
🔓

Save your learning progress

Sign in with a free account to save your practice answers and quiz results, then return to them when you revise.