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NEET • PHYSICS • GRAVITATION

Gravitation practice · Mixed difficulty

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Showing 241-255 of 255 questions
241 Easy
241 of 255
Application +4 | -1
A mass of 3×1024 kg3 \times {10^{24}}\,kg is to be compressed in a sphere in such a way that the escape velocity from the sphere is 9×108 m/s.9 \times {10^8}\,m/s. Radius of the sphere should be [G=6.67×10−11 Nm2/kg2]\left[ {G = 6.67 \times {{10}^{ - 11}}\,N{m^2}/k{g^2}} \right]
View answer and explanation
Correct option: D
0.49 mm
Option D (0.49 mm) is correct. The radius is calculated using the escape velocity formula.
242 Medium
242 of 255
Application +4 | -1
Radius of orbit of satellite of earth is R. Its kinetic energy is proportional to
View answer and explanation
Correct option: A
1R\frac{1}{R}
Option A (1R\frac{1}{R}) is correct because the kinetic energy of a satellite is inversely proportional to the radius of its orbit.
243 Medium
243 of 255
Understanding +4 | -1
A small satellite is in elliptical orbit around Earth as shown. If L denotes the magnitude of its angular momentum and K denotes kinetic energy
View answer and explanation
Correct option: D
L2=L1 and K2>K1{L_2} = {L_1}\,and\,{K_2} > {K_1}
Option D (L2=L1 and K2>K1{L_2} = {L_1}\,and\,{K_2} > {K_1}) is correct because angular momentum is conserved in an elliptical orbit, and kinetic energy is greater at the point closest to Earth.
244 Easy
244 of 255
Application +4 | -1
A body of mass 60 g experiences a gravitational force of 3.0 N, when placed at a particular point. The magnitude of the gravitational field intensity at that point is:
View answer and explanation
Correct option: C
50 N/kg
Option C (50 N/kg) is correct because gravitational field intensity is force per unit mass, calculated as 3.0 N / 0.06 kg = 50 N/kg.
245 Medium
245 of 255
Analyzing +4 | -1
A clock S is based on oscillation of a spring and a clock P is based on pendulum motion. Both clocks run at the same rate on earth. On a planet having the same density as earth but twice the radius
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Correct option: B
PP will run faster than SS
Option B (PP will run faster than SS) is correct because the time period of a spring clock is independent of gravity, while the time period of a pendulum clock depends on gravity, which changes on the new planet.
246 Medium
246 of 255
Application +4 | -1
If earth were to suddenly contract to half its present size, without change in its mass, the duration of the day will be
View answer and explanation
Correct option: C
6 hours
Option C (6 hours) is correct because the duration of the day is inversely proportional to the square of the radius.
247 Easy
247 of 255
Understanding +4 | -1
When a body is taken from the equator to the poles, its weight
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Correct option: B
increases
Option B (increases) is correct because the weight of a body increases when taken from the equator to the poles due to the decrease in the centrifugal force and the slightly smaller radius at the poles.
248 Easy
248 of 255
Understanding +4 | -1
The velocity with which a projectile must be fired so that it escapes Earth's gravitation does not depend on
View answer and explanation
Correct option: B
mass of the projectile
Option B (mass of the projectile) is correct because the escape velocity formula does not include the mass of the projectile.
249 Easy
249 of 255
Understanding +4 | -1
The gravitational field due to a uniform solid sphere of mass MM and radius aa at the centre of the sphere is
View answer and explanation
Correct option: D
zero
Option D (zero) is correct because the gravitational field at the center of a uniform solid sphere is zero due to symmetry.
250 Easy
250 of 255
Application +4 | -1
What should be the period of rotation of earth so as to make any object on the equator weigh half of its present value?
View answer and explanation
Correct option: A
2
The correct answer is A (2) because it represents the factor by which the Earth's rotation period must decrease to halve the effective weight at the equator.
251 Easy
251 of 255
Analyzing +4 | -1
Assertion: For a satellite revolving very near to earth's surface the time period of revolution is given by 1 h 24 min1 \mathrm{~h} \, 24 \mathrm{~min}.
Reason: The period of revolution of a satellite depends only upon its height above the earth's surface.
View answer and explanation
Correct option: A
Assertion is True, Reason is True; Reason is a correct explanation for Assertion
Option B (7R) is correct because the distance of a geostationary satellite from the center of the Earth is approximately 42,000 km, which is about 7 times the Earth's radius.
252 Medium
252 of 255
Understanding +4 | -1
The centripetal force acting on a satellite orbiting round the earth and the gravitational force of earth acting on the satellite both equal FF. The net force on the satellite is
View answer and explanation
Correct option: B
FF
Option B (FF) is correct because the net force is the gravitational force, which provides the centripetal force.
253 Hard
253 of 255
Analyzing +4 | -1
Pertaining to two planets, the ratio of escape velocities from respective surfaces is 1:21: 2, the ratio of the time period of the same simple pendulum at their respective surfaces is 2:12: 1 (in same order). Then the ratio of their average densities is
View answer and explanation
Correct option: C
1:41: 4
Option C is correct. Given the ratio of escape velocities and time periods, we can derive the ratio of their average densities using the formulas for escape velocity and the time period of a simple pendulum.
254 Easy
254 of 255
Analyzing +4 | -1
Earth orbiting satellite will escape, if
View answer and explanation
Correct option: C
Both (a) and (b) are correct
Option C (Both (a) and (b) are correct) is correct because increasing the speed by 41% or doubling the kinetic energy will allow the satellite to reach escape velocity.
255 Medium
255 of 255
Understanding +4 | -1
For a body to escape from earth, angle at which it should be fired is?
View answer and explanation
Correct option: D
any angle
Option D is correct because the escape velocity is independent of the angle at which the body is projected. The body needs to have sufficient kinetic energy to overcome the gravitational potential energy, regardless of the angle.
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