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JEE • PHYSICS • GRAVITATION

Gravitation practice · Medium

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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 151-180 of 392 questions

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151 Medium
151 of 392
Remembering +4 | -1
Kepler's law states that, square of the time period of any planet moving around the sun in an elliptical orbit of semi-major axis (R)(R) is directly proportional to
View answer and explanation
Correct option: C
R3{R}^3
Option C (R3{R}^3) is correct because Kepler's third law states that the square of the period is proportional to the cube of the semi-major axis.
152 Medium
152 of 392
Application +4 | -1
An object is allowed to fall from a height RR above the earth, where R\mathrm{R} is the radius of earth. Its velocity when it strikes the earth's surface, ignoring air resistance, will be
View answer and explanation
Correct option: C
gR\sqrt{g R}
Option C (gR\sqrt{g R}) is correct because the velocity is derived from the conservation of energy, considering the change in gravitational potential energy.
153 Medium
153 of 392
Analyzing +4 | -1
Assertion: A body kept inside a spherical shell does not experience any gravitational force.
Reason: The body inside a spherical shell is protected from the gravitational attraction of bodies outside the shell.
View answer and explanation
Correct option: D
If both Assertion & Reason are false
Option D is correct because a body inside a spherical shell experiences no net gravitational force, and this is because of the symmetrical distribution of mass, not protection from external bodies.
154 Medium
154 of 392
Understanding +4 | -1
A satellite of mass mm moves along an elliptical path around earth. Its areal velocity is
View answer and explanation
Correct option: C
Independent of its mass
Option C (Independent of its mass) is correct because the areal velocity of a satellite depends only on its angular momentum and is independent of its mass.
155 Medium
155 of 392
Understanding +4 | -1
Assertion: The universal gravitational constant is same as acceleration due to gravity.
Reason: Gravitational constant and acceleration due to gravity have same dimensional formula.
View answer and explanation
Correct option: D
If both assertion and reason are false
Option D (If both assertion and reason are false) is correct because the universal gravitational constant and acceleration due to gravity are different physical quantities with different dimensions.
156 Medium
156 of 392
Application +4 | -1
Gravitational acceleration on the surface of a planet is 611  g,  \frac{{\sqrt 6 }}{{11}}\mathrm{\;}g,\;, where gg is the gravitational acceleration on the surface of earth. The average mass density of the planet is 23\frac{2}{3} times
that of the earth. If the escape speed on the surface of the earth is taken on be 11  kms−1,11\mathrm{\;km}{\mathrm{s}^{ - 1}},
the escape speed on the surface of the planet in kms−1  \mathrm{km}{\mathrm{s}^{ - 1}}\mathrm{\;} will be
View answer and explanation
Correct option: C
3
Option C (3) is correct because the escape speed is proportional to the square root of gR, and using the given ratios, the escape speed on the planet is 3 km/s.
157 Medium
157 of 392
Application +4 | -1
Find the velocity of escape at the earth given that its radius is 6.4×106 m6.4 \times 10^6 \mathrm{~m} and the value of gg at its surface is 9.8 ms−29.8 \mathrm{~ms}^{-2}.
View answer and explanation
Correct option: A
11.2 kms−111.2 \mathrm{~kms}^{-1}
Option A (11.2 kms−111.2 \mathrm{~kms}^{-1}) is correct because it is the result of applying the escape velocity formula using the given values of Earth's radius and gravitational acceleration.
158 Medium
158 of 392
Understanding +4 | -1
A man starts walking from a point on the surface of earth (assumed smooth) and reaches diagonally opposite point. What is the work done by him?
View answer and explanation
Correct option: A
Zero
Option A (Zero) is correct because the gravitational potential energy at the starting and ending points are the same, resulting in zero net work done.
159 Medium
159 of 392
Understanding +4 | -1
A satellite is orbiting around the earth with orbital radius RR and time period TT. The quantity which remain constant is
View answer and explanation
Correct option: D
T2R3\frac{{{T^2}}}{{{R^3}}}
Option D (T2R3\frac{{{T^2}}}{{{R^3}}}) is correct because Kepler's third law states that the square of the orbital period is proportional to the cube of the orbital radius.
160 Medium
160 of 392
Application +4 | -1
Two point masses 2 g2\,g and 10 g10\,g are separated by 10cm10cm. The gravitational potential energy of system is
View answer and explanation
Correct option: B
−G5000\frac{{ - G}}{{5000}}
Option B (-G/5000) is correct because it correctly applies the formula for gravitational potential energy with the correct sign and unit conversion.
161 Medium
161 of 392
Understanding +4 | -1
The acceleration due to gravity at height hh above the earth if h<<Rh < < R (Radius of earth) is given by
View answer and explanation
Correct option: C
g′=g(1−2hR)g^{\prime}=g\left(1-\frac{2 h}{R}\right)
Option C (g′=g(1−2hR)g^{\prime}=g\left(1-\frac{{2 h}}{{R}}\right)) is correct because it represents the approximate acceleration due to gravity at a height h << R above the Earth's surface.
162 Medium
162 of 392
Application +4 | -1
The largest and the shortest distance of the earth from the sun are r1 and r2{r_1}\,and\,{r_2}, its distance from the sun when it is at the perpendicular to the major axis of the orbit drawn from the sun
View answer and explanation
Correct option: C
2r1r2r1+r2\frac{{2{r_1}{r_2}}}{{{r_1} + {r_2}}}
The correct answer is C (2r1r2r1+r2\frac{{2{r_1}{r_2}}}{{{r_1} + {r_2}}}) because it correctly represents the semi-latus rectum of the elliptical orbit.
163 Medium
163 of 392
Understanding +4 | -1
A particle is projected from earth with speed less than the escape speed. The sum of KEKE and PEPE of particle is
View answer and explanation
Correct option: B
Negative and constant
Option B (Negative and constant) is correct because when a particle is projected with a speed less than the escape speed, its total mechanical energy (KE + PE) is negative and remains constant due to conservation of energy.
164 Medium
164 of 392
Understanding +4 | -1
A spring balance is graduated on sea level. If a body is weighed with this balance at consecutively increasing heights from earth’s surface, the weight indicated by the balance
View answer and explanation
Correct option: B
Will go on decreasing continuously
Option B (Will go on decreasing continuously) is correct because the value of g decreases with increasing height, thus the weight indicated by the balance will decrease.
165 Medium
165 of 392
Understanding +4 | -1
An interstellar spacecraft far away from the influence of any star or planet is moving at high speed under the influence of fusion rockets (due to thrust exerted by fusion rockets, the spacecrafts is accelerating). Suddenly the engine malfunctions and stops. The spacecraft will
View answer and explanation
Correct option: D
Keeps moving forever with constant speed
Option D (Keeps moving forever with constant speed) is correct because, according to Newton's first law, an object in motion will stay in motion with constant velocity unless acted upon by an external force.
166 Medium
166 of 392
Analyzing +4 | -1
Assertion: If value of acceleration due to gravity of a planet is lesser than earth then escape velocity from planet is lesser too.
Reason: Escape velocity just depends only on acceleration due to gravity.
View answer and explanation
Correct option: D
If both Assertion & Reason are false
Option D is correct because both the assertion and the reason are false. Escape velocity depends on both the acceleration due to gravity and the radius of the planet.
167 Medium
167 of 392
Analyzing +4 | -1
In an experiment, a boy draws graph between v2(y−v^2(y- axis) and a2{a}^2 (x-axis) (where v={v}= velocity and a={a}= tangential acceleration) for a simple pendulum. The graph is found to be a straight line of negative slope making an angle of 30∘30^{\circ} (with x{x}-axis) when experiment was done on the ground and 60∘60^{\circ} (with x{x} axis) when experiment was done at height hh above the ground. Then hh must be (R=({R}= radius of earth ))
View answer and explanation
Correct option: C
0.73R0.73{R}
Option C (0.73R0.73{R}) is correct because it is derived from the relationship between tangential acceleration, velocity, and the radius of the Earth, considering the change in gravity with height.
168 Medium
168 of 392
Application +4 | -1
The geostationary orbit of the earth is at a distance of about 36000 km36000 \mathrm{~km} from the earth's surface. Find the weight of a 120−kg120-\mathrm{kg} equipment placed in a geostationary satellite. The radius of the earth is 6400 km6400 \mathrm{~km}.
View answer and explanation
Correct option: A
27 N27 \mathrm{~N}
Option A (27 N) is correct because the weight of the equipment in the geostationary satellite is calculated using the reduced gravitational acceleration at that altitude.
169 Medium
169 of 392
Application +4 | -1
An artificial satellite circled around the earth at a distance of 3400 km3400 \mathrm{~km}. Calculate its orbital velocity and period of revolution. Radius of earth =6400 km=6400 \mathrm{~km} and g=9.8 ms−2g=9.8 \mathrm{~ms}^{-2}.
View answer and explanation
Correct option: A
6400 ms−1,9621 s6400 \mathrm{~ms}^{-1}, 9621 \mathrm{~s}
Option A (6400 ms−1,9621 s6400 \mathrm{~ms}^{-1}, 9621 \mathrm{~s}) is correct because it is calculated using the formulas for orbital velocity and period of revolution.
170 Medium
170 of 392
Understanding +4 | -1
A satellite in its orbit around earth is weightless on account of its
View answer and explanation
Correct option: D
acceleration
Option D is correct because the satellite's weightlessness is due to its continuous freefall, resulting in a centripetal acceleration.
171 Medium
171 of 392
Understanding +4 | -1
Find the unit of the ratio of magnitude of gravitational intensity and gravitational force:-
View answer and explanation
Correct option: B
1kg\frac{1}{{kg}}
Option B (1kg\frac{1}{{kg}}) is correct because gravitational intensity is force per unit mass, so the ratio is inverse of mass.
172 Medium
172 of 392
Understanding +4 | -1
For a satellite moving in an orbit around the planet, the ratio of magnitude of potential energy to kinetic energy is
View answer and explanation
Correct option: A
2:12: 1
Option A (2:12: 1) is correct because the magnitude of potential energy is twice the kinetic energy for a satellite in orbit.
173 Medium
173 of 392
Understanding +4 | -1
Assertion: Planet is heavenly body revolving round the sun.

Reason: Star is luminous body made of gaseous material.
View answer and explanation
Correct option: B
If both assertion and reason are true but reason is not correct explanation of the assertion.
Option B (If both assertion and reason are true but reason is not correct explanation of the assertion) is correct because both statements are true, but the reason doesn't explain the assertion.
174 Medium
174 of 392
Understanding +4 | -1
If the distance between two masses is doubled, the gravitational attraction between them
View answer and explanation
Correct option: D
Is reduced to a quarter
Option D (Is reduced to a quarter) is correct because gravitational force is inversely proportional to the square of the distance between the masses.
175 Medium
175 of 392
Application +4 | -1
The escape velocity from the earth is 11  km/s11\,\,km/s . The escape velocity from a planet having twice the radius and same density as that of the earth is (in km/skm/s )
View answer and explanation
Correct option: A
22
Option A (22) is correct because escape velocity is proportional to radius times the square root of density, and with twice the radius and same density, it doubles.
176 Medium
176 of 392
Application +4 | -1
The acceleration due to gravity increase by 0.5%0.5\% when we go from the equator to the poles. What will be the time period of the pendulum at the equator which beats seconds at the poles?
View answer and explanation
Correct option: D
2.005 s
Option D (2.005 s) is correct because the time period of a pendulum is inversely proportional to the square root of the acceleration due to gravity, and the change in time period is calculated based on the given percentage change in gravity.
177 Medium
177 of 392
Application +4 | -1
A particle of mass 1 kg1 \mathrm{~kg} is kept on the surface of a uniform sphere of mass 20 kg20 \mathrm{~kg} and radius 1.0 m1.0 \mathrm{~m}. Find the work to be done against the gravitational force between them to take the particle away from the sphere.
View answer and explanation
Correct option: A
1.334×10−9 J1.334 \times 10^{-9} \mathrm{~J}
Option A (1.334×10−9 J1.334 \times 10^{-9} \mathrm{~J}) is correct because it calculates the work done against gravity to move the particle away from the sphere.
178 Medium
178 of 392
Understanding +4 | -1
Comets moves around the Sun in a highly elliptical orbit. The gravitational force on the comet due to the Sun is not normal to the comet’s velocity in general. Yet the work done by the gravitational force over every orbit of the comet is zero. Why?
View answer and explanation
Correct option: A
The gravitational force is conservative, hence work done is zero
Option A is correct because gravitational force is a conservative force, and the work done by a conservative force over a closed path (like an orbit) is zero.
179 Medium
179 of 392
Analyzing +4 | -1
A rocket is intended to leave the Earth’s gravitational field. The
fuel in its main engine is a little less than the amount that is
necessary and an auxiliary engine, (only capable of operating
for a short time) has to be used as well. When is it best to switch
on: the auxiliary engine?
View answer and explanation
Correct option: A
At take-off
Option A (At take-off) is correct because the auxiliary engine should provide maximum energy at the start to overcome Earth's gravitational field most effectively.
180 Medium
180 of 392
Application +4 | -1
A body of mass MM is divided into two parts of masses mm and (M−m)(M-m). If they are kept at a constant distance, then the relation between mm and MM for maximum force of attraction between them, will be
View answer and explanation
Correct option: D
m=M2m=\frac{M}{2}
Option D (m=M2m=\frac{M}{2}) is correct because the gravitational force is maximized when the product of the two masses is maximized, which occurs when the masses are equal.
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