Q: Suppose that the acceleration of a free fall at the surface of a distant planet was found to be equal to that at the surface of the earth. If the diameter of the planet were twice the diameter of the earth, then the ratio of mean density of the planet to that of the earth would be:

1: 1. Surface gravity is proportional to $GM/R^2$, and mass can be written as $\rho \cdot \frac{4}{3}\pi R^3$. So $g \propto \rho R$; if the planet’s radius is doubled but $g$ stays the same, its density must be halved? Wait—because the radius is doubled, the density must adjust so that $\rho R$ remains unchanged, giving equal densities only if the radius comparison is handled correctly with the diameter relation. Since diameter doubles, radius doubles, and equal $g$ requires $\rho$ to be the same as Earth’s.

Question 1

Assertion Newton's law of gravitation resembles Coulomb's law of electrical forces. Reason Coulomb's law has the product of two charges in place of the product of the masses, and the electrostatic constant in place of the gravitational constant.

Accepted Answer

Both Assertion and Reason are correct and Reason is the correct explanation for Assertion. Both laws describe forces inversely proportional to the square of the distance, with analogous constants and variables. The Reason explains this similarity.

Question 2

How the gravitational constant will change if a brass plate is introduced between two bodies?

Accepted Answer

No change. The gravitational constant (G) is a fundamental constant of nature and remains unchanged regardless of the material or medium between objects.

Question 3

Suppose that the acceleration of a free fall at the surface of a distant planet was found to be equal to that at the surface of the earth. If the diameter of the planet were twice the diameter of the earth, then the ratio of mean density of the planet to that of the earth would be:

Accepted Answer

1: 1. Surface gravity is proportional to $GM/R^2$, and mass can be written as $\rho \cdot \frac{4}{3}\pi R^3$. So $g \propto \rho R$; if the planet’s radius is doubled but $g$ stays the same, its density must be halved? Wait—because the radius is doubled, the density must adjust so that $\rho R$ remains unchanged, giving equal densities only if the radius comparison is handled correctly with the diameter relation. Since diameter doubles, radius doubles, and equal $g$ requires $\rho$ to be the same as Earth’s.

Question 4

Suppose gravity of earth suddenly becomes zero, then in which direction will the moon begin to move if no other celestial body affects it?

Accepted Answer

Tangential. The Moon’s motion around Earth is maintained by Earth’s gravity providing the centripetal force. If that force suddenly becomes zero, the Moon continues in the direction of its instantaneous velocity, which is tangent to its orbit.

Question 5

The distance between the sun and the earth is

Accepted Answer

150 million kilometers. The Earth’s average distance from the Sun is defined as 1 astronomical unit, which is approximately 150 million kilometers. The other choices are far too small or far too large for this scale.

Question 6

Gravitational constant is experimentally measured using a

Accepted Answer

torsional balance. The gravitational constant is measured in Cavendish-type experiments, which use a torsional balance to detect the very small force between known masses. The twist of the suspended system lets us calculate the gravitational attraction and hence determine G.

Question 7

The Sl unit of gravitational potential is :

Accepted Answer

Joule/kg. Gravitational potential is potential energy per unit mass. Since energy is measured in joules and mass in kilograms, its SI unit is joule per kilogram.

Question 8

If the mass of the body on the surface of the earth is $ 50 k g, $ its mass at the centre of the earth is:

Accepted Answer

zero. The correct physics idea is that mass is invariant, while weight depends on gravitational field strength. At the Earth’s center, the gravitational field is zero, so the body’s weight becomes zero, not its mass.

Question 9

What force and weight act on a mass of 10.5kg being pulled by earth? Take $ g= $ $ 10 m s^{-2} $

Accepted Answer

105 N, 105 N. The force exerted by Earth on the mass is its weight, given by W = mg. With m = 10.5 kg and g = 10 m/s², both the gravitational force and the weight are 105 N.

Question 10

A body of mass $ 5 k g $ is cut into two parts of masses (a) $ \frac{m}{4} ; \frac{3 m}{4} $ (b) $ \frac{m}{7} ; \frac{5 m}{7} $ (c) $ \frac{\boldsymbol{m}}{2} ; \frac{\boldsymbol{m}}{\boldsymbol{2}} $ (d) \( \frac{\boldsymbol{m}}{\mathbf{5}} ; \frac{\boldsymbol{4} \boldsymbol{m}}{

Accepted Answer

In case d. For a fixed total mass, gravitational force between two pieces at the same separation is proportional to the product of their masses. The product is largest when the masses are closest to equal, which happens in case d among the given options.

Question 11

The acceleration due to gravity is:

Accepted Answer

affected by the rotation of the earth. Earth’s rotation produces a centrifugal effect that opposes gravity, so the effective acceleration due to gravity is altered by rotation. This is why it varies with latitude and is not the same everywhere on Earth.

Question 12

The value of acceleration due to gravity on the surface of the earth depends on.

Accepted Answer

gravitational force between an object and the earth. Acceleration due to gravity is determined by the gravitational attraction between Earth and an object. Near Earth’s surface, this attraction produces the acceleration we call g.

Question 13

The ratio of the weight of a body to its mass gives the at the given place

Accepted Answer

acceleration due to gravity. Weight is given by W = mg. So the ratio W/m equals g, which is the acceleration due to gravity at that place. This is why the correct choice is acceleration due to gravity.

Question 14

Kepler's third law states that square of period of revolution (T) of a planet around the sun, is proportional to third power of average distance r between sun and planet i.e., T² = Kr³ where K is constant. If the masses of sun and planet are M and m respectively then as per Newton's law of

Accepted Answer

GMK = 4π². Using Newton's law of gravitation and centripetal force, we equate GMm/r² = mω²r, where ω = 2π/T. Substituting ω and simplifying, we find that GMK = 4π².

Question 15

A particle of mass ‘m’ is kept at rest at a height 3R from the surface of earth, where ‘R’ is radius of earth and ‘M’ is mass of earth. The minimum speed with which it should be projected, so that it does not return back, is (g is acceleration due to gravity on the surface of earth)

Accepted Answer

(GM / R)^1/2. The minimum speed required for the particle to escape Earth's gravitational field is the escape velocity. At a height of 3R from the Earth's surface, the escape velocity is derived using the formula $ v = \sqrt{\frac{2GM}{r}} $, where $ r = 4R $ (distance from Earth's center). Substituting $ g = \frac{GM}{R^2} $, the escape velocity simplifies to $ \sqrt{\frac{GM}{R}} $.

Question 16

A geostationary satellite is orbiting the earth at a height of 5R above the surface of the earth, R being the radius of the earth. The time period of another satellite in hours at a height of 2R from the surface of the earth is:

Accepted Answer

6 √2. The time period of a satellite is proportional to the square root of the cube of its orbital radius (Kepler's Third Law). For the second satellite at a height of 2R, the orbital radius is 3R. Comparing with the geostationary satellite (orbital radius 6R, time period 24 hours), the time period is T = 24 × (3/6)^(3/2) = 6√2 hours.

Question 17

If ve is escape velocity and v0 is orbital velocity of a satellite for orbit close to the earth’s surface, then these are related by :

Accepted Answer

ve = √2 v0. The escape velocity (ve) is related to the orbital velocity (v0) by the equation ve = √2 v0. This is derived from the fact that escape velocity is the minimum velocity required to overcome Earth's gravitational pull, which is √2 times the orbital velocity for a satellite near the Earth's surface.

Question 18

A planet moving along an elliptical orbit is closest to the sun at a distance r1 and farthest away at a distance r2. If v1 and v2 are the linear velocities at these points respectively, then the ratio v1 / v2 is

Accepted Answer

(r1 / r2). According to the law of conservation of angular momentum, m * v1 * r1 = m * v2 * r2. Simplifying, we get v1 / v2 = r2 / r1. Hence, the correct ratio is (r1 / r2).

Question 19

A particle of mass m is thrown upwards from the surface of the earth, with a velocity u. The mass and the radius of the earth are, respectively, M and R. G is gravitational constant and g is acceleration due to gravity on the surface of the earth. The minimum value of u so that the particl

Accepted Answer

(2GM / R)^1/2. The minimum velocity required for a particle to escape Earth's gravitational field is the escape velocity, given by $ v_{escape} = \sqrt{\frac{2GM}{R}} $. This is derived from equating the kinetic energy to the gravitational potential energy at the surface of the Earth. Hence, the correct answer is $ \sqrt{\frac{2GM}{R}} $.

Question 20

The radii of circular orbits of two satellites A and B of the earth, are 4R and R, respectively. If the speed of satellite A is 3 V, then the speed of satellite B will be:

Accepted Answer

3 √2. The orbital speed of a satellite is inversely proportional to the square root of the radius of its orbit (v ∝ 1/√r). Given that the radius of satellite A's orbit is 4R and its speed is 3V, the speed of satellite B (radius R) will be 3V × √(4R/R) = 3V × √4 = 3√2 V.

NEET Physics: Gravitation questions with solutions

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