If $R$ is the radius of the earth and the acceleration due to gravity on the surface of earth is $g=\pi^2 \mathrm{~m} / \mathrm{s}^2$, then the length of the second's pendulum at a height $h=2 R$ from the surface of earth will be,:

Given below are two statements: one is labelled as Assertion $A$ and the other is labelled as Reason $R$.

Assertion $A:$ A pendulum clock when taken to Mount Everest becomes fast.

Reason $R:$ The value of $g$ (acceleration due to gravity) is less at Mount Everest than its value on the surface of earth.

In the light of the above statements, choose the most appropriate answer from the options given below

The acceleration due to gravity near the surface of a planet of radius $R$ and density $d$ is proportional to

If a man at the equator would weigh $(3/5)^{th}$ of his weight, the angular speed of the earth is

Let $\omega$ be the angular velocity of the earth’s rotation about its axis. Assume that the acceleration due to gravity on the earth’s surface has the same value at the equator and the poles. An object weighed at the equator gives the same reading as a reading taken at a depth d below earth’s surface at a pole $(d < < R)$ The value of $d$ is

Given below are two statements:

Statement $I:$ Rotation of the earth shows effect on the value of acceleration due to gravity $(g)$.

Statement $II:$ The effect of rotation of the earth on the value of $g$ at the equator is minimum and that at the pole is maximum.

In the light of the above statements, choose the correct answer from the options given below.