At a height $R$ above the earth's surface,the gravitational acceleration is ($R$ = radius of earth,$g$ = acceleration due to gravity on earth's surface).

If the radius of the Earth is $R$,then the height $h$ at which the value of $g$ becomes one-fourth is:

Given below are two statements:
Statement-$I$: Acceleration due to gravity is different at different places on the surface of Earth.
Statement-$II$: Acceleration due to gravity increases as we go down below the Earth's surface.
In the light of the above statements,choose the correct answer from the options given below:

The mass and diameter of a planet are twice those of earth. The period of oscillation of a pendulum on this planet will be (if it is a second's pendulum on earth).

Assuming the earth to be a sphere of uniform mass density, a body weighs $300 \,N$ on the surface of the earth. How much would it weigh at a depth of $R/4$ below the surface of the earth (in $\,N$)?

The mass of a planet and its diameter are three times those of the Earth. Then the acceleration due to gravity on the surface of the planet is ....... $m/s^2$.