The density of a newly discovered planet is twice that of Earth. The acceleration due to gravity at the surface of the planet is equal to that at the surface of the Earth. If the radius of the Earth is $R$,the radius of the planet would be

The value of acceleration due to gravity at a height of $10 \,km$ from the surface of the Earth is $x$. At what depth inside the Earth is the value of the acceleration due to gravity the same value $x$ (in $\,km$)?

The acceleration of a body due to the attraction of the earth (radius $R$) at a distance $2R$ from the surface of the earth is ($g =$ acceleration due to gravity at the surface of the earth).

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.

At what depth $d$ below the surface of the Earth does the acceleration due to gravity become $\frac{1}{n}$ times its value at the surface? ($R$ = radius of the Earth)

Derive the equation for the variation of acceleration due to gravity $g$ with depth $d$ below the surface of the Earth.