$A$ mine is located at depth $\frac{R}{3}$ below the earth's surface. The acceleration due to gravity at that depth in the mine is ($R = \text{radius of earth}$,$g = \text{acceleration due to gravity at surface}$).

$A$ pendulum is oscillating with frequency $n$ on the surface of the Earth. If it is taken to a depth $d = R/3$ below the surface of the Earth,what is the new frequency of oscillation? ($R$ is the radius of the Earth)

$A$ newly discovered planet has a density eight times the density of the earth and a radius twice the radius of the earth. The time taken by a $2\, kg$ mass to fall freely through a distance $S$ near the surface of the earth is $1$ second. Then the time taken for a $4\, kg$ mass to fall freely through the same distance $S$ near the surface of the new planet is ....... $\sec$.

$T$ is the time period of a simple pendulum on the Earth's surface. Its time period becomes $xT$ when taken to a height $R$ (equal to the Earth's radius) above the Earth's surface. Then,the value of $x$ will be:

Which of the following graphs represents the variation of acceleration due to gravity $(g)$ with distance $r$ from the centre of the 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 $\,N$)?