$A$ 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)

Given below are two statements: one is labelled as Assertion $(A)$ and the other is labelled as Reason $(R).$
Assertion $(A):$ Time period of a simple pendulum is longer at the top of a mountain than that at the base of the mountain.
Reason $(R):$ Time period of a simple pendulum decreases with increasing value of acceleration due to gravity and vice-versa.
In the light of the above statements,choose the most appropriate answer from the options given below:

The radii of two planets $A$ and $B$ are $R$ and $4R$ and their densities are $\rho$ and $\rho/3$ respectively. The ratio of acceleration due to gravity at their surfaces $(g_A : g_B)$ will be

The weight of a body on the earth is $400\,N$. Then weight of the body when taken to a depth half of the radius of the earth will be ............ $N$.

When a body is taken from the equator to the poles,its weight

Two planets $A$ and $B$ have the same mass $M$ and radius $R$. The variation of density $\rho$ of the planets with distance $r$ from the center is shown in the following diagrams. The ratio of acceleration due to gravity at the surface of the planets $A$ and $B$ will be: