$A$ 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)

The percentage decrease in the weight of a body when taken to a height of $48 \text{ km}$ above the surface of the earth is (Radius of the earth is $6400 \text{ km}$) (in $\%$)

If the radius of the Earth shrinks by $1.5 \%$ (mass remaining same),then the value of gravitational acceleration changes by ......... $\%$

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

Assertion $(A)$: $A$ particle of mass $m$ dropped into a hole made along the diameter of the Earth from one end to the other possesses simple harmonic motion.
Reason $(R)$: Gravitational force between any two particles is inversely proportional to the square of the distance between them.

$A$ tunnel is dug along a diameter of the Earth. If $M_e$ and $R_e$ are the mass and radius of the Earth respectively,then the force on a particle of mass $m$ placed in the tunnel at a distance $r$ from the center is: