Weight of a body of mass $m$ decreases by $1\%$ when it is raised to height $h$ above the Earth's surface. If the body is taken to a depth $h$ in a mine,then its weight will

$A$ pendulum is oscillating with frequency $n$ on the surface of the Earth. If it is taken to a depth $d = \frac{R}{4}$ below the surface of the Earth,what is the new frequency of oscillation? ($R =$ radius of the Earth)

If the density of the earth is doubled keeping the radius constant,find the new acceleration due to gravity (in $m/s^2$)? $(g = 9.8 \ m/s^2)$

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.

At what height above the earth's surface does the acceleration due to gravity fall to $1\%$ of its value at the earth's surface (in $,R$)?

If $R$ is the radius of the earth and $g$ is the acceleration due to gravity on the earth's surface,the mean density of the earth is