The period of a seconds pendulum on a planet,whose mass and radius are three times that of Earth,is

The time period of a simple pendulum on the surface of the earth is $T$. The height above the surface of the earth at which the time period of the pendulum becomes $2T$ is (Radius of the earth $= 6400 \text{ km}$) (in $\text{ km}$)

$A$ 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)

$A$ body weighs $500 \, N$ on the surface of the earth. How much would it weigh halfway below the surface of the earth (in $, N$)?

If the mass and radius of a planet are double those of the Earth,what will be the acceleration due to gravity on this planet compared to the acceleration due to gravity on Earth?

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.