The time period of a simple pendulum increases by a factor of $\sqrt{2}$ at a height $h$ from the surface of the Earth. Then the value of $h$ is:

Assume that the Earth is a solid sphere of uniform density and a tunnel is dug along its diameter throughout the Earth. It is found that when a particle is released in this tunnel, it executes a simple harmonic motion. The mass of the particle is $100 \, g$. The time period of the motion of the particle will be (approximately) (take $g = 10 \, m/s^2$, radius of Earth $R = 6400 \, km$):

The height at which the acceleration due to gravity becomes $\frac{g}{9}$ (where $g$ = the acceleration due to gravity on the surface of the earth) in terms of $R$,the radius of the earth,is

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.

The depth $d$ at which the value of acceleration due to gravity becomes $\frac{1}{n}$ times the value at the surface is [$R =$ radius of the earth].

The difference in the acceleration due to gravity at the pole and equator is ( $g=$ acceleration due to gravity,$R=$ radius of earth,$\theta=$ latitude,$\omega=$ angular velocity,$\cos 0^{\circ}=1, \cos 90^{\circ}=0$ ).