The ratio of gravitational acceleration at height $3R$ to that at height $4R$ from the surface of the earth is (where $R$ is the radius of the earth):

Consider Earth to be a sphere of radius $R_e$ rotating about its own axis with angular speed $\omega$. If $g_{E}$ and $g_{P}$ are the accelerations due to gravity at the equator and the poles respectively,then $(g_{P}-g_{E})$ is given by $\left[\cos (0^{\circ})=\sin (\frac{\pi}{2})=1, \sin (0^{\circ})=\cos (\frac{\pi}{2})=0\right]$

The time taken by a simple pendulum for one oscillation is $T$ on the earth's surface. Its time period becomes $xT$ when taken to a height $R$ (equal to the earth's radius) above the earth's surface. The value of $x$ is

The acceleration due to gravity on the planet $A$ is $9$ times the acceleration due to gravity on planet $B$. $A$ man jumps to a height of $2 \ m$ on the surface of $A$. What is the height of jump by the same person on the planet $B$?

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.

$T$ is the time period of a simple pendulum on the Earth's surface. Its time period becomes $xT$ when taken to a height $R$ (equal to the Earth's radius) above the Earth's surface. Then,the value of $x$ will be: