Gravitational acceleration on the surface of a planet is $\frac{\sqrt 6}{11}g$ , where $g$ is the gravitational acceleration on the surface of the earth. The average mass  density of the planet is $\frac{2}{3}\, times$ that of the earth. If the escape speed on the surface of the earth is taken to be $11\, kms^{-1}$, the escape speed on the surface of  the  planet in $kms^{-1}$ will be

The height ${ }^{\prime} h ^{\prime}$ at which the weight of a body will be the same as that at the same depth $'h'$ from the surface of the earth is (Radius of the earth is $R$ and effect of the rotation of the earth is neglected):

Given below are two statements: One is labelled as Assertion $A$ and the other is labelled as Reason $R$.

Assertion $A$: If we move from poles to equator, the direction of acceleration due to gravity of earth always points towards the center of earth without any variation in its magnitude.

Reason $R $: At equator, the direction of acceleration due to the gravity is towards the center of earth. In the light of above statements, choose the correct answer from the options given below

What is the depth at which the value of acceleration due to gravity becomes $\frac{1}{n}$ times the value that at the surface of earth? (radius of earth $=R$ )

Assuming the earth to be a sphere of uniform density the acceleration due to gravity