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

Given below are two statements :

Statement$-I:$ Acceleration due to gravity is different at different places on the surface of earth.

Statement$-II:$ Acceleration due to gravity increases as we go down below the earth's surface.

In the light of the above statements, choose the correct answer from the options given below

Two masses $m_1$ and $m_2\, (m_1 < m_2)$ are released from rest from a finite distance. They start under their mutual gravitational attraction

At what distance above and below the surface of the earth a body will have same weight, (take radius of earth as $R$.)

The acceleration due to gravity near the surface of a planet of radius $R$ and density $d$ is proportional to

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