If the radius of earth shrinks by $2 \%$ while its mass remains the same,the acceleration due to gravity on the earth's surface will approximately:

If a body takes time $t$ to reach the ground when dropped from a height $h$ on Earth,how much time will it take to reach the ground when dropped from the same height $h$ on the Moon?

$A$ simple pendulum is oscillating with frequency $F$ on the surface of the earth. It is taken to a depth $R/3$ below the surface of the earth ($R =$ radius of earth). The frequency of oscillation at depth $R/3$ is

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,$\cos 0^{\circ}=1, \cos 90^{\circ}=0$ ).

Given below are two statements:
Statement $I$: The law of gravitation holds good for any pair of bodies in the universe.
Statement $II$: The weight of any person becomes zero when the person is at the centre of the earth.
In the light of the above statements,choose the correct answer from the options given below.

The approximate height from the surface of earth at which the weight of the body becomes $\frac{1}{3}$ of its weight on the surface of earth is $.......... \, km$ : [Radius of earth $R = 6400 \, km$ and $\sqrt{3} = 1.732$]