The radius of the Earth is $6400 \,km$ and the acceleration due to gravity is $g=10 \,ms^{-2}$. For the weight of a body of mass $5 \,kg$ to be zero at the equator,the rotational angular velocity of the Earth must be (in $rad/s$):

$A$ pendulum is oscillating with frequency $n$ on the surface of the Earth. If it is taken to a depth $d = \frac{R}{4}$ below the surface of the Earth,what is the new frequency of oscillation? ($R =$ radius of the Earth)

Acceleration due to gravity at the surface of a planet is equal to that at the surface of the Earth,and its density is $1.5$ times that of the Earth. If the radius of the Earth is $R$,the radius of the planet is:

Given below are two statements: one is labelled as Assertion $A$ and the other is labelled as Reason $R$.
Assertion $A:$ $A$ pendulum clock when taken to Mount Everest becomes fast.
Reason $R:$ The value of $g$ (acceleration due to gravity) is less at Mount Everest than its value on the surface of earth.
In the light of the above statements,choose the most appropriate answer from the options given below.

The height of any point $P$ above the surface of the Earth is equal to the diameter of the Earth. The value of acceleration due to gravity at point $P$ will be: (Given $g = $ acceleration due to gravity at the surface of the Earth)

Starting from the centre of the earth having radius $R$,the variation of $g$ (acceleration due to gravity) is shown by: