$A$ pendulum is oscillating with frequency $n$ on the surface of the Earth. If it is taken to a depth $d = \frac{R}{2}$ below the surface of the Earth,where $R$ is the radius of the Earth,what is the new frequency of oscillations at this depth?

The density of a newly discovered planet is twice that of Earth. The acceleration due to gravity at the surface of the planet is equal to that at the surface of the Earth. If the radius of the Earth is $R$,the radius of the planet would be

An object is taken to a height above the surface of the Earth at a distance of $\frac{5}{4} R$ from the center of the Earth,where the radius of the Earth is $R = 6400 \, km$. The percentage decrease in the weight of the object will be $....\%$.

If the earth were to cease rotating about its own axis,the increase in the value of $g$ in the $C.G.S.$ system at a place of latitude of $45^{\circ}$ will be ........ $cm/sec^{2}$.

At what height over the earth's pole,the free fall acceleration decreases by one percent? (Assume the radius of earth to be $6400 \, km$)

Given below are two statements. One is labelled as Assertion $(A)$ and the other is labelled as Reason $(R)$.
Assertion $(A) :$ $A$ simple pendulum is taken to a planet of mass and radius,$4$ times and $2$ times,respectively,than the Earth. The time period of the pendulum remains the same on Earth and the planet.
Reason $(R) :$ The mass of the pendulum remains unchanged at Earth and the other planet. In the light of the above statements,choose the correct answer from the options given below $:$