$A$ planet has the same density as that of the Earth,and the universal gravitational constant $G$ is twice that of the Earth. The ratio of the acceleration due to gravity on the planet to that on the Earth is:

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

The value of acceleration due to gravity at Earth's surface is $9.8\, m\,s^{-2}$. The altitude above its surface at which the acceleration due to gravity decreases to $4.9\, m\,s^{-2}$,is close to: (Radius of Earth $= 6.4 \times 10^6\, m$)

$A$ simple pendulum is placed at a place where its distance from the earth's surface is equal to the radius of the earth. If the length of the string is $4 \ m$,then the time period of small oscillations will be . . . . . . $s$. [Take $g = \pi^2 \ ms^{-2}$]

The mass of a spherical planet is $4$ times the mass of the earth, but its radius $(R)$ is the same as that of the earth. How much work is done in lifting a body of mass $5 \,kg$ through a distance of $2 \,m$ on the planet (in $\,J$)? (Take $g = 10 \,ms^{-2}$ for Earth)

Given below are two statements: one is labelled as Assertion $(A)$ and the other is labelled as Reason $(R).$
Assertion $(A):$ Time period of a simple pendulum is longer at the top of a mountain than that at the base of the mountain.
Reason $(R):$ Time period of a simple pendulum decreases with increasing value of acceleration due to gravity and vice-versa.
In the light of the above statements,choose the most appropriate answer from the options given below: