$A$ newly discovered planet has a density eight times the density of the earth and a radius twice the radius of the earth. The time taken by a $2\, kg$ mass to fall freely through a distance $S$ near the surface of the earth is $1$ second. Then the time taken for a $4\, kg$ mass to fall freely through the same distance $S$ near the surface of the new planet is ....... $\sec$.

At what height above the earth's surface does the acceleration due to gravity fall to $1\%$ of its value at the earth's surface (in $,R$)?

$A$ body is taken to a height of $n R$ from the surface of the earth. The ratio of the acceleration due to gravity on the surface to that at the altitude is

The radii of two planets $A$ and $B$ are $R$ and $4R$ and their densities are $\rho$ and $\rho/3$ respectively. The ratio of acceleration due to gravity at their surfaces $(g_A : g_B)$ will be

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 $:$

The ratio of the accelerations due to gravity at heights $1280 \ km$ and $3200 \ km$ above the surface of the earth is (Radius of the earth $= 6400 \ km$)