$A$ simple pendulum has a time period $T_1$ when on the earth's surface and $T_2$ when taken to a height $R$ above the earth's surface,where $R$ is the radius of the earth. The value of $T_2/T_1$ is

The value of $g$ on the Earth's surface is $980 \, cm/s^2$. Its value at a height of $64 \, km$ from the Earth's surface is ........ $cm/s^2$ (Radius of the Earth $R = 6400 \, km$).

Assume that the Earth is a solid sphere of uniform density and a tunnel is dug along its diameter throughout the Earth. It is found that when a particle is released in this tunnel, it executes a simple harmonic motion. The mass of the particle is $100 \, g$. The time period of the motion of the particle will be (approximately) (take $g = 10 \, m/s^2$, radius of Earth $R = 6400 \, km$):

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:

Two planets $A$ and $B$ of radii $R$ and $1.5 R$ have densities $\rho$ and $\rho / 2$ respectively. The ratio of acceleration due to gravity at the surface of $B$ to $A$ is:

If $g_E$ and $g_M$ are the accelerations due to gravity on the surfaces of the earth and the moon respectively and if Millikan's oil drop experiment could be performed on the two surfaces,one will find the ratio (electronic charge on the moon / electronic charge on the earth) to be