$T$ is the time period of a simple pendulum on the Earth's surface. Its time period becomes $xT$ when taken to a height $R$ (equal to the Earth's radius) above the Earth's surface. Then,the value of $x$ will be:

If a planet has a mass and radius both half that of the Earth,the acceleration due to gravity at its surface would be ......... $m/s^2$ ($g$ on Earth $= 9.8\, m/s^2$)

$A$ planet of radius $R = \frac{1}{10} \times$ (radius of Earth) has the same mass density as Earth. Scientists dig a well of depth $\frac{R}{5}$ on it and lower a wire of the same length and of linear mass density $\lambda = 10^{-3} \ kg \ m^{-1}$ into it. If the wire is not touching anywhere,the force applied at the top of the wire by a person holding it in place is (take the radius of Earth $= 6 \times 10^6 \ m$ and the acceleration due to gravity on Earth $g_e = 10 \ m \ s^{-2}$) (in $N$)

The acceleration due to gravity at a height $(1/20)^{th}$ of the radius of the Earth above the Earth's surface is $9 m s^{-2}$. Its value at an equal depth below the surface of the Earth is: (in $m s^{-2}$)

The radius of the Earth is approximately $6000 \, km$. The weight of a body at a height of $6000 \, km$ from the Earth's surface becomes:

Two planets have radii $R_1$ and $R_2$ and densities $\rho_1$ and $\rho_2$ respectively. Find the ratio of the acceleration due to gravity on these planets.