The time period of a simple pendulum on the surface of the earth is $T$. If the pendulum is taken to a height equal to half of the radius of the earth,then its time period is

An astronaut inside a small spaceship orbiting around the earth cannot detect gravity. If the space station orbiting around the earth has a large size,can he hope to detect gravity?

If the angular velocity of earth's spin is increased such that the bodies at the equator start floating,the duration of the day would be approximately ........ minutes
(Take: $g = 10 \, m/s^2$,the radius of earth,$R = 6400 \times 10^3 \, m$,Take $\pi = 3.14$)

If the radius of the earth were to shrink by $1\%$ with its mass remaining the same,the acceleration due to gravity on the earth's surface would

Choose the correct alternative:
$(a)$ Acceleration due to gravity increases/decreases with increasing altitude.
$(b)$ Acceleration due to gravity increases/decreases with increasing depth (assume the Earth to be a sphere of uniform density).
$(c)$ Acceleration due to gravity is independent of mass of the Earth/mass of the body.
$(d)$ The formula $-G M m(1 / r_{2}-1 / r_{1})$ is more/less accurate than the formula $m g(r_{2}-r_{1})$ for the difference of potential energy between two points $r_{2}$ and $r_{1}$ distance away from the centre of the Earth.

Assuming the earth to be a sphere of uniform mass density, a body weighs $300 \,N$ on the surface of the earth. How much would it weigh at a depth of $R/4$ below the surface of the earth (in $\,N$)?