$A$ body weighs $200 \; N$ on the surface of the earth. How much will it weigh halfway down to the centre of the earth (in $; N$)?

If all objects on the equator of the earth feel weightless,then the duration of the day will nearly become ....... $hr$.

Let $g$ be the acceleration due to gravity at the Earth's surface and $K$ be the rotational kinetic energy of the Earth. Suppose the Earth's radius decreases by $2 \%$ keeping its mass the same,then:

If $R_{E}$ is the radius of the Earth,then the ratio between the acceleration due to gravity at a depth $r$ below and a height $r$ above the Earth's surface is: (Given: $r < R_{E}$)

$A$ 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 $g = \pi^2 \ ms^{-2}$]

Earth is assumed to be a sphere of radius $R$. If $g_{\phi}$ is the value of effective acceleration due to gravity at latitude $30^{\circ}$ and $g$ is the value at the equator,then the value of $|g - g_{\phi}|$ is ($\omega$ is the angular velocity of rotation of the Earth,$\cos 30^{\circ} = \frac{\sqrt{3}}{2}$).