The radius of the earth is $6400\, km$ and $g = 10\,m/{\sec ^2}$. In order that a body of $5 \,kg$ weighs zero at the equator, the angular speed of the earth is

A box weighs $196 \;\mathrm{N}$ on a spring balance at the north pole. Its weight recorded on the same balance if it is shifted to the equator is close to ....... $N$

(Take $\mathrm{g}=10\; \mathrm{ms}^{-2}$ at the north pole and the radius of the earth $=6400\; \mathrm{km}$)

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

Define acceleration due to gravity. Give the magnitude of $g$ on the surface of earth.

If a man at the equator would weigh $(3/5)^{th}$ of his weight, the angular speed of the earth is

In the reported figure of earth, the value of acceleration due to gravity is same at point $A$ and $C$ but it is smaller than that of its value at point $B$ (surface of the earth). The value of OA $: AB$ will be $x : y .$ The value of $x$ is $\ldots \ldots \ldots$