What is the change in mass of a body, when taken $64 \ km$ below the surface of the earth? $[$Take radius of the earth as $6400 \ km]$

The speed with which the earth would have to rotate about its axis so that a person on the equator would weigh $\frac{1}{6}$ th as much as at present is ($g=$ gravitational acceleration,$R=$ equatorial radius of the earth).

The density of a newly discovered planet is twice that of Earth. The acceleration due to gravity at the surface of the planet is equal to that at the surface of the Earth. If the radius of the Earth is $R$,the radius of the planet would be

Find the angular speed of rotation of the Earth,so that the apparent $g$ at the equator becomes $(1/6)$th of its original value. $(R = 6.4 \times 10^6 \ m)$

The density of a new planet is twice that of earth. The acceleration due to gravity at the surface of the planet is equal to that at the surface of earth. If $R$ is the radius of earth,then the radius of the planet would be

The time period of a simple pendulum is $T_1$ when on the Earth's surface and $T_2$ when taken to a height $h = 2R$ above the Earth's surface, where $R$ is the radius of the Earth. The ratio $T_1 : T_2$ is: