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 $10^{-3} \ kgm ^{-1}$ 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 is $10 \ ms ^{-2}$ )

A certain planet completes one rotation about its axis in time $T$. The weight of an object placed at the equator on the planet's surface is a fraction $f(f$ is close to unity) of its weight recorded at a latitude of $60^{\circ}$. The density of the planet (assumed to be a uniform perfect sphere) is given by

A body weight $ W $ newton at the surface of the earth. Its weight at a height equal to half the radius of the earth will be

At a height of $10 \,km$ above the surface of earth, the value of acceleration due to gravity is the same as that of a particular depth below the surface of earth. Assuming uniform mass density for the earth, the depth is …………. $km$

A mass falls from a helght $h$ and its time of fall $t$ is recorded in terms of time period $T$ of a simple pendulum. On the surface of earth it is found that $t =2 T$. The entre setup is taken on the surface of another planet whose mass is half of that of earth and radius the same. Same experiment is repeated and corresponding times noted as $t'$ and $T'$.