At what height above the Earth's surface is the acceleration due to gravity half of its value at the surface? (R is the radius of Earth)

At what height from above the surface of earth, then value of acceleration due to gravity reduces to half of its present value (radius of earth - $6400 \mathrm{~km}$)

Planet X has a mass $\frac{1}{10}$ that of Earth and a radius half that of Earth. If a pendulum has a period T on Earth, what would its period be on Planet X?

At what height $(h)$ from the surface of Earth, the value of acceleration due to gravity is one fourth of the value at a height $R$ above the surface? $(R=$ radius of earth)

The acceleration due to the gravity on the moon is one sixth that on earth. If the average density of moon is three fifth that of earth, the moon's radius in terms of earth's radius $\mathrm{R}_{\mathrm{e}}$ is

The frictional force due to air resistance on a body of mass $0.25 \mathrm{~kg}$ falling downwards with acceleration $9.2 \mathrm{~m} / \mathrm{s}^2$ is

The value of the acceleration due to gravity is ${g_1}$ at a height $h = \frac{R}{2}$ (where, $R =$ radius of the earth) from the surface of the earth. It is again equal to ${g_1}$ at a depth $d$ below the surface of the earth. The ratio $\left( {\frac{d}{R}} \right)$ equals

A smooth wedge $\mathrm{A}$ is fitted in a chamber hanging from a fixed ceiling near the earth's surface. A block $\mathrm{B}$ placed at the top of the wedge takes time $\mathrm{T}$ to slide down the length of the wedge and the cable supporting the chamber is broken at the same instant, the block will be

The acceleration due to gravity at height $h$ above the earth if $h < < R$ (Radius of earth) is given by

If the mass of a body on the earth surface is $M$, then its mass on the moon surface, is

Two planets of mean distance ${d_1}\,and\,{d_2}$ from the sun and their frequencies are ${n_1}\,and\,{n_2}$ respectively then