The height at which the acceleration due to gravity becomes $\frac{g}{9}$ (where $g$ = the acceleration due to gravity on the surface of the earth) in terms of $R$, the radius of the earth, is

The variation of acceleration due to gravity $g$ with distance $d$ from centre of the earth is best represented by ($R =$ Earth's radius)

What should be the velocity of earth due to rotation about its own axis so that the weight at equator become $3/5$ of initial value. Radius of earth on equator is $ 6400\, km$

If density of a planet is double that of the earth and the radius $1.5$ times that of the earth, the acceleration due to gravity on the surface of the planet is ........

The mass density of a spherical body is given by $\rho \left( r \right) = \frac{k}{r}$ for $r \leq R\,\,$ and $\rho \left( r \right) = 0\,$ for $r > R$ , where $r$ is the distance from the centre. The correct graph that describes qualitatively the acceleration, $a$, of a test particle as a function of $r$ is

A simple pendulum doing small oscillations at a place $\mathrm{R}$ height above earth surface has time period of $T_1=4 \mathrm{~s}$. $T_2$ would be it's time period if it is brought to a point which is at a height $2 R$ from earth surface. Choose the correct relation $[R=$ radius of Earth]: