A simple pendulum has a time period ${T_1}$ when on the earth’s surface and ${T_2}$ when taken to a height $R$ above the earth’s surface, where $R$ is the radius of the earth. The value of ${T_2}/{T_1}$ is

Write the difference between $G$ and $g$.

$Assertion$ : A balloon filled with hydrogen will fall with acceleration $\frac{g}{6}$ of the moon

$Reason$ : Moon has no atmosphere.

The radii of two planets are respectively ${R_1}$ and ${R_2}$ and their densities are respectively ${\rho _1}$ and ${\rho _2}$. The ratio of the accelerations due to gravity at their surfaces is

If the change in the value of $‘g’$ at a height $h$ above the surface of the earth is the same as at a depth $x$ below it, then (both $x$ and $h$ being much smaller than the radius of the earth)

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]: