If $R$ is the radius of the earth and the acceleration due to gravity on the surface of earth is $g=\pi^2 \mathrm{~m} / \mathrm{s}^2$, then the length of the second's pendulum at a height $h=2 R$ from the surface of earth will be,:

If the Earth losses its gravity, then for a body

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

The mass of a planet and its diameter are three times those of earth's. Then the  acceleration due to gravity on the surface of the planet is ……. $m/s^2$

$Assertion$ : Space rocket are usually launched in the equatorial line from west to east
$Reason$ : The acceleration due to gravity is minimum at the equator.