The time period of a simple pendulum is $T_1$ when on the Earth's surface and $T_2$ when taken to a height $h = 2R$ above the Earth's surface, where $R$ is the radius of the Earth. The ratio $T_1 : T_2$ is:

$A$ 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 speed with which the earth would have to rotate about its axis so that a person on the equator would weigh $\frac{3}{5}$ th as much as at present weight is ($g=$ gravitational acceleration,$R=$ equatorial radius of the earth).

The depth $d$ at which the value of acceleration due to gravity becomes $\frac{1}{n}$ times the value at the earth's surface is $(R = \text{radius of the earth})$

Which graph correctly presents the variation of acceleration due to gravity with the distance from the centre of the earth (radius of the earth $= R_E$)?

$A$ spherical planet far out in space has a mass $M_0$ and diameter $D_0$. $A$ particle of mass $m$ falling freely near the surface of this planet will experience an acceleration due to gravity which is equal to