The rotation of the earth having radius $R$ about its axis speeds up to a value such that a man at latitude angle $60^{\circ}$ feels weightless. The duration of the day in such case will be:

The time taken by a simple pendulum for one oscillation is $T$ on the earth's surface. Its time period becomes $xT$ when taken to a height $R$ (equal to the earth's radius) above the earth's surface. The value of $x$ is

Given below are two statements. One is labelled as Assertion $(A)$ and the other is labelled as Reason $(R)$.
Assertion $(A) :$ $A$ simple pendulum is taken to a planet of mass and radius,$4$ times and $2$ times,respectively,than the Earth. The time period of the pendulum remains the same on Earth and the planet.
Reason $(R) :$ The mass of the pendulum remains unchanged at Earth and the other planet. In the light of the above statements,choose the correct answer from the options given below $:$

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:

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)

State the values of $g$ and $G$ at the center of the Earth. Identify which of $g$ and $G$ is a vector and which is a scalar.