$T$ is the time period of simple pendulum on the earth's surface. Its time period becomes $x T$ when taken to a height $R$ (equal to earth's radius) above the earth's surface. Then, the value of $x$ will be:
$4$
$2$
$\frac{1}{2}$
$\frac{1}{4}$
Assuming the earth to be a sphere of uniform density, the acceleration due to gravity inside the earth at a distance of $r$ from the centre is proportional to
The radius of the earth is $6400\, km$ and $g = 10\,m/{\sec ^2}$. In order that a body of $5 \,kg$ weighs zero at the equator, the angular speed 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)
If the earth stops rotating, the value of $‘g’$ at the equator will
A body weight $500 \,N$ on the surface of the earth. How much would it weigh half way below the surface of the earth ....... $N$