The acceleration due to gravity at a height $1\, km$ above the earth is the same as at a depth $d$ below the surface of earth. Then $d = $ ......... $km$

The time period of a simple pendulum on the surface of the earth is $T$. If the pendulum is taken to a height equal to half of the radius of the earth,then its time period is

The angular velocity of the Earth at present is $\omega$. With what angular velocity should it rotate so that the weight of a body at the equator appears to be zero (in $\omega$)? (Express the answer as a multiple of $\omega$)

Imagine Earth to be a solid sphere of mass $M$ and radius $R$. If the value of acceleration due to gravity at a depth $d$ below Earth's surface is the same as its value at a height $h$ above its surface and equal to $\frac{g}{4}$ (where $g$ is the value of acceleration due to gravity on the surface of Earth),the ratio of $\frac{h}{d}$ will be

Where will it be profitable to purchase $1$ kilogram of sugar?

Which of the following graphs represents the variation of acceleration due to gravity $(g)$ with distance $r$ from the centre of the Earth?