Assuming the earth to be a sphere of uniform density, its acceleration due to gravity acting on a body:

A body weighs \(72~\text{N}\) on the surface of the earth. What is the gravitational force on it at a height equal to half the radius of the earth?
1. \(32~\text{N}\)
2. \(30~\text{N}\)
3. \(24~\text{N}\)
4. \(48~\text{N}\)

A body weighs \(200\) N on the surface of the earth. How much will it weigh halfway down the centre of the earth?

A mass falls from a height '\(h\)' and its time of fall '\(t\)' is recorded in terms of time period \(T\) of a simple pendulum. On the surface of the earth, it is found that \(t=2T\). The entire setup is taken on the surface of another planet whose mass is half of that of the Earth and whose radius is the same. The same experiment is repeated and corresponding times are noted as '\(t\)' and '\(T\)'. Then we can say:
1. \(t' = \sqrt{2}T\)
2. \(t'>2T'\)
3. \(t'<2T'\)
4. \(t' = 2T'\)

If the mass of the sun were ten times smaller and the universal gravitational constant were ten times larger in magnitude, which of the following statements would not be correct?

The acceleration due to gravity at a height \(1~\text{km}\) above the earth's surface is the same as at a depth \(d\) below the surface of the earth. Then:

Starting from the centre of the earth, having radius \(R,\) the variation of \(g\) (acceleration due to gravity) is shown by:

1.		2.
3.		4.

The height at which the weight of a body becomes \(\left ( \frac{1}{16} \right )^\mathrm{th}\) of its weight on the surface of the earth (radius \(R\)) is:
1. \(5R\)
2. \(15R\)
3. \(3R\)
4. \(4R\)