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 density of a newly discovered planet is twice that of the earth. The acceleration due to gravity at the surface of the planet is equal to that at the surface of the earth. If the radius of the earth is \(R,\) the radius of the planet would be:

Radii and densities of two planets are ${\mathrm{R}}_1, {\mathrm{R}}_2$ and ${\mathrm{\rho }}_1, {\mathrm{\rho }}_2$ respectively. The ratio of accelerations due to gravity on their surfaces is:

1.  $\frac{{\mathrm{\rho }}_1}{{\mathrm{R}}_1} : \frac{{\mathrm{\rho }}_2}{{\mathrm{R}}_2}$

2.  $\frac{{\mathrm{\rho }}_1}{{\mathrm{R}}_1^2} : \frac{{\mathrm{\rho }}_2}{{\mathrm{R}}_2^2}$

3.  ${\mathrm{\rho }}_1{\mathrm{R}}_1 : {\mathrm{\rho }}_2{\mathrm{R}}_2$

4.  $\frac{1}{{\mathrm{\rho }}_1{\mathrm{R}}_1} : \frac{1}{{\mathrm{\rho }}_2{\mathrm{R}}_2}$

Imagine a new planet having the same density as that of the Earth but 3 times bigger than the Earth in size. If the acceleration due to gravity on the surface of the earth is g and that on the surface of the new planet is g', then:

The height of a point vertically above the earth’s surface, at which the acceleration due to gravity becomes 1% of its value at the surface is: (Radius of the earth = R)

1. 8R           

2. 9R

3. 10R         

4. 20R

1 kg of sugar has maximum weight:
1. at the pole.
2. at the equator.
3. at a latitude of 45$°$.
4. in India.

 An object weighs 72 N on earth. Its weight at a height R/2 from the surface of the earth will be:

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

Acceleration due to gravity is: