If $R$ is the radius of the Earth and $g$ is the acceleration due to gravity on the Earth's surface,then the mean density of the Earth is:

  • A
    $\frac{4 \pi G}{3 g R}$
  • B
    $\frac{3 \pi R}{4 g G}$
  • C
    $\frac{3 g}{4 \pi R G}$
  • D
    $\frac{\pi R G}{12 g}$

Explore More

Similar Questions

Let $g$ be the acceleration due to gravity at the Earth's surface and $K$ be the rotational kinetic energy of the Earth. Suppose the Earth's radius decreases by $2 \%$ keeping its mass the same,then:

Difficult
View Solution

The density of a new planet is twice that of earth. The acceleration due to gravity at the surface of the planet is equal to that at the surface of earth. If $R$ is the radius of earth,then the radius of the planet would be

$A$ tunnel is dug along a diameter of the Earth. If $M_e$ and $R_e$ are the mass and radius of the Earth respectively,then the force on a particle of mass $m$ placed in the tunnel at a distance $r$ from the center is:

Difficult
View Solution

Two planets $A$ and $B$ have densities $\varrho_1$ and $\varrho_2$ and have radii $r_1$ and $r_2$ respectively. The ratio of acceleration due to gravity on $A$ to that of $B$ is:

$A$ body of weight $W$ is projected vertically upwards from the Earth's surface to reach a height above the Earth which is equal to nine times the radius of the Earth. The weight of the body at that height will be

Vedclass Products

For Students

Vedclass Test Series

Mock tests in real JEE/NEET style with performance analysis. 5-day free trial.

Start Free Trial
For Teachers

Exam Paper Generator

Generate Set A/B/C/D exam papers from 7.5L+ questions in 2 minutes. 3 chapters free.

Try Free
For Institutes

Online Exam Module

Live online exams with unlimited students, 360° analytics & white-label branding.

See Demo