Q. The density of earth in terms of acceleration due to gravity $(g),$ radius of earth $(R)$ and universal gravitational constant $(G)$ is
A) $\frac{4\pi RG}{3g}$
B) $\frac{3\pi RG}{4g}$
C) $\frac{4g}{3\pi RG}$
D) $\frac{3g}{4\pi RG}$
Explanation
(d) Acceleration due to gravity
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ g = \frac{GM}{R^{2}} = \frac{G}{R^{2}} \times \frac{4}{3}\pi R^{3}\rho$
$\therefore\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \rho = \frac{3g}{4\pi GR}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.