Considering Earth to be a sphere of radius $R$ having uniform density $\rho$,the value of acceleration due to gravity $g$ in terms of $R$,$\rho$,and $G$ is:
- A$g=\sqrt{\frac{3 \pi R}{\rho G}}$
- B$g=\sqrt{\frac{4}{3} \pi \rho GR}$
- C$g=\frac{4}{3} \pi \rho GR$
- D$g=\frac{GM}{\rho R^2}$
Explore More
Similar Questions
If the angular speed of the earth is doubled,the value of acceleration due to gravity $(g)$ at the north pole:
Easy
View Solution$A$ body weighs $200 \; N$ on the surface of the earth. How much will it weigh halfway down to the centre of the earth (in $; N$)?
MediumNEET 2019
View Solution$g_e$ and $g_p$ denote the acceleration due to gravity on the surface of the Earth and another planet whose mass and radius are twice that of the Earth. Then:
Easy
View SolutionAt what altitude will the acceleration due to gravity be $25\%$ of that at the earth's surface (given radius of earth is $R$)?
Medium
View SolutionThe value of the acceleration due to gravity is $g_{1}$ at a height $h = \frac{R}{2}$ ($R$ = radius of the earth) from the surface of the earth. It is again equal to $g_{1}$ at a depth $d$ below the surface of the earth. The ratio $\left(\frac{d}{R}\right)$ equals
DifficultJEE MAIN 2020
View SolutionVedclass Products
For Students
Vedclass Test Series
Mock tests in real JEE/NEET style with performance analysis. 5-day free trial.
Start Free TrialFor Teachers
Exam Paper Generator
Generate Set A/B/C/D exam papers from 7.5L+ questions in 2 minutes. 3 chapters free.
Try FreeFor Institutes
Online Exam Module
Live online exams with unlimited students, 360° analytics & white-label branding.
See Demo