The depth $d$ below the surface of the earth where the value of acceleration due to gravity becomes $\left(\frac{1}{n}\right)$ times the value at the surface of the earth is ($R$ = radius of the earth).

  • A
    $R\left(\frac{n-1}{n}\right)$
  • B
    $R\left(\frac{n}{n+1}\right)$
  • C
    $\frac{R}{n}$
  • D
    $\frac{R}{n^2}$

Explore More

Similar Questions

What is the change in mass of a body, when taken $64 \ km$ below the surface of the earth? $[$Take radius of the earth as $6400 \ km]$

$A$ point mass of $10 \,kg$ is placed at the centre of the Earth. The weight of the point mass is

By approximately how many meters is the radius of the Earth at the equator greater than the radius of the Earth at the poles (in $,000 \ m$)?

If the earth were to cease rotating about its own axis,the increase in the value of $g$ in the $C.G.S.$ system at a place of latitude of $45^{\circ}$ will be ........ $cm/sec^{2}$.

$A$ narrow but tall cabin is falling freely near the earth's surface. Inside the cabin,two small stones $A$ and $B$ are released from rest (relative to the cabin). Initially $A$ is much above the centre of mass and $B$ is much below the centre of mass of the cabin. $A$ close observation of the motion of $A$ and $B$ will reveal that

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