Question

The depth at which the value of acceleration due to gravity becomes $\frac{1}{n}$ times the value at the surface is (R be the radius of the earth):

• A. $\frac{R}{n}$
• B. $\frac{R}{n^2}$
• C. $\frac{R(n-1)}{n}$
• D. $\frac{Rn}{(n-1)}$

Answer 1

Answer: (C)

$\frac{R(n-1)}{n}$

option C

$g^1=g[1-\frac{d}{R}]$

$⟹\frac{g}{n}=g[1-\frac{d}{R}]⟹d=\left[\frac{n-1}{n}\right]R$