Q. The depth $d$ at which the value of acceleration due to gravity becomes $1/n$ times the value of the surface, is [ R = radius of the earth]
A) $\frac{R}{n}$
B) $R\left( \frac{n - 1}{n} \right)$
C) $\frac{R}{n^{2}}$
D) $R\left( \frac{n}{n + 1} \right)$
Explanation
(b) $g^{'} = g\left( 1 - \frac{d}{R} \right) \Rightarrow \frac{g}{n} = g\left( 1 - \frac{d}{R} \right) \Rightarrow d = \left( \frac{n - 1}{n} \right)R$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.