Question

The depth ${}^{\prime }{d}^{\prime }$ at which the value of acceleration due to gravity becomes $\frac{1}{n}$ times the value at the earth's surface is $(R=$ radius of earth)

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

Answer 1

The correct option is B $d=R\left(\frac{n-1}{n}\right)$

Acceleration due to gravity at a depth $d$ under the earth surface $g^{\prime }=g_s(1-\frac{d}{R})$

Given : $g^{\prime }=\frac{g_s}{n}$

$\therefore$ $\frac{g_s}{n}=g_s(1-\frac{d}{R})$

Or $\frac{1}{n}=(1-\frac{d}{R})$

$⟹$ $d=R(\frac{n-1}{n})$