Question

The depth $d$ at which the value of acceleration due to gravity becomes $\frac{1}{n}$ times the value at the surface, is:
($R=\text{Radius of earth})$

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

Answer 1

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

Variation of $g$ due to depth is given by,

$g^{\prime }=g(1-\frac{d}{R})$

But, $g^{\prime }=\frac{g}{n}$

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

$\Rightarrow \frac{d}{R}=1-\frac{1}{n}$

$\Rightarrow d=R\left(\frac{n-1}{n}\right)$

Hence, option $\left(B\right)$ is the correct answer.

Why this Question? Note: Variation of $g$ due to depth is given by, $g^{\prime }=g(1-\frac{d}{R})$