Question

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

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

Answer 1

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

The acceleration of due to gravity at a depth $d$ below the surface of the Earth is given by $$g_d=g(1-\frac{d}{R})$$ As per the question,

$g_d=\frac{g}{n}$

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

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

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

Hence, option (b) is the correct answer.

Why this Question? Tip - The result $g_d=(1-\frac{d}{R})$ is valid for all values of d, small or large.