Question

The depth at which acceleration due to gravity becomes $g/n$ is ($R=$ radius of earth, $g=$ acceleration due to gravity, $n=$ integer)

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

Hint:

Inside the Earth, $g$ decreases linearly with depth until it reaches zero at the center.

Answer 1

The Correct Option is A

Step 1: Formula
Acceleration due to gravity at depth $d$ is $g_d = g(1 - \frac{d}{R})$.

Step 2: Analysis
Given $g_d = g/n$.
$g/n = g(1 - \frac{d}{R}) \implies \frac{1}{n} = 1 - \frac{d}{R}$.

Step 3: Calculation
$\frac{d}{R} = 1 - \frac{1}{n} = \frac{n-1}{n}$
$d = \frac{R(n-1)}{n}$.

Step 4: Conclusion
Hence, the depth is $\frac{R(n-1)}{n}$. Final Answer: (A)