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: Understanding the Question:
The acceleration due to gravity decreases as we go below the surface of the earth. We need to find the depth \( d \) where gravity is \( 1/n \) times its value at the surface.

Step 2: Key Formula or Approach:
The formula for acceleration due to gravity at depth \( d \) is:
\[ g' = g \left( 1 - \frac{d}{R} \right) \]

Step 3: Detailed Explanation:
Given:
\[ g' = \frac{g}{n} \] Substitute this into the depth formula:
\[ \frac{g}{n} = g \left( 1 - \frac{d}{R} \right) \] Divide both sides by \( g \):
\[ \frac{1}{n} = 1 - \frac{d}{R} \] Rearrange to solve for \( \frac{d}{R} \):
\[ \frac{d}{R} = 1 - \frac{1}{n} \] \[ \frac{d}{R} = \frac{n-1}{n} \] Therefore, depth \( d \):
\[ d = R \left( \frac{n-1}{n} \right) \]

Step 4: Final Answer:
The depth is \( \frac{R(n-1)}{n} \).