The depth 'd' at which the value of acceleration due to gravity becomes `(1)/(n)` times the value at the earth's surface is (R = radius of earth)

Options

A. `d = R((n)/(n-1))`
B. `d = R((n-1)/(2n))`
C. `d = R((n-1)/(n))`
D. `d = R^2((n-1)/(n))`

Answer

To solve the problem, we need to find the depth \( d \) at which the acceleration due to gravity \( g_d \) becomes \( \frac{g}{n} \), where \( g \) is the acceleration due to gravity at the surface of the Earth and \( R \) is the radius of the Earth.

### Step-by-Step Solution:

1. **Understanding the Relationship**:
The acceleration due to gravity at a depth \( d \) below the Earth's surface is given by the formula:
\[ g_d = g \left(1 - \frac{d}{R}\right) \]
where \( g \) is the acceleration due to gravity at the surface and \( R \) is the radius of the Earth.

2. **Setting Up the Equation**:
We want to find the depth \( d \) where:
\[ g_d = \frac{g}{n} \]
Substituting this into the equation gives:
\[ \frac{g}{n} = g \left(1 - \frac{d}{R}\right) \]

3. **Canceling \( g \)**:
Since \( g \) is common on both sides (assuming \( g \neq 0 \)), we can cancel it out:
\[ \frac{1}{n} = 1 - \frac{d}{R} \]

4. **Rearranging the Equation**:
Rearranging the equation to solve for \( \frac{d}{R} \):
\[ \frac{d}{R} = 1 - \frac{1}{n} \]

5. **Finding \( d \)**:
Now, multiplying both sides by \( R \) to isolate \( d \):
\[ d = R \left(1 - \frac{1}{n}\right) \]
This can be simplified to:
\[ d = R \left(\frac{n-1}{n}\right) \]

### Final Answer:
Thus, the depth \( d \) at which the value of acceleration due to gravity becomes \( \frac{1}{n} \) times the value at the Earth's surface is:
\[ d = R \frac{n-1}{n} \]