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

Options

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

Explanation

To solve the problem, we need to find the depth \( d \) at which the acceleration due to gravity \( g' \) becomes \( \frac{1}{n} \) times the value at the surface \( g \).

### Step-by-Step Solution:

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

2. **Setting up the equation**:
According to the problem, we want \( g' \) to be \( \frac{1}{n} g \). Therefore, we can set up the equation:
\[ g \left(1 - \frac{d}{R}\right) = \frac{1}{n} g \]

3. **Dividing both sides by \( g \)**:
Since \( g \) is not zero, we can divide both sides of the equation by \( g \):
\[ 1 - \frac{d}{R} = \frac{1}{n} \]

4. **Rearranging the equation**:
Rearranging the equation gives us:
\[ \frac{d}{R} = 1 - \frac{1}{n} \]
Simplifying this further, we find:
\[ d = R \left(1 - \frac{1}{n}\right) \]

5. **Expressing in terms of \( n \)**:
We can express \( 1 - \frac{1}{n} \) as \( \frac{n-1}{n} \):
\[ d = R \cdot \frac{n-1}{n} \]

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