What is the depth at which the value of acceleration due to gravity become 1/n times the value that at the surface of earth ? (radius of earth = R)

Options

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

Explanation

To find the depth at which the acceleration due to gravity becomes \( \frac{1}{n} \) times the value at the surface of the Earth, we can follow these steps:

### Step 1: Understand 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 depth \( D \),
- \( g \) is the acceleration due to gravity at the surface of the Earth,
- \( R \) is the radius of the Earth.

### Step 2: Set up the equation for the given condition
According to the problem, we want to find the depth \( D \) where the acceleration due to gravity \( g' \) becomes \( \frac{g}{n} \). Therefore, we can set up the equation:
\[ g \left(1 - \frac{D}{R}\right) = \frac{g}{n} \]

### Step 3: Simplify the equation
We can cancel \( g \) from both sides (assuming \( g \neq 0 \)):
\[ 1 - \frac{D}{R} = \frac{1}{n} \]

### Step 4: Rearrange the equation to solve for \( D \)
Now, we can rearrange the equation to isolate \( D \):
\[ \frac{D}{R} = 1 - \frac{1}{n} \]
\[ D = R \left(1 - \frac{1}{n}\right) \]

### Step 5: Final expression for depth
To express this in a more simplified form:
\[ D = R \left(\frac{n-1}{n}\right) \]

### Conclusion
Thus, the depth \( D \) at which the acceleration due to gravity becomes \( \frac{1}{n} \) times the value at the surface of the Earth is:
\[ D = R \left(\frac{n-1}{n}\right) \]