The depth at which the value of `g` becomes `25%` of that at the surface of the earth is (in `KM`)

Options

A. `4800`
B. `2400`
C. `3600`
D. `1200`

Explanation

To find the depth at which the value of gravitational acceleration \( g \) becomes 25% of that at the surface of the Earth, we can follow these steps:

### Step-by-Step Solution:

1. **Understand the formula for gravitational acceleration at depth**:
The gravitational acceleration at a depth \( d \) below the surface of the Earth is given by:
\[ g_d = g \left(1 - \frac{d}{R}\right) \]
where \( g \) is the gravitational acceleration at the surface and \( R \) is the radius of the Earth.

2. **Set up the equation for 25% of surface gravity**:
We need to find the depth \( d \) where \( g_d \) is 25% of \( g \):
\[ g_d = \frac{g}{4} \]

3. **Substitute into the formula**:
Substitute \( g_d \) into the formula:
\[ \frac{g}{4} = g \left(1 - \frac{d}{R}\right) \]

4. **Cancel \( g \) from both sides**:
Since \( g \) is not zero, we can cancel it:
\[ \frac{1}{4} = 1 - \frac{d}{R} \]

5. **Rearrange the equation**:
Rearranging gives:
\[ \frac{d}{R} = 1 - \frac{1}{4} = \frac{3}{4} \]

6. **Solve for \( d \)**:
Multiply both sides by \( R \):
\[ d = \frac{3}{4} R \]

7. **Substitute the radius of the Earth**:
The average radius of the Earth \( R \) is approximately 6400 km. Thus:
\[ d = \frac{3}{4} \times 6400 \text{ km} \]

8. **Calculate the depth**:
\[ d = 4800 \text{ km} \]

### Final Answer:
The depth at which the value of \( g \) becomes 25% of that at the surface of the Earth is **4800 km**.