At a certain depth "d" below surface of earth,value of acceleration due to gravity becomes four times that of its value at a height `3R` above earth surface.Where `R` is Radius of earth (Take `R=6400km)`.The depth `d` is equal to

Options

A. `4800km`
B. `640km`
C. `2560km`
D. `5260km`

Answer

To solve the problem, we need to find the depth \( d \) below the Earth's surface where the acceleration due to gravity becomes four times that at a height of \( 3R \) above the Earth's surface. Given that \( R = 6400 \) km, we can follow these steps:

### Step-by-Step Solution:

1. **Understanding the Problem**:
- Let \( g \) be the acceleration due to gravity at the Earth's surface.
- The value of gravity at a height \( h \) above the surface is given by the formula:
\[ g_h = \frac{g}{(1 + \frac{h}{R})^2} \]
- For \( h = 3R \):
\[ g_{3R} = \frac{g}{(1 + 3)^2} = \frac{g}{16} \]

2. **Setting Up the Equation**:
- We need to find the depth \( d \) such that:
\[ g_d = 4 \cdot g_{3R} = 4 \cdot \frac{g}{16} = \frac{g}{4} \]
- The formula for gravity at a depth \( d \) below the surface is:
\[ g_d = g \left(1 - \frac{d}{R}\right) \]

3. **Equating the Two Expressions**:
- Set the two expressions for \( g_d \) equal to each other:
\[ g \left(1 - \frac{d}{R}\right) = \frac{g}{4} \]

4. **Solving for Depth \( d \)**:
- Cancel \( g \) from both sides (assuming \( g \neq 0 \)):
\[ 1 - \frac{d}{R} = \frac{1}{4} \]
- Rearranging gives:
\[ \frac{d}{R} = 1 - \frac{1}{4} = \frac{3}{4} \]
- Therefore:
\[ d = \frac{3}{4} R \]

5. **Substituting the Value of \( R \)**:
- Substitute \( R = 6400 \) km:
\[ d = \frac{3}{4} \times 6400 \text{ km} = 4800 \text{ km} \]

### Final Answer:
The depth \( d \) is equal to **4800 km**.