At a place, value of acceleration due to gravity g is reduced by 2% of its value on the surface of the earth (Radius of earth = 6400 km). The place is

Options

A. 64 km below the surface of the earth
B. 64 km above the surface of the earth
C. 32 km above the surface of the earth
D. 32 km below the surface of the earth.

Answer

To solve the problem, we need to determine where the acceleration due to gravity \( g \) is reduced by 2% of its value on the surface of the Earth. We will analyze the changes in \( g \) both above and below the Earth's surface.

### Step-by-Step Solution:

1. **Understand the Formula for Gravity**:
The acceleration due to gravity at a distance \( r \) from the center of the Earth is given by:
\[ g' = \frac{GM}{r^2} \]
where \( G \) is the gravitational constant, and \( M \) is the mass of the Earth.

2. **Calculate the Surface Gravity**:
The acceleration due to gravity at the surface of the Earth (\( g \)) is:
\[ g = \frac{GM}{R^2} \]
where \( R \) is the radius of the Earth (6400 km).

3. **Determine the Reduced Gravity**:
A reduction of 2% means:
\[ g' = g - 0.02g = 0.98g \]

4. **Consider the Effects of Height and Depth**:
- **Above the Surface**: For a height \( h \) above the surface, the formula becomes:
\[ g' = \frac{gR^2}{(R + h)^2} \]
- **Below the Surface**: For a depth \( d \) below the surface, the formula is:
\[ g' = g \left(1 - \frac{d}{R}\right) \]

5. **Set Up the Equations**:
- For height \( h \):
\[ 0.98g = \frac{gR^2}{(R + h)^2} \]
Simplifying gives:
\[ 0.98(R + h)^2 = R^2 \]
- For depth \( d \):
\[ 0.98g = g \left(1 - \frac{d}{R}\right) \]
Simplifying gives:
\[ 0.98 = 1 - \frac{d}{R} \implies \frac{d}{R} = 0.02 \implies d = 0.02R \]

6. **Calculate Depth**:
Since \( R = 6400 \, \text{km} \):
\[ d = 0.02 \times 6400 \, \text{km} = 128 \, \text{km} \]

7. **Check for Height**:
For height, we can solve the equation \( 0.98(R + h)^2 = R^2 \) to find \( h \). However, since we already found a solution for depth, we can conclude that the place is 128 km below the surface of the Earth.

### Conclusion:
The place where the value of acceleration due to gravity \( g \) is reduced by 2% of its value on the surface of the Earth is **128 km below the surface of the Earth**.