At a place, value of g is less by 1% than 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. 30 km above the surface of the Earth
D. 32 km below the surface of the Earth.

Explanation

To solve the problem, we need to determine the height or depth at which the value of acceleration due to gravity (g) is reduced by 1% from its value on the surface of the Earth.

### Step-by-step Solution:

1. **Understanding the Problem**:
- The value of g at a certain place is 1% less than the value of g on the surface of the Earth.
- We denote the value of g on the surface as \( g \). Therefore, at the place in question, the value of g is:
\[ g' = g - 0.01g = 0.99g \]

2. **Using the Formula for Gravity at Height**:
- The formula for the acceleration due to gravity at a height \( h \) above the Earth's surface is given by:
\[ g' = g \left(1 - \frac{2h}{R}\right) \]
- Here, \( R \) is the radius of the Earth (6400 km).

3. **Setting Up the Equation**:
- We set \( g' = 0.99g \):
\[ 0.99g = g \left(1 - \frac{2h}{R}\right) \]
- Dividing both sides by \( g \):
\[ 0.99 = 1 - \frac{2h}{R} \]

4. **Solving for Height \( h \)**:
- Rearranging the equation gives:
\[ \frac{2h}{R} = 1 - 0.99 = 0.01 \]
- Multiplying both sides by \( R \):
\[ 2h = 0.01R \]
- Dividing by 2:
\[ h = \frac{0.01R}{2} = 0.005R \]

5. **Substituting the Radius of the Earth**:
- Now substituting \( R = 6400 \) km:
\[ h = 0.005 \times 6400 = 32 \text{ km} \]

6. **Understanding the Context**:
- The height \( h \) we calculated is above the surface of the Earth. However, the question asks for a place where g is less than 1% on the surface.
- We need to find the depth \( d \) below the Earth's surface where the same condition applies.

7. **Using the Formula for Gravity at Depth**:
- The formula for the acceleration due to gravity at a depth \( d \) below the Earth's surface is:
\[ g' = g \left(1 - \frac{d}{R}\right) \]
- Setting \( g' = 0.99g \):
\[ 0.99g = g \left(1 - \frac{d}{R}\right) \]
- Dividing both sides by \( g \):
\[ 0.99 = 1 - \frac{d}{R} \]

8. **Solving for Depth \( d \)**:
- Rearranging gives:
\[ \frac{d}{R} = 1 - 0.99 = 0.01 \]
- Multiplying both sides by \( R \):
\[ d = 0.01R = 0.01 \times 6400 = 64 \text{ km} \]

### Final Answer:
The place is **64 km below the surface of the Earth**.

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