A point mass when enters d depth below the earth's surface such that the g reduces by 1% then the value of d is [Take `R_(e )=6400` km]

Options

A. 64 km
B. 32 km
C. 128 km
D. 256 km

Answer

To solve the problem, we need to determine the depth \( d \) below the Earth's surface at which the acceleration due to gravity \( g \) reduces by 1%.

### Step-by-Step Solution:

1. **Understanding the Problem**:
We know that the acceleration due to gravity at a depth \( d \) below the Earth's surface can be expressed as:
\[ g_d = g \left(1 - \frac{d}{R}\right) \]
where \( g \) is the acceleration due to gravity at the surface, \( R \) is the radius of the Earth, and \( g_d \) is the acceleration due to gravity at depth \( d \).

2. **Setting Up the Equation**:
Since the problem states that \( g \) reduces by 1%, we can express this mathematically as:
\[ g_d = 0.99g \]

3. **Substituting into the Equation**:
Now, we can substitute \( g_d \) into the equation:
\[ 0.99g = g \left(1 - \frac{d}{R}\right) \]

4. **Cancelling \( g \)**:
Since \( g \) is common on both sides (and is not zero), we can cancel it out:
\[ 0.99 = 1 - \frac{d}{R} \]

5. **Rearranging the Equation**:
Rearranging the equation gives:
\[ \frac{d}{R} = 1 - 0.99 = 0.01 \]

6. **Solving for \( d \)**:
Now, we can solve for \( d \):
\[ d = 0.01R \]

7. **Substituting the Value of \( R \)**:
Given that \( R = 6400 \) km, we substitute this value into the equation:
\[ d = 0.01 \times 6400 \text{ km} = 64 \text{ km} \]

8. **Conclusion**:
Therefore, the depth \( d \) below the Earth's surface at which the acceleration due to gravity reduces by 1% is:
\[ d = 64 \text{ km} \]

### Final Answer:
The value of \( d \) is **64 km**.

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