At what depth from the surface of earth the time period of a simple pendulum is `0.5%` more than that on the surface of the Earth? (Radius of earth is `6400 km`)

Options

A. `32 km`
B. `64 km`
C. `96 km`
D. `128 km`

Explanation

To solve the problem, we need to find the depth \(d\) from the surface of the Earth at which the time period of a simple pendulum is 0.5% more than that on the surface of the Earth.

### Step-by-Step Solution:

1. **Understanding the Time Period of a Pendulum**:
The time period \(T\) of a simple pendulum is given by the formula:
\[ T = 2\pi \sqrt{\frac{L}{g}} \]
where \(L\) is the length of the pendulum and \(g\) is the acceleration due to gravity.

2. **Time Period at the Surface**:
Let the time period at the surface of the Earth be \(T_1\) and the acceleration due to gravity at the surface be \(g\). Thus,
\[ T_1 = 2\pi \sqrt{\frac{L}{g}} \]

3. **Time Period at Depth \(d\)**:
At a depth \(d\), the acceleration due to gravity \(g'\) is given by:
\[ g' = g \left(1 - \frac{d}{R}\right) \]
where \(R\) is the radius of the Earth (6400 km).

4. **Time Period at Depth**:
Let the time period at depth \(d\) be \(T_2\):
\[ T_2 = 2\pi \sqrt{\frac{L}{g'}} \]

5. **Given Condition**:
According to the problem, \(T_2\) is 0.5% more than \(T_1\):
\[ T_2 = T_1 + 0.005 T_1 = 1.005 T_1 \]

6. **Setting Up the Equation**:
We can relate \(T_1\) and \(T_2\) using the expressions for \(g\) and \(g'\):
\[ \frac{T_1}{T_2} = \sqrt{\frac{g'}{g}} \]
Substituting \(T_2\):
\[ \frac{T_1}{1.005 T_1} = \sqrt{\frac{g \left(1 - \frac{d}{R}\right)}{g}} \]
This simplifies to:
\[ \frac{1}{1.005} = \sqrt{1 - \frac{d}{R}} \]

7. **Squaring Both Sides**:
Squaring both sides gives:
\[ \left(\frac{1}{1.005}\right)^2 = 1 - \frac{d}{R} \]

8. **Rearranging the Equation**:
Rearranging gives:
\[ \frac{d}{R} = 1 - \left(\frac{1}{1.005}\right)^2 \]

9. **Calculating \(d\)**:
Now substituting \(R = 6400 \times 10^3\) meters:
\[ d = R \left(1 - \left(\frac{1}{1.005}\right)^2\right) \]
Calculate \(\left(\frac{1}{1.005}\right)^2\):
\[ \left(\frac{1}{1.005}\right)^2 \approx 0.990025 \]
Thus,
\[ d = 6400 \times 10^3 \left(1 - 0.990025\right) = 6400 \times 10^3 \times 0.009975 \approx 63.84 \times 10^3 \text{ meters} \]
Converting to kilometers:
\[ d \approx 63.84 \text{ km} \]

### Final Answer:
The depth from the surface of the Earth at which the time period of a simple pendulum is 0.5% more than that on the surface of the Earth is approximately **63.84 km**.