The period of oscillation of a simple pendulum of constant length at surface of the earth is T. Its time period inside mine will be

Options

A. Increases
B. Decreases
C. No change
D. None

Explanation

To solve the problem, we need to determine the time period of a simple pendulum at a certain depth inside the Earth, given that its time period at the surface is T.

### Step-by-Step Solution:

1. **Understand the Formula for Time Period**:
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,
- \( g \) is the acceleration due to gravity at the surface of the Earth.

2. **Identify the Time Period at Depth**:
When the pendulum is taken to a depth \( d \) inside the Earth, the effective acceleration due to gravity \( g' \) changes. The formula for the effective gravity at a depth \( d \) is:
\[ g' = g \left(1 - \frac{d}{R}\right) \]
where \( R \) is the radius of the Earth.

3. **Substitute into the Time Period Formula**:
The time period at depth \( d \) can be expressed as:
\[ T_d = 2\pi \sqrt{\frac{L}{g'}} \]
Substituting \( g' \) into this equation gives:
\[ T_d = 2\pi \sqrt{\frac{L}{g \left(1 - \frac{d}{R}\right)}} \]

4. **Compare Time Periods**:
Since \( g' < g \) (because \( 1 - \frac{d}{R} < 1 \)), it follows that:
\[ T_d > T \]
This means that the time period at depth \( d \) is greater than the time period at the surface.

5. **Conclusion**:
Therefore, the time period of the pendulum inside the mine (at depth) will be greater than \( T \).

### Final Answer:
The time period inside the mine will be greater than \( T \).

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