For a given length of a pendulum, the time period is maximum

Options

A. on the surface of the earth
B. on the surface of the moon
C. at the centre of the earth
D. none of these

Explanation

To determine where the time period of a pendulum is maximum for a given length, we need to analyze the formula for the time period of a simple pendulum and how the acceleration due to gravity (g) affects it.

### Step-by-Step Solution:

1. **Understanding the Time Period Formula**:
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. **Identifying the Variables**:
- The length \(L\) of the pendulum is constant as per the question.
- The variable that affects the time period is \(g\), the acceleration due to gravity.

3. **Comparing Different Locations**:
- **On the Surface of the Earth**: The value of \(g\) is approximately \(9.8 \, \text{m/s}^2\).
- **On the Surface of the Moon**: The value of \(g\) is about \(1.6 \, \text{m/s}^2\), which is significantly less than that on Earth.
- **At the Center of the Earth**: The value of \(g\) is \(0\) because gravitational forces from all sides cancel out.

4. **Calculating Time Periods**:
- **On Earth**:
\[ T_e = 2\pi \sqrt{\frac{L}{g}} \quad (g \approx 9.8) \]
- **On the Moon**:
\[ T_m = 2\pi \sqrt{\frac{L}{g/6}} = 2\pi \sqrt{\frac{6L}{g}} \quad (g \approx 1.6) \]
This shows that \(T_m\) is greater than \(T_e\) because \(g\) is smaller.
- **At the Center of the Earth**:
\[ T_c = 2\pi \sqrt{\frac{L}{0}} \quad \text{(undefined)} \]
This means the pendulum does not oscillate at all.

5. **Conclusion**:
Since the time period is directly proportional to \(1/\sqrt{g}\), the smaller the value of \(g\), the larger the time period. Therefore, the time period is maximum on the surface of the Moon compared to the Earth and is undefined at the center of the Earth.

### Final Answer:
The time period is maximum on the surface of the Moon.

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