A pendulum clock that keeps the correct time on the earth is taken to the moon. It will run

Options

A. at correct rate
B. 6 times faster
C. `sqrt6` times faster
D. `sqrt6` slower

Answer

To solve the problem of how a pendulum clock behaves when taken from Earth to the Moon, we need to analyze the relationship between the time period of the pendulum and the acceleration due to gravity.

### Step-by-Step Solution:

1. **Understand 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. **Identify the Values of \( g \)**:
- On Earth, the acceleration due to gravity \( g \) is approximately \( 9.8 \, \text{m/s}^2 \).
- On the Moon, the acceleration due to gravity \( g' \) is about \( \frac{g}{6} \) (approximately \( 1.63 \, \text{m/s}^2 \)).

3. **Calculate the Time Period on Earth**:
Using the formula, the time period on Earth \( T_{earth} \) is:
\[ T_{earth} = 2\pi \sqrt{\frac{L}{g}} \]

4. **Calculate the Time Period on the Moon**:
Substitute \( g' = \frac{g}{6} \) into the time period formula for the Moon:
\[ T_{moon} = 2\pi \sqrt{\frac{L}{g'}} = 2\pi \sqrt{\frac{L}{\frac{g}{6}}} = 2\pi \sqrt{\frac{6L}{g}} = \sqrt{6} \cdot 2\pi \sqrt{\frac{L}{g}} = \sqrt{6} \cdot T_{earth} \]

5. **Compare the Time Periods**:
From the calculations, we find that:
\[ T_{moon} = \sqrt{6} \cdot T_{earth} \]
This means that the time period of the pendulum clock on the Moon is \( \sqrt{6} \) times greater than that on Earth.

6. **Determine the Effect on Timekeeping**:
Since the time period on the Moon is longer, the clock will take more time to complete one oscillation. Therefore, the pendulum clock will run slower on the Moon.

7. **Conclusion**:
The clock will run \( \sqrt{6} \) times slower than it does on Earth.

### Final Answer:
The pendulum clock will run **root 6 times slower** on the Moon.

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