A second pendulum is moved to moon where acceleration dur to gravity is 1/6 times that of the earth, the length of the second pendulum on moon would be

Options

A. 6 times
B. 12 times
C. `(1)/(6)` times
D. `(1)/(12)` times

Explanation

To solve the problem of finding the length of a second pendulum on the Moon, we can follow these steps:

### Step-by-Step Solution:

1. **Understand 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. **Identify the Time Period of a Second Pendulum**:
A second pendulum has a time period of 2 seconds. Therefore, for Earth:
\[ T_e = 2 \text{ seconds} \]

3. **Set Up the Equation for Earth**:
Using the time period formula for Earth, we have:
\[ 2 = 2\pi \sqrt{\frac{L_e}{g_e}} \]
Squaring both sides gives:
\[ 4 = 4\pi^2 \frac{L_e}{g_e} \]
Simplifying this, we find:
\[ L_e = \frac{g_e}{\pi^2} \]

4. **Consider the Moon's Gravity**:
The acceleration due to gravity on the Moon \( g_m \) is given as:
\[ g_m = \frac{g_e}{6} \]

5. **Set Up the Equation for the Moon**:
The time period on the Moon is the same as on Earth (2 seconds), so we can write:
\[ 2 = 2\pi \sqrt{\frac{L_m}{g_m}} \]
Squaring both sides gives:
\[ 4 = 4\pi^2 \frac{L_m}{g_m} \]
Simplifying this, we find:
\[ L_m = \frac{g_m}{\pi^2} \]

6. **Substituting for Moon's Gravity**:
Substitute \( g_m = \frac{g_e}{6} \) into the equation for \( L_m \):
\[ L_m = \frac{\frac{g_e}{6}}{\pi^2} = \frac{g_e}{6\pi^2} \]

7. **Relate Lengths on Earth and Moon**:
Now, we can relate \( L_m \) to \( L_e \):
\[ L_m = \frac{1}{6} L_e \]

8. **Conclusion**:
Therefore, the length of the second pendulum on the Moon is:
\[ L_m = \frac{1}{6} L_e \]

### Final Answer:
The length of the second pendulum on the Moon would be \( \frac{1}{6} \) times that of the length on Earth.

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