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 need to understand the relationship between the length of the pendulum, the acceleration due to gravity, and the time period of the pendulum.

### 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. **Given Information**:
- The time period of the second pendulum is 2 seconds on Earth.
- The acceleration due to gravity on the Moon is \( \frac{1}{6} \) of that on Earth, i.e., \( g_m = \frac{g_e}{6} \).

3. **Setting Up the Equation for Earth**:
For Earth, we can express the time period as:
\[ T_e = 2\pi \sqrt{\frac{L_e}{g_e}} \]
Since \( T_e = 2 \) seconds, we have:
\[ 2 = 2\pi \sqrt{\frac{L_e}{g_e}} \]

4. **Squaring Both Sides**:
Squaring both sides gives:
\[ 4 = 4\pi^2 \frac{L_e}{g_e} \]
Simplifying this, we find:
\[ L_e = \frac{g_e}{\pi^2} \]

5. **Setting Up the Equation for the Moon**:
For the Moon, the time period is given by:
\[ T_m = 2\pi \sqrt{\frac{L_m}{g_m}} \]
Since \( T_m = 2 \) seconds as well, we have:
\[ 2 = 2\pi \sqrt{\frac{L_m}{g_m}} \]

6. **Squaring Both Sides Again**:
Squaring both sides gives:
\[ 4 = 4\pi^2 \frac{L_m}{g_m} \]
Simplifying this, we find:
\[ L_m = \frac{g_m}{\pi^2} \]

7. **Substituting for \( g_m \)**:
Since \( g_m = \frac{g_e}{6} \), we substitute this into the equation for \( L_m \):
\[ L_m = \frac{\frac{g_e}{6}}{\pi^2} = \frac{L_e}{6} \]

8. **Conclusion**:
Thus, the length of the second pendulum on the Moon is:
\[ L_m = \frac{L_e}{6} \]
This means the length of the second pendulum on the Moon is \( \frac{1}{6} \) times that of the Earth.

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