The length of a second's pendulum on the surface of the moon, where g is `1//6^(th)` if the value of g on the surface of the

Options

A. `1//36m`
B. `1//6m`
C. `6m`
D. `36m`

Explanation

To find the length of a second's pendulum on the surface of the moon, we can follow these steps:

### Step 1: Understand the relationship between period, length, and gravity
The formula for the period \( T \) of a simple pendulum is given by:
\[ T = 2\pi \sqrt{\frac{L}{g}} \]
where \( L \) is the length of the pendulum and \( g \) is the acceleration due to gravity.

### Step 2: Set up the equations for Earth and Moon
Let \( g_e \) be the acceleration due to gravity on Earth and \( g_m \) be the acceleration due to gravity on the Moon. According to the problem, we have:
\[ g_m = \frac{g_e}{6} \]

For a second's pendulum, the period \( T \) is 2 seconds (1 second for each half swing). Therefore, we can write the equations for the period on Earth and the Moon as follows:

**On Earth:**
\[ T = 2\pi \sqrt{\frac{L_e}{g_e}} = 2 \quad \text{(for a second's pendulum)} \]

**On Moon:**
\[ T = 2\pi \sqrt{\frac{L_m}{g_m}} = 2 \]

### Step 3: Equate the periods
Since the period of the pendulum is the same on both Earth and the Moon, we can set the two equations equal to each other:
\[ 2\pi \sqrt{\frac{L_e}{g_e}} = 2\pi \sqrt{\frac{L_m}{g_m}} \]

### Step 4: Simplify the equation
We can cancel \( 2\pi \) from both sides:
\[ \sqrt{\frac{L_e}{g_e}} = \sqrt{\frac{L_m}{g_m}} \]

### Step 5: Square both sides
Squaring both sides gives us:
\[ \frac{L_e}{g_e} = \frac{L_m}{g_m} \]

### Step 6: Substitute \( g_m \)
Substituting \( g_m = \frac{g_e}{6} \) into the equation:
\[ \frac{L_e}{g_e} = \frac{L_m}{\frac{g_e}{6}} \]

### Step 7: Rearrange the equation
This can be rearranged to:
\[ \frac{L_e}{g_e} = \frac{6L_m}{g_e} \]

### Step 8: Cancel \( g_e \)
Since \( g_e \) is common on both sides, we can cancel it:
\[ L_e = 6L_m \]

### Step 9: Solve for \( L_m \)
Now, if we know the length of the second's pendulum on Earth \( L_e \) (which is approximately 1 meter), we can find \( L_m \):
\[ L_m = \frac{L_e}{6} = \frac{1 \text{ m}}{6} \approx 0.1667 \text{ m} \]

### Final Result
Thus, the length of a second's pendulum on the surface of the Moon is approximately:
\[ L_m \approx 0.1667 \text{ m} \text{ or } 16.67 \text{ cm} \]

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