The time period of a simple pendulum on the surface of the earth is 4s. Its time period on the surface of the moon is

Options

A. 4s
B. 8s
C. 10s
D. 12s

Explanation

To find the time period of a simple pendulum on the surface of the moon, we can follow these steps:

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

### Step 2: Identify the known values
From the question, we know:
- The time period on the surface of the Earth (\( T_e \)) is 4 seconds.
- The acceleration due to gravity on Earth (\( g_e \)) is approximately \( 9.81 \, \text{m/s}^2 \).
- The acceleration due to gravity on the Moon (\( g_m \)) is approximately \( \frac{1}{6} g_e \).

### Step 3: Set up the ratio of the time periods
Since the length of the pendulum (\( L \)) remains the same on both the Earth and the Moon, we can set up the ratio of the time periods:
\[ \frac{T_e}{T_m} = \sqrt{\frac{g_m}{g_e}} \]

### Step 4: Substitute known values into the equation
We know that:
\[ g_m = \frac{1}{6} g_e \]
Substituting this into the ratio gives:
\[ \frac{T_e}{T_m} = \sqrt{\frac{\frac{1}{6} g_e}{g_e}} = \sqrt{\frac{1}{6}} \]

### Step 5: Solve for \( T_m \)
Now we can rearrange the equation to solve for \( T_m \):
\[ T_m = T_e \cdot \sqrt{6} \]
Substituting \( T_e = 4 \, \text{s} \):
\[ T_m = 4 \cdot \sqrt{6} \]

### Step 6: Calculate the numerical value
Calculating \( \sqrt{6} \) gives approximately \( 2.45 \):
\[ T_m \approx 4 \cdot 2.45 \approx 9.8 \, \text{s} \]
Rounding this gives us approximately \( 10 \, \text{s} \).

### Conclusion
The time period of the simple pendulum on the surface of the moon is approximately **10 seconds**.

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