If a Seconds pendulum is moved to a planet where acceleration due to gravity is 4 times, the length of the second's pendulum on the planet should be made

Options

A. 2 times
B. 4 times
C. 8 times
D. 15 times

Answer

To solve the problem of determining the new length of a seconds pendulum on a planet where the acceleration due to gravity is four times that of Earth, 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 the Seconds Pendulum:**
A seconds pendulum has a time period of 2 seconds. Therefore, we can set:
\[ T = 2 \text{ seconds} \]

3. **Set Up the Equation for Earth:**
For the seconds pendulum on Earth, we can express the time period as:
\[ 2 = 2\pi \sqrt{\frac{L}{g}} \]
Rearranging this gives:
\[ \sqrt{\frac{L}{g}} = \frac{1}{\pi} \]
Squaring both sides results in:
\[ \frac{L}{g} = \frac{1}{\pi^2} \]
Therefore, we can express the length \( L \) as:
\[ L = \frac{g}{\pi^2} \]

4. **Set Up the Equation for the New Planet:**
On the new planet, the acceleration due to gravity \( g' \) is four times that of Earth:
\[ g' = 4g \]
We want the time period to remain the same (2 seconds), so we set up the equation for the new length \( L' \):
\[ 2 = 2\pi \sqrt{\frac{L'}{g'}} \]
Rearranging gives:
\[ \sqrt{\frac{L'}{g'}} = \frac{1}{\pi} \]
Squaring both sides results in:
\[ \frac{L'}{g'} = \frac{1}{\pi^2} \]
Substituting \( g' = 4g \) into the equation gives:
\[ \frac{L'}{4g} = \frac{1}{\pi^2} \]
Therefore, we can express the new length \( L' \) as:
\[ L' = \frac{4g}{\pi^2} \]

5. **Relate the New Length to the Original Length:**
From the previous step, we have:
\[ L' = 4 \left(\frac{g}{\pi^2}\right) = 4L \]
This shows that the new length \( L' \) must be four times the original length \( L \).

### Conclusion:
To maintain the time period of a seconds pendulum at 2 seconds on a planet where the acceleration due to gravity is four times that of Earth, the length of the pendulum should be made **4 times the original length**.