The time period of a simple pendulum is 2s. If its length is increased by 4 times, then its period becomes
(a) `16s` (b) `12s` (c) `8s` (d) `4s`

Options

A. `16s`
B. `12s`
C. `8s`
D. `4s`

Answer

To solve the problem, we need to determine the new time period of a simple pendulum when its length is increased by four times. We will use the formula for the time period of a simple pendulum:

### Step-by-Step Solution:

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 \( L \) is the length of the pendulum and \( g \) is the acceleration due to gravity.

2. **Identify the initial conditions:**
We are given that the initial time period \( T_1 \) is 2 seconds. Therefore:
\[ T_1 = 2 \text{ s} \]

3. **Determine the initial length:**
Let the initial length of the pendulum be \( L \).

4. **Calculate the new length:**
If the length is increased by four times, the new length \( L_2 \) will be:
\[ L_2 = 4L \]

5. **Express the new time period using the new length:**
The new time period \( T_2 \) can be expressed as:
\[ T_2 = 2\pi \sqrt{\frac{L_2}{g}} = 2\pi \sqrt{\frac{4L}{g}} \]

6. **Simplify the expression for the new time period:**
We can simplify \( T_2 \):
\[ T_2 = 2\pi \sqrt{4} \sqrt{\frac{L}{g}} = 2\pi \cdot 2 \sqrt{\frac{L}{g}} = 4 \cdot (2\pi \sqrt{\frac{L}{g}}) \]
Since \( 2\pi \sqrt{\frac{L}{g}} = T_1 \), we can substitute:
\[ T_2 = 4 \cdot T_1 \]

7. **Substitute the value of \( T_1 \):**
Now substituting \( T_1 = 2 \text{ s} \):
\[ T_2 = 4 \cdot 2 \text{ s} = 8 \text{ s} \]

8. **Final answer:**
Therefore, the new time period \( T_2 \) is:
\[ T_2 = 8 \text{ s} \]
The correct option is (c) 8s.