The time period of a pendulum is proportiona to the square rot of the length of the pendulum.Consider the following statements:
1. If the length of the pendulum is doubled, then the time period is also doubled.
2. If the length is halved, then time period becomes one fourth of the original time period.
The correct assertions are

Options

A. 1
B. 2
C. neigher 1 nor 2
D. both 1 and 2

Answer

To solve the problem, we need to analyze the relationship between the length of the pendulum and its time period based on the given statements.

### Given:
The time period \( T \) of a pendulum is proportional to the square root of its length \( L \):
\[ T \propto \sqrt{L} \]
This can be expressed as:
\[ T = k \sqrt{L} \]
where \( k \) is a proportionality constant.

### Analyzing the Statements:

1. **If the length of the pendulum is doubled:**
- If \( L \) is doubled, then:
\[ L' = 2L \]
- The new time period \( T' \) becomes:
\[ T' = k \sqrt{L'} = k \sqrt{2L} = k \sqrt{2} \sqrt{L} = \sqrt{2} k \sqrt{L} = \sqrt{2} T \]
- Since \( \sqrt{2} \) is approximately 1.414, the time period does not double; it increases by a factor of \( \sqrt{2} \).

2. **If the length is halved:**
- If \( L \) is halved, then:
\[ L' = \frac{1}{2}L \]
- The new time period \( T' \) becomes:
\[ T' = k \sqrt{L'} = k \sqrt{\frac{1}{2}L} = k \sqrt{\frac{1}{2}} \sqrt{L} = \frac{1}{\sqrt{2}} k \sqrt{L} = \frac{1}{\sqrt{2}} T \]
- This means the time period becomes \( \frac{1}{\sqrt{2}} \) times the original time period, not one fourth.

### Conclusion:
Both statements are incorrect:
1. Doubling the length does not double the time period; it increases it by a factor of \( \sqrt{2} \).
2. Halving the length does not make the time period one fourth; it reduces it to \( \frac{1}{\sqrt{2}} \) of the original time period.

### Final Answer:
Neither statement 1 nor statement 2 is correct.

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