Time period T of a simple pendulum of length l is given by T =`2pisqrt((l)/(g))` . If the length is increased by `2%` then an approximate change in the time period is

Options

A. 0.02
B. `1%`
C. `(1)/(2)%`
D. none of these

Answer

To solve the problem, we need to find the approximate change in the time period \( T \) of a simple pendulum when the length \( l \) is increased by \( 2\% \). The formula for the time period \( T \) is given by:

\[ T = 2\pi \sqrt{\frac{l}{g}} \]

### Step-by-Step Solution:

1. **Understanding the Formula**:
The time period \( T \) depends on the length \( l \) of the pendulum and the acceleration due to gravity \( g \).

2. **Differentiate the Time Period**:
To find the change in \( T \) with respect to a change in \( l \), we can differentiate \( T \) with respect to \( l \). We can use the formula:
\[ \frac{dT}{T} = \frac{1}{2} \frac{dl}{l} \]
This comes from implicit differentiation of the formula for \( T \).

3. **Substituting the Change in Length**:
We know that the length \( l \) is increased by \( 2\% \). Therefore, we can express this as:
\[ \frac{dl}{l} = \frac{2}{100} = 0.02 \]

4. **Calculate the Change in Time Period**:
Substitute \( \frac{dl}{l} \) into the differentiated equation:
\[ \frac{dT}{T} = \frac{1}{2} \times 0.02 = 0.01 \]

5. **Convert to Percentage Change**:
To find the percentage change in the time period, we multiply by \( 100 \):
\[ \text{Percentage change in } T = \frac{dT}{T} \times 100 = 0.01 \times 100 = 1\% \]

### Final Answer:
The approximate change in the time period \( T \) when the length is increased by \( 2\% \) is \( 1\% \).

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