While measuring the acceleration due to gravity by a simple pendulum , a student makes a positive error of `1%` in the length of the pendulum and a negative error of `3%` in the value of time period . His percentage error in the measurement of `g` by the relation ` g = 4 pi^(2) ( l // T^(2))` will be

Options

A. `2 %`
B. `4 %`
C. ` 7%`
D. `10%`

Explanation

To solve the problem of finding the percentage error in the measurement of acceleration due to gravity \( g \) using the formula \( g = 4 \pi^2 \frac{l}{T^2} \), we will follow these steps:

### Step-by-Step Solution

1. **Identify the formula**: The formula for \( g \) is given as:
\[ g = 4 \pi^2 \frac{l}{T^2} \]
Here, \( l \) is the length of the pendulum and \( T \) is the time period.

2. **Determine the errors**: The problem states that there is a positive error of \( 1\% \) in the length \( l \) and a negative error of \( 3\% \) in the time period \( T \).

3. **Express the errors in terms of relative errors**:
- The relative error in \( l \) (length) is:
\[ \frac{\Delta l}{l} = 1\% \]
- The relative error in \( T \) (time period) is:
\[ \frac{\Delta T}{T} = -3\% \]

4. **Apply the formula for percentage error**: The formula for the percentage error in \( g \) based on the errors in \( l \) and \( T \) is given by:
\[ \frac{\Delta g}{g} = \frac{\Delta l}{l} + 2 \frac{\Delta T}{T} \]
Here, the factor of \( 2 \) comes from the \( T^2 \) in the denominator.

5. **Substitute the values**:
- Substitute \( \frac{\Delta l}{l} = 1\% \) and \( \frac{\Delta T}{T} = -3\% \):
\[ \frac{\Delta g}{g} = 1\% + 2(-3\%) \]

6. **Calculate the total percentage error**:
\[ \frac{\Delta g}{g} = 1\% - 6\% = -5\% \]

7. **Convert to percentage**: Since we are interested in the absolute percentage error, we take the magnitude:
\[ \text{Percentage error in } g = | -5\% | = 5\% \]

### Final Answer
The percentage error in the measurement of \( g \) is \( 5\% \).

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