While measuring acceleration due to gravity by simpe 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 the time period. His percentage error in the measurement of the value of g will be -

Options

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

Explanation

To find the percentage error in the measurement of the acceleration due to gravity (g) using a simple pendulum, we can follow these steps:

### Step-by-Step Solution:

1. **Understand the Formula**: The time period (T) of a simple pendulum is given by the formula:
\[ T = 2\pi \sqrt{\frac{L}{g}} \]
Rearranging this formula to express \(g\) in terms of \(T\) and \(L\):
\[ g = \frac{4\pi^2 L}{T^2} \]

2. **Identify Errors**: We are given:
- Positive error in length (L): \( +1\% \)
- Negative error in time period (T): \( -3\% \)

3. **Calculate the Percentage Error in g**: The formula for the percentage error in \(g\) based on the errors in \(L\) and \(T\) is:
\[ \frac{\Delta g}{g} \times 100 = \frac{\Delta L}{L} \times 100 + 2 \times \frac{\Delta T}{T} \times 100 \]
Here, the error in \(L\) contributes directly, while the error in \(T\) contributes twice because \(T\) is squared in the formula for \(g\).

4. **Substituting the Values**:
- Error in \(L\): \( +1\% \)
- Error in \(T\): \( -3\% \)
\[ \frac{\Delta g}{g} \times 100 = 1 + 2 \times (-3) \]

5. **Calculating the Result**:
\[ \frac{\Delta g}{g} \times 100 = 1 - 6 = -5\% \]

6. **Conclusion**: The percentage error in the measurement of \(g\) is \( -5\% \).

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