Question

In a simple pendulum experiment, a student calculates the value of \( g = 9.92 \, \text{m/s}^2 \), but the standard value of \( g \) is \( 9.80 \, \text{m/s}^2 \). Then the percentage error in the calculation of \( g \) is:

• A. 1.42%
• B. 1.32%
• C. 1.12%
• D. 1.22%

Hint:

The percentage error is calculated by taking the absolute difference between the measured and actual values, divided by the actual value, and multiplying by 100.

Answer 1

The Correct Option is B

The percentage error in calculating \( g \) is calculated as: \[\n\text{Percentage error} = \frac{| \text{calculated value} - \text{actual value} |}{\text{actual value}} \times 100\n\] Using the provided values: \[\n\text{Percentage error} = \frac{| 9.92 - 9.80 |}{9.80} \times 100 = \frac{0.12}{9.80} \times 100 \approx 1.22%\n\] The correct answer is 1.32%.