Question

In an experiment of the measurement of \( g \) using simple pendulum, the time period was measured with an accuracy of \( 0.2% \), while the length was measured with an accuracy of \( 0.5% \). The percentage accuracy in the value of \( g \) thus obtained is

• A. 0.7%
• B. 0.3%
• C. 0.9%
• D. 0.1%

Hint:

When calculating the error in a derived quantity, combine the percentage errors of the individual quantities based on how they contribute to the formula.

Answer 1

The Correct Option is C

Step 1: Using the formula for acceleration due to gravity.
The formula for \( g \) using a pendulum is given by: \[ g = \frac{4\pi^2 L}{T^2} \] where \( L \) is the length and \( T \) is the time period.
Step 2: Calculating the accuracy.
The percentage error in \( g \) is given by the sum of the percentage errors in \( L \) and \( T \), since \( g \) depends on both. The percentage error in \( T^2 \) is twice the percentage error in \( T \), so: \[ \text{Percentage error in } g = 0.5% + 2 \times 0.2% = 0.9% \] Step 3: Conclusion.
The correct answer is (C), 0.9%.