In an experiment to find $g$ with a simple pendulum, the length of the pendulum is measured as $100.0$ cm using a metre scale of least count $1$ mm, and the time for $20$ oscillations is measured as $40.0$ s using a stopwatch of least count $0.1$ s. The maximum percentage error in the value of $g$ is
Answer: (D) $0.6\%$
Concept. $g = \dfrac{4\pi^2L}{T^2}$, so the maximum relative error is $\dfrac{\Delta g}{g} = \dfrac{\Delta L}{L} + 2\dfrac{\Delta T}{T}$. Timing many oscillations gives the same relative error in $T$ as in the total time.
Step 1. Error in $L$.
$$\frac{\Delta L}{L} = \frac{0.1\ \text{cm}}{100.0\ \text{cm}} = 0.1\%$$
Step 2. Error in $T$. $T = \dfrac{t}{20}$, so
$$\frac{\Delta T}{T} = \frac{\Delta t}{t} = \frac{0.1}{40.0} = 0.25\%$$
Step 3. Error in $g$.
$$\frac{\Delta g}{g} = 0.1\% + 2(0.25\%) = 0.6\%$$
Answer. $0.6\%$ (Option D). Forgetting the factor 2 gives $0.35\%$.
Solution by Sreeraj P, M.Sc Physics