The length of a pendulum is measured as $1.01\ \text{m}$ and the time for 30 oscillations is measured as one minute 3 seconds. The error in length is $0.01\ \text{m}$ and the error in time is $3\ \text{s}$. The percentage error in the measurement of acceleration due to gravity is
Answer: (C) $10$
From $T = 2\pi\sqrt{L/g}$, we get $g = \dfrac{4\pi^2 L}{T^2}$, so
$$\frac{\Delta g}{g}\times 100 = \frac{\Delta L}{L}\times 100 + 2\,\frac{\Delta t}{t}\times 100$$
The time for 30 oscillations is $t = 63\ \text{s}$ with error $3\ \text{s}$. (Using total time or time period gives the same fractional error.)
$$\frac{0.01}{1.01}\times 100 + 2\times\frac{3}{63}\times 100 \approx 0.99\% + 9.52\% \approx 10.5\%$$
The closest option is $10\%$.
Solution by Sreeraj P, M.Sc Physics