Q 11-01-125JEE MainJEE Main 2024 (31 Jan, Shift 2)Medium
The measured value of the length of a simple pendulum is $20\ \text{cm}$ with $2\ \text{mm}$ accuracy. The time for 50 oscillations was measured to be 40 seconds with 1 second resolution. From these measurements, the accuracy in the measurement of acceleration due to gravity is $N\%$. The value of $N$ is:
Answer: (C) $6$
$g = \dfrac{4\pi^2L}{T^2}$, so
$$\frac{\Delta g}{g}\times100 = \frac{0.2}{20}\times100 + 2\times\frac{1}{40}\times100 = 1 + 5 = 6\%$$
Solution by Sreeraj P, M.Sc Physics