Q 11-01-177JEE MainJEE Main 2021 (24 Feb, Shift 2)Easy
The period of oscillation of a simple pendulum is $T = 2\pi\sqrt{\dfrac{L}{g}}$. Measured value of $L$ is 1.0 m from meter scale having a minimum division of 1 mm and time of one complete oscillation is 1.95 s measured from stopwatch of 0.01 s resolution. The percentage error in the determination of $g$ will be:
Answer: (D) 1.13%
$g = \dfrac{4\pi^2L}{T^2}$, so
$$\frac{\Delta g}{g}\times100 = \frac{\Delta L}{L}\times100 + 2\frac{\Delta T}{T}\times100 = \frac{0.001}{1.0}\times100 + 2\times\frac{0.01}{1.95}\times100$$
$$= 0.10\% + 1.03\% \approx 1.13\%$$
Solution by Sreeraj P, M.Sc Physics