Q 11-01-194JEE MainJEE Main 2020 (8 Jan, Shift 2)Medium
A simple pendulum is being used to determine the value of gravitational acceleration $g$ at a certain place. The length of the pendulum is $25.0$ cm and a stopwatch with $1$ s resolution measures the time taken for $40$ oscillations to be $50$ s. The accuracy in $g$ is:
Answer: (C) $4.40\%$
From $T = 2\pi\sqrt{L/g}$ we get $g = \dfrac{4\pi^2 L}{T^2}$, so
$$\frac{\Delta g}{g} = \frac{\Delta L}{L} + 2\,\frac{\Delta T}{T}$$
Length $25.0$ cm is read to $0.1$ cm: $\dfrac{\Delta L}{L} = \dfrac{0.1}{25.0} = 0.4\%$.
The total time $t = 50$ s is measured to $1$ s, and $T = t/40$, so $\dfrac{\Delta T}{T} = \dfrac{\Delta t}{t} = \dfrac{1}{50} = 2\%$.
$$\frac{\Delta g}{g} = 0.4\% + 2(2\%) = 4.4\%$$
Solution by Sreeraj P, M.Sc Physics