Q 11-01-219JEE MainJEE Main 2019 (8 Apr, Shift 2)Easy
In a simple pendulum experiment for determination of acceleration due to gravity ($g$), time taken for 20 oscillations is measured by using a watch of 1 second least count. The mean value of time taken comes out to be $30\ \text{s}$. The length of the pendulum is measured by using a meter scale of least count $1\ \text{mm}$ and the value obtained is $55.0\ \text{cm}$. The percentage error in the determination of $g$ is close to
Answer: (B) $6.8\%$
$g = \dfrac{4\pi^2l}{T^2}$, so
$$\frac{\Delta g}{g} = \frac{\Delta l}{l} + 2\frac{\Delta t}{t} = \frac{0.1}{55.0} + 2\times\frac{1}{30} = 0.0018 + 0.0667 = 0.0685$$
The percentage error is about $6.8\%$.
Solution by Sreeraj P, M.Sc Physics