Q 11-01-095JEE MainJEE Main 2024 (5 Apr, Shift 1)Easy
Time periods of oscillation of the same simple pendulum measured using four different measuring clocks were recorded as $4.62\ \text{s}$, $4.632\ \text{s}$, $4.6\ \text{s}$ and $4.64\ \text{s}$. The arithmetic mean of these readings in the correct significant figures is
Answer: (C) $4.6\ \text{s}$
$$\bar T = \frac{4.62 + 4.632 + 4.6 + 4.64}{4} = \frac{18.492}{4} = 4.623\ \text{s}$$
In addition, the result keeps as many decimal places as the least precise reading. $4.6\ \text{s}$ has one decimal place, so the mean is $4.6\ \text{s}$.
Solution by Sreeraj P, M.Sc Physics