Q 11-13-120JEE MainJEE Main 2021 (26 Aug, Shift 2)Easy
If the length of the pendulum in pendulum clock increases by $0.1\%$, then the error in time per day is:
Answer: (A) $43.2$ s
$T \propto \sqrt{l}$, so $\dfrac{\Delta T}{T} = \dfrac{1}{2}\dfrac{\Delta l}{l} = 0.05\% = 5\times10^{-4}$.
The clock loses this fraction of time: error per day $= 86400\times5\times10^{-4} = 43.2$ s.
Solution by Sreeraj P, M.Sc Physics