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Q. The acceleration due to gravity is found upto an accuracy of $4 \%$ on a planet. The energy supplied to a simple pendulum to known mass ' $m$ ' to undertake oscillations of time period $T$ is being estimated. If time period is measured to an accuracy of $3 \%$, the accuracy to which $E$ is known as ...........$\%$

JEE MainJEE Main 2021Gravitation

Solution:

$T =2 \pi \sqrt{\frac{\ell}{ g }}$
$ \Rightarrow \ell=\frac{ T ^{2} g }{4 \pi^{2}}$
$E = mg \ell \frac{\theta^{2}}{2}$
$= mg ^{2} \frac{ T ^{2} \theta^{2}}{8 \pi^{2}}$
$\frac{ dE }{ E }=2\left(\frac{ dg }{ g }+\frac{ dT }{ T }\right)$
$=(4+3)=14 \%$