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 $\mathrm{T}$ is being estimated. If time period is measured to an accuracy of $3 \%$, the accuracy to which $\mathrm{E}$ is known as .........\%
$\mathrm{T}=2 \pi \sqrt{\frac{\ell}{\mathrm{g}}} \Rightarrow \ell=\frac{\mathrm{T}^{2} \mathrm{~g}}{4 \pi^{2}}$
$\mathrm{E}=\mathrm{mg} \ell \frac{\theta^{2}}{2}=\mathrm{mg}^{2} \frac{\mathrm{T}^{2} \theta^{2}}{8 \pi^{2}}$
$\frac{\mathrm{dE}}{\mathrm{E}}=2\left(\frac{\mathrm{dg}}{\mathrm{g}}+\frac{\mathrm{dT}}{\mathrm{T}}\right)$
$=(4+3)=14 \%$
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