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(D) The time period of a simple pendulum is given by $T = 2\pi \sqrt{\frac{\ell}{g}}$,which implies $\ell = \frac{T^2 g}{4\pi^2}$.The energy $E$ of a

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$14$

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(D) The time period of a simple pendulum is given by $T = 2\pi \sqrt{\frac{\ell}{g}}$,which implies $\ell = \frac{T^2 g}{4\pi^2}$.The energy $E$ of a simple pendulum for small oscillations is given by $E = \frac{1}{2} m g \ell 	heta^2$,where $	heta$ is the amplitude.Substituting the expression for $\ell$,we get $E = \frac{1}{2} m g \left( \frac{T^2 g}{4\pi^2} ight) 	heta^2 = \frac{m g^2 T^2 	heta^2}{8\pi^2}$.Assuming mass $m$ and amplitude $	heta$ are constant,the relative error in $E$ is given by $\frac{\Delta E}{E} = 2 \frac{\Delta g}{g} + 2 \frac{\Delta T}{T}$.Given $\frac{\Delta g}{g} = 4\%$ and $\frac{\Delta T}{T} = 3\%$,we have $\frac{\Delta E}{E} = 2(4\%) + 2(3\%) = 8\% + 6\% = 14\%$.Thus,the accuracy to which $E$ is known is $14\%$.

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