MCQ
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 $..........\,\%$
  • A
    $85$
  • B
    $31$
  • C
    $24$
  • $14$

Answer

Correct option: D.
$14$
d
${T}=2 \pi \sqrt{\frac{\ell}{{g}}} \Rightarrow \ell=\frac{{T}^{2} {g}}{4 \pi^{2}}$

${E}={mg} \ell \frac{\theta^{2}}{2}={mg}^{2} \frac{{T}^{2} \theta^{2}}{8 \pi^{2}}$

$\frac{{dE}}{{E}}=2\left(\frac{{dg}}{{g}}+\frac{{dT}}{{T}}\right)$

$=(4+3)=14\, \%$

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