JEE Mains · Physics · STD 11 - 7. gravitation
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 \(24\)
- D \(14\)
Answer & Solution
Correct Answer
(D) \(14\)
Step-by-step Solution
Detailed explanation
\({T}=2 \pi \sqrt{\frac{\ell}{{g}}} \Rightarrow \ell=\frac{{T}^{2} {g}}{4 \pi^{2}}\) \({E}={mg} \ell \frac{\theta^{2}}{2}={mg}^{2} \frac{{T}^{2} \theta^{2}}{8 \pi^{2}}\) \(\frac{{dE}}{{E}}=2\left(\frac{{dg}}{{g}}+\frac{{dT}}{{T}}\right)\) \(=(4+3)=14\, \%\)
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