JEE Mains · Physics · STD 11 - Units,dimensions and measurement
- JEE Mains 2021
- Hard
- Numerical
The time period of a simple pendulum is given by \(T =2 \pi \sqrt{\frac{\ell}{ g }}\). The measured value of the length of pendulum is \(10\, cm\) known to a \(1\, mm\) accuracy. The time for \(200\) oscillations of the pendulum is found to be \(100\) second using a clock of \(1s\) resolution. The percentage accuracy in the determination of \('g'\) using this pendulum is \('x'\). The value of \('x'\) to the nearest integer is ...........\(\%\)
Pick an option to check your answer.
Answer & Solution
Correct Answer
(B) \(3\)
Step-by-step Solution
Detailed explanation
\(g=\frac{4 \pi^{2} \ell}{T^{2}}\)
\(\frac{\Delta g}{g}=\frac{\Delta \ell}{\ell}+2 \frac{\Delta T}{T}=\frac{0.1}{10}+2\left(\frac{\frac{1}{200}}{0.5}\right)\)
\(\frac{\Delta g}{g}=\frac{1}{100}+\frac{1}{50}\)
\(\frac{\Delta g}{g} \times 100=3 \%\)
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