JEE Mains · Physics · STD 11 - Units,dimensions and measurement
- JEE Mains 2022
- Medium
- Numerical
In an experiment to find acceleration due to gravity \((g)\) using simple pendulum, time period of \(0.5\,s\) is measured from time of \(100\) oscillation with a watch of \(1\;s\) resolution. If measured value of length is \(10\; cm\) known to \(1\; mm\) accuracy. The accuracy in the determination of \(g\) is found to be \(x \%\). The value of \(x\) is
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Answer & Solution
Correct Answer
(B) \(5\)
Step-by-step Solution
Detailed explanation
\(T=2 \pi \sqrt{\frac{\ell}{g}}\)
\(g=\frac{1}{4 \pi^{2}} \frac{T^{2}}{\ell}\)
\(\frac{\Delta g}{g}=\frac{2 \Delta T }{ T }+\frac{\Delta \ell}{\ell}\)
\(\frac{\Delta g}{g}=2 \cdot \frac{1}{100 \times 0.5}+\frac{1\,mm }{10\,cm }\)
\(\frac{\Delta g}{g}=\frac{5}{100}\)
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