JEE Mains · Physics · STD 11 - Units,dimensions and measurement
- JEE Mains 2021
- Medium
- MCQ
The period of oscillation of a simple pendulum is \(T =2 \pi \sqrt{\frac{ L }{ g }} .\) Measured value of \( L \) is \(1.0\, m\) from meter scale having a minimum division of \(1 \,mm\) and time of one complete oscillation is \(1.95\, s\) measured from stopwatch of \(0.01 \,s\) resolution. The percentage error in the determination of \(g\) will be ..... \(\%.\)
Pick an option to check your answer.
Answer & Solution
Correct Answer
(A) \(1.13\)
Step-by-step Solution
Detailed explanation
\(T =2 \pi \sqrt{\frac{\ell}{ g }}\)
\(g =\frac{4 \pi^{2} \ell}{ T ^{2}}\)
\(\frac{\Delta g }{ g }=\frac{\Delta \ell}{\ell}+\frac{2 \Delta T }{ T }\)
\(\frac{\Delta g }{ g }=\frac{1 \times 10^{-3}}{1}+2 \times \frac{0.01}{1.95}\)
\(\frac{\Delta g }{ g }=0.0113\) or \(1.13\, \%\)
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