JEE Mains · Physics · STD 11 - 1. units,dimensions and measurement
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 ..... \(\%.\)
- A \(1.13\)
- B \(1.03\)
- C \(1.33\)
- D \(1.30\)
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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