JEE Mains · Physics · STD 11 - Units,dimensions and measurement
- JEE Mains 2020
- Medium
- MCQ
A simple pendulum is being used to determine the value of gravitational acceleration \(\mathrm{g}\) at a certain place. The length of the pendulum is \(25.0\; \mathrm{cm}\) and a stop watch with \(1\; \mathrm{s}\) resolution measures the time taken for \(40\) oscillations to be \(50\; s\). The accuracy in \(g\) is ....... \(\%\)
Pick an option to check your answer.
Answer & Solution
Correct Answer
(C) \(4.40 \)
Step-by-step Solution
Detailed explanation
\(\mathrm{T}=2 \pi \sqrt{\frac{\ell}{\mathrm{g}}}\)
\(\mathrm{g}=\frac{4 \pi^{2} \ell}{\mathrm{T}^{2}}\)
\(\frac{\Delta g}{g}=\frac{\Delta \ell}{\ell}+\frac{2 \Delta T}{T}\)
\(=\frac{0.1}{25}+\frac{2 \times 1}{50}\)
\(\frac{\Delta g}{g}=4.4 \%\)
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