MHT CET · Physics · Oscillations
In an experiment of the measurement of 'g' using simple pendulum, the time period was measured with an accuracy of \(0 \cdot 2 \%\) while the length was measured with an accuracy of \(0.5 \%\). The percentage accuracy in the value of ' \(\mathrm{g}\) ' thus obtained is
- A \(0 \cdot 7 \%\)
- B \(0 \cdot 3 \%\)
- C \(0 \cdot 9 \%\)
- D \(0 \cdot 1 \%\)
Answer & Solution
Correct Answer
(C) \(0 \cdot 9 \%\)
Step-by-step Solution
Detailed explanation
\( g = \frac{4\pi^2 L}{T^2} \) \( \frac{\Delta g}{g} \times 100\% = \frac{\Delta L}{L} \times 100\% + 2 \frac{\Delta T}{T} \times 100\% \)
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