MHT CET · Physics · Mathematics in Physics
The period of oscillating simple pendulum is \(T=2 \pi \sqrt{\frac{\ell}{g}}\) where length ' \(\ell\) ' is 100 cm with error 1 mm. Period is 2 second. The time of 100 oscillations is measured by a stopwatch of least count 0.1 s. The percentage error in gravitational acceleration ' g ' is
- A \(0 \cdot 2 \%\)
- B \(0 \cdot 1\) %
- C \(1 \%\)
- D 2 %
Answer & Solution
Correct Answer
(A) \(0 \cdot 2 \%\)
Step-by-step Solution
Detailed explanation
\(g = 4 \pi^2 \frac{\ell}{T^2}\) \(\frac{\Delta g}{g} = \frac{\Delta \ell}{\ell} + 2 \frac{\Delta T}{T}\)
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