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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

  1. A \(0 \cdot 2 \%\)
  2. B \(0 \cdot 1\) %
  3. C \(1 \%\)
  4. D 2 %
Verified Solution

Answer & Solution

Correct Answer

(A) \(0 \cdot 2 \%\)

Step-by-step Solution

Detailed explanation

\(g = 4 \pi^2 \frac{\ell}{T^2}\) \(\frac{\Delta g}{g} = \frac{\Delta \ell}{\ell} + 2 \frac{\Delta T}{T}\)