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JEE Mains · Physics · STD 11 - 1. units,dimensions and measurement

In an experiment to find acceleration due to gravity \((g)\) using simple pendulum, time period of \(0.5\,s\) is measured from time of \(100\) oscillation with a watch of \(1\;s\) resolution. If measured value of length is \(10\; cm\) known to \(1\; mm\) accuracy. The accuracy in the determination of \(g\) is found to be \(x \%\). The value of \(x\) is

  1. A \(4\)
  2. B \(5\)
  3. C \(3\)
  4. D \(2\)
Verified Solution

Answer & Solution

Correct Answer

(B) \(5\)

Step-by-step Solution

Detailed explanation

\(T=2 \pi \sqrt{\frac{\ell}{g}}\) \(g=\frac{1}{4 \pi^{2}} \frac{T^{2}}{\ell}\) \(\frac{\Delta g}{g}=\frac{2 \Delta T }{ T }+\frac{\Delta \ell}{\ell}\) \(\frac{\Delta g}{g}=2 \cdot \frac{1}{100 \times 0.5}+\frac{1\,mm }{10\,cm }\) \(\frac{\Delta g}{g}=\frac{5}{100}\)
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