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JEE Advanced · Physics · 29. Experimental Physics

Students I, II and III perform an experiment for measuring the acceleration due to gravity \((g)\) using a simple pendulum. They use different lengths of the pendulum and/or record time for different number of oscillations. The observations are shown in the table.
Least count for length \(=0.1 \mathrm{~cm}\)
Least count for time \(=0.1 \mathrm{~s}\)

If \(E_{\mathrm{I}}, E_{\mathrm{II}}\) and \(E_{\mathrm{III}}\) are the percentage errors in \(g\), i.e., \(\left(\frac{\Delta g}{g} \times 100\right)\) for students I, II and III, respectively.

  1. A \(E_{\mathrm{I}}=0\)
  2. B \(E_{\mathrm{I}}\) is minimum
  3. C \(E_{\mathrm{I}}=E_{\mathrm{II}}\)
  4. D \(E_{\mathrm{II}}\) is maximum
Verified Solution

Answer & Solution

Correct Answer

(B) \(E_{\mathrm{I}}\) is minimum

Step-by-step Solution

Detailed explanation

\(T =2 \pi \sqrt{\frac{l}{g}} \quad \text { or } \quad \frac{t}{n}=2 \pi \sqrt{\frac{l}{g}} \)
\( \therefore \quad g =\frac{\left(4 \pi^2\right)\left(n^2\right) l}{t^2}\)
\(\%\) error in \(g=\frac{\Delta g}{g} \times 100=\left(\frac{\Delta l}{l}+\frac{2 \cdot \Delta t}{t}\right) \times 100\)
\(E_{\mathrm{I}}=\left(\frac{0.1}{64}+\frac{2 \times 0.1}{128}\right) \times 100=0.3125 \% \)
\( E_{\mathrm{II}}=\left(\frac{0.1}{64}+\frac{2 \times 0.1}{64}\right) \times 100=0.46875 \% \quad\)\(E_{\mathrm{III}}=\left(\frac{0.1}{20}+\frac{2 \times 0.1}{36}\right) \times 100=1.055 \%\)
Hence, \(E_{\mathrm{I}}\) is minimum.
\(\therefore\) correct option is (b).
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