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JEE Advanced · Physics · 1. Math in Physics

A student performs an experiment for determination of \(g\left(=\frac{4 \pi^2 l}{T^2}\right), l \approx 1 \mathrm{~m}\), and he commits an error of \(\Delta l\). For \(T\) he takes the time of \(n\) oscillations with the stop watch of least count \(\Delta T\) and he commits a human error of \(0.1 \mathrm{~s}\). For which of the following data, the measurement of \(g\) will be most accurate?

  1. A \(\Delta L=0.5, \Delta T=0.1, n=20\)
  2. B \(\Delta L=0.5, \Delta T=0.1, n=50\)
  3. C \(\Delta L=0.5, \Delta T=0.01, n=20\)
  4. D \(\Delta L=0.1, \Delta T=0.05, n=50\)
Verified Solution

Answer & Solution

Correct Answer

(D) \(\Delta L=0.1, \Delta T=0.05, n=50\)

Step-by-step Solution

Detailed explanation

\(\frac{\Delta g}{g}=\frac{\Delta l}{l}+2 \frac{\Delta T}{n}\)
In option (d) error in \(\Delta g\) is minimum and number of observations made are maximum. Hence, in this case error in \(g\) will be minimum.
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