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?
- A \(\Delta L=0.5, \Delta T=0.1, n=20\)
- B \(\Delta L=0.5, \Delta T=0.1, n=50\)
- C \(\Delta L=0.5, \Delta T=0.01, n=20\)
- D \(\Delta L=0.1, \Delta T=0.05, n=50\)
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.
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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