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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?
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The correct answer is:
$\Delta L=0.1, \Delta T=0.05, n=50$
$\Delta L=0.1, \Delta T=0.05, n=50$
$\frac{\Delta g}{g}=\frac{\Delta l}{l}+2 \frac{\Delta T}{n}$
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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