Three students S_{1}, S_{2} and S_{3} perform | vedclass

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(C) The formula for the time period of a simple pendulum is $T = 2\pi \sqrt{\frac{\ell}{g}}$.Squaring and rearranging,we get $g = \frac{4\pi^2 \ell}{T

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$1$

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(C) The formula for the time period of a simple pendulum is $T = 2\pi \sqrt{\frac{\ell}{g}}$.Squaring and rearranging,we get $g = \frac{4\pi^2 \ell}{T^2}$.The relative error in $g$ is given by $\frac{\Delta g}{g} = \frac{\Delta \ell}{\ell} + 2 \frac{\Delta T}{T}$.Here,the time period $T$ is calculated as $T = \frac{t}{n}$,where $t$ is the total time and $n$ is the number of oscillations.The error in the time period is $\Delta T = \frac{\Delta t}{n}$,where $\Delta t$ is the least count of the stopwatch.Substituting this,the relative error becomes $\frac{\Delta g}{g} = \frac{\Delta \ell}{\ell} + 2 \frac{\Delta t}{n \cdot T} = \frac{\Delta \ell}{\ell} + 2 \frac{\Delta t}{t}$.Given $\Delta \ell = 0.1 \, cm$ and $\Delta t = 0.1 \, s$ for all students:For student $1$: $\frac{\Delta g}{g} = \frac{0.1}{64.0} + 2 \frac{0.1}{128.0} = 0.00156 + 0.00156 = 0.00312$.For student $2$: $\frac{\Delta g}{g} = \frac{0.1}{64.0} + 2 \frac{0.1}{64.0} = 0.00156 + 0.00312 = 0.00468$.For student $3$: $\frac{\Delta g}{g} = \frac{0.1}{20.0} + 2 \frac{0.1}{36.0} = 0.005 + 0.0055 = 0.0105$.Comparing the values,the minimum error is obtained by student $1$.

Student No.	Length of pendulum $(cm)$	No. of oscillations $(n)$	Total time for oscillations $(s)$	Time period $(s)$
$1.$	$64.0$	$8$	$128.0$	$16.0$
$2.$	$64.0$	$4$	$64.0$	$16.0$
$3.$	$20.0$	$4$	$36.0$	$9.0$

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