Students I,II,and III perform an experiment f | vedclass

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(B) The formula for acceleration due to gravity is $g = 4\pi^2 \frac{\ell}{T^2}$.Taking relative error: $\frac{\Delta g}{g} = \frac{\Delta \ell}{\ell}

Vedclass · Accepted Answer

$E_I$ is minimum

Vedclass · Answer

(B) The formula for acceleration due to gravity is $g = 4\pi^2 \frac{\ell}{T^2}$.Taking relative error: $\frac{\Delta g}{g} = \frac{\Delta \ell}{\ell} + 2\frac{\Delta T}{T}$.Since $T = \frac{t}{n}$,where $t$ is total time and $n$ is number of oscillations,$\Delta T = \frac{\Delta t}{n}$.Thus,$\frac{\Delta T}{T} = \frac{\Delta t/n}{t/n} = \frac{\Delta t}{t}$.So,$\frac{\Delta g}{g} = \frac{\Delta \ell}{\ell} + 2\frac{\Delta t}{t}$.Given $\Delta \ell = 0.1 	ext{ cm}$ and $\Delta t = 0.1 	ext{ s}$.For student $I$: $\frac{\Delta g}{g} = \frac{0.1}{64.0} + 2(\frac{0.1}{128.0}) = 0.00156 + 0.00156 = 0.00312$.For student $II$: $\frac{\Delta g}{g} = \frac{0.1}{64.0} + 2(\frac{0.1}{64.0}) = 0.00156 + 0.00312 = 0.00468$.For student $III$: $\frac{\Delta g}{g} = \frac{0.1}{20.0} + 2(\frac{0.1}{36.0}) = 0.005 + 0.0055 = 0.0105$.Comparing the values,$E_I$ is the minimum.

Student	Length $(cm)$	Oscillations $(n)$	Total Time $(s)$	Time Period $(s)$
$I$	$64.0$	$8$	$128.0$	$16.0$
$II$	$64.0$	$4$	$64.0$	$16.0$
$III$	$20.0$	$4$	$36.0$	$9.0$

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