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(D) The relative error in $g$ is given by $\frac{\Delta g}{g} = \frac{\Delta \ell}{\ell} + 2 \frac{\Delta T}{T}$.Here,the total error in time measurem

Vedclass · Accepted Answer

$\Delta \ell = 1 \, mm, \Delta T = 0.1 \, s, n = 50$

Vedclass · Answer

(D) The relative error in $g$ is given by $\frac{\Delta g}{g} = \frac{\Delta \ell}{\ell} + 2 \frac{\Delta T}{T}$.Here,the total error in time measurement is $\Delta T_{total} = \Delta T + 0.1 \, s$.The time period $T$ is measured as $T = \frac{t}{n}$,where $t$ is the time for $n$ oscillations.Thus,$\frac{\Delta T}{T} = \frac{\Delta T_{total}}{t} = \frac{\Delta T + 0.1}{n 	imes T_0}$,where $T_0 \approx 2 \, s$ for a $1 \, m$ pendulum.Substituting values:$(A)$ $\frac{\Delta g}{g} = \frac{5 	imes 10^{-3}}{1} + 2 \left( \frac{0.2 + 0.1}{10 	imes 2} ight) = 0.005 + 0.03 = 0.035$$(B)$ $\frac{\Delta g}{g} = \frac{5 	imes 10^{-3}}{1} + 2 \left( \frac{0.2 + 0.1}{20 	imes 2} ight) = 0.005 + 0.015 = 0.020$$(C)$ $\frac{\Delta g}{g} = \frac{5 	imes 10^{-3}}{1} + 2 \left( \frac{0.1 + 0.1}{10 	imes 2} ight) = 0.005 + 0.02 = 0.025$$(D)$ $\frac{\Delta g}{g} = \frac{1 	imes 10^{-3}}{1} + 2 \left( \frac{0.1 + 0.1}{50 	imes 2} ight) = 0.001 + 0.004 = 0.005$Since the relative error is minimum in case $(D)$,the measurement is most accurate.

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