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(B) The formula for the internal resistance $r$ of a cell using a potentiometer is given by $r = R \left( \frac{l_1}{l_2} - 1 \right)$,where $l_1 = 60

Vedclass · Accepted Answer

$4.2$

Vedclass · Answer

(B) The formula for the internal resistance $r$ of a cell using a potentiometer is given by $r = R \left( \frac{l_1}{l_2} - 1 ight)$,where $l_1 = 60 	ext{ cm}$ and $l_2 = 50 	ext{ cm}$.Given $\Delta l = 1 	ext{ mm} = 0.1 	ext{ cm}$ and $\frac{\Delta R}{R} 	imes 100 = 2 \%$.Taking the logarithmic derivative of $r = R \left( \frac{l_1}{l_2} - 1 ight)$,we have $\frac{\Delta r}{r} = \frac{\Delta R}{R} + \frac{\Delta (l_1/l_2)}{(l_1/l_2) - 1}$.Since $\frac{l_1}{l_2} = \frac{60}{50} = 1.2$,then $\frac{l_1}{l_2} - 1 = 0.2$.The relative error in $\frac{l_1}{l_2}$ is $\frac{\Delta (l_1/l_2)}{(l_1/l_2)} = \frac{\Delta l_1}{l_1} + \frac{\Delta l_2}{l_2} = \frac{0.1}{60} + \frac{0.1}{50} = \frac{1}{600} + \frac{1}{500} = \frac{11}{3000}$.Thus,$\Delta (l_1/l_2) = 1.2 	imes \frac{11}{3000} = 0.0044$.The error in $r$ is $\frac{\Delta r}{r} 	imes 100 = \left( \frac{\Delta R}{R} + \frac{\Delta (l_1/l_2)}{(l_1/l_2) - 1} ight) 	imes 100 = 2 \% + \left( \frac{0.0044}{0.2} ight) 	imes 100 = 2 \% + 2.2 \% = 4.2 \%$.

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