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(C) The maximum voltage that can be measured by a potentiometer is equal to the potential drop across the entire length of the potentiometer wire $AB$

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$5$

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(C) The maximum voltage that can be measured by a potentiometer is equal to the potential drop across the entire length of the potentiometer wire $AB$.First,calculate the total resistance of the wire $AB$:Length of wire $AB = 10 \, m = 1000 \, cm$.Resistance per unit length = $0.1 \, \Omega/cm$.Total resistance $R_{AB} = 1000 \, cm 	imes 0.1 \, \Omega/cm = 100 \, \Omega$.Now,calculate the potential drop across $AB$ using the voltage divider rule:The circuit consists of a $6 \, V$ battery,an internal resistance of $20 \, \Omega$,and the potentiometer wire resistance $R_{AB} = 100 \, \Omega$ in series.The current in the primary circuit is $I = \frac{V}{R_{total}} = \frac{6 \, V}{20 \, \Omega + 100 \, \Omega} = \frac{6}{120} \, A = 0.05 \, A$.The potential drop across $AB$ is $V_{AB} = I 	imes R_{AB} = 0.05 \, A 	imes 100 \, \Omega = 5 \, V$.Thus,the maximum emf that can be measured is $5 \, V$.

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