In the given figure, there is a circuit of potentiometer of length $A B = 10 m$ . The resistance
per unit length is $0.1 Ω$ per cm. Across AB, battery of emf E and internal resistance 'r' is connected. The maximum value of emf measured by this potentiometer is :

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6 V

2.75 V

2.25 V

5 V

answer is A.

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Detailed Solution

Max. voltage that can be measured by this potentiometer will be equal to potential
drop across AB $R_{A B} = 10 \times 0.1 \times 100 = 100 ohm$

$∴ V_{A B} = \frac{6}{20 + 100} \times 100 = 6 \times \frac{100}{120} = 5 V$

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