In the given figure, there is a circuit of potentiometer of length $A B=10 \,{m}$. The resistance per unit length is $0.1 \,\Omega$ per ${cm}$. Across ${AB}$, a battery of emf ${E}$ and internal resistance ' ${r}^{\prime}$ is connected. The maximum value of emf measured by this potentiometer is : (In $V$)
JEE MAIN 2021, Medium
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Max. voltage that can be measured by this potentiometer will be equal to potential drop across AB
$R _{ AB }=10 \times 0.1 \times 100=100 \Omega$
$\therefore V _{ AB }=\frac{6}{20+100} \times 100=6 \times \frac{100}{120}=5 V$
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