JEE Mains · Physics · STD 12 - Current electricity
- JEE Mains 2021
- Medium
- Numerical
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 \(V\))

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Answer & Solution
Correct Answer
(C) \(5\)
Step-by-step Solution
Detailed explanation
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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