Physics · Current Electricity

JEE Main 2026 — 28 January, Morning Shift — Question 27

In the potentiometer, when the cell in the secondary circuit is shunted with $4 \Omega$ resistance, the balance is obtained at the length 120 cm of wire. Now when the same cell in shunted with $12 \Omega$ resistance, the balance is shifted to a length of 180 cm . The internal resistance of cell is $\_\_\_\_$ $\Omega$ .

Option A:
3
Option B:
4
Correct
Option C:
12
Option D:
6

Answer: B

Step-by-step solution

Let E is emf and r is internal resistance of cell. $\frac{\mathrm{E} \cdot 4}{\mathrm{r}+4}=120 \mathrm{~K}$ $\frac{\mathrm{E} \cdot 12}{\mathrm{r}+12}=180 \mathrm{~K}$ $\Rightarrow \frac{1}{3} \frac{\mathrm{r}+12}{\mathrm{r}+4}=\frac{2}{3}$ $\mathrm{r}+12=2(\mathrm{r}+4)$ $\Rightarrow \mathrm{r}=4$

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Exam: JEE Main 2026
Paper: 28 January, Morning Shift
Subject: Physics
Chapter: Current Electricity
Topic: Electrical Measuring Instruments

Answer: B

Step-by-step solution

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