JEE Main 2021 · Physics · Current Electricity · medium

In a potentiometer experiment a cell of emf E balances at 300 cm of the wire. When the cell is connected across a 15 Ω resistor, its terminal voltage…

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In a potentiometer experiment a cell of emf

E

balances at

300\ \mathrm{cm}

of the wire. When the cell is connected across a

15\ \Omega

resistor, its terminal voltage balances at

250\ \mathrm{cm}

. The internal resistance of the cell is

Show answer and solution

Answer: (C) $3\ \Omega$

The balance length is proportional to potential difference. The open-circuit emf corresponds to

l_1=300

cm and the terminal voltage to

l_2=250

cm. With

V=E\dfrac{R}{R+r}

we get

r=R\left(\dfrac{l_1}{l_2}-1\right)=15\left(\dfrac{300}{250}-1\right)=3\ \Omega

. The wire's potential gradient is not needed.

Original question written in the JEE Main 2021 pattern and checked twice for the correct answer. Not an official question paper. How we write and check questions · Corrections policy · Report a mistake