JEE Advanced 2023 Paper 2 · Physics · Current Electricity · easy

In a potentiometer experiment the driver cell has emf 3 V and negligible internal resistance and is connected in series with a 10 Ω resistance to the…

Single correct option

In a potentiometer experiment the driver cell has emf

3\ \mathrm{V}

and negligible internal resistance and is connected in series with a

10\ \Omega

resistance to the potentiometer wire, which is

10\ \mathrm{m}

long and has resistance

20\ \Omega

. A test cell with its positive terminal at the high-potential end is balanced at

6.0\ \mathrm{m}

on open circuit. When a

5\ \Omega

resistor is connected across the test cell, the balance length becomes

5.0\ \mathrm{m}

. The test cell is now shunted by a

2\ \Omega

resistor instead. The new balance length is

Show answer and solution

Answer: (B) $4.0\ \mathrm{m}$

Current in the wire is

I=\dfrac{3}{20+10}=0.1\ \mathrm{A}

, so the potential gradient is

\dfrac{0.1\times20}{10}=0.2\ \mathrm{V\,m^{-1}}

and the emf of the test cell is

E=0.2\times6.0=1.2\ \mathrm{V}

. With a shunt

R

the balance length measures the terminal voltage

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

, so

r=R\left(\dfrac{l_1}{l_2}-1\right)=5\left(\dfrac{6}{5}-1\right)=1\ \Omega

. For

R=2\ \Omega

V=1.2\times\dfrac{2}{3}=0.8\ \mathrm{V}

, and the balance length is

\dfrac{0.8}{0.2}=4.0\ \mathrm{m}

. The value

2.0\ \mathrm{m}

comes from using

E\,r/(R+r)

4.2\ \mathrm{m}

from the wrong formula

r=R(l_1-l_2)/l_1

, and

6.0\ \mathrm{m}

from ignoring internal resistance.

Original question written in the JEE Advanced 2023 Paper 2 pattern and checked twice for the correct answer. Not an official question paper. How we write and check questions · Corrections policy · Report a mistake