Physics · Current Electricity

JEE Main 2026 — 22 January, Morning Shift — Question 30

A meter bridge with two resistances $\mathrm{R}_{1}$ and $\mathrm{R}_{2}$ as shown in figure was balanced (null point) at 40 cm from the point P . The null point changed to 50 cm from the point P , when $16 \Omega$ resistance is connected in parallel to $R_{2}$ . The values of resistances $R_{1}$ and $\mathrm{R}_{2}$ are $\_\_\_\_$ .

Option A:
$\mathrm{R}_{2}=16 \Omega, \mathrm{R}_{1}=\frac{16}{3} \Omega$
Option B:
$\mathrm{R}_{2}=4 \Omega, \mathrm{R}_{1}=\frac{4}{3} \Omega$
Option C:
$\mathrm{R}_{2}=8 \Omega, \mathrm{R}_{1}=\frac{16}{3} \Omega$
Correct
Option D:
$\mathrm{R}_{2}=12 \Omega, \mathrm{R}_{1}=\frac{12}{3} \Omega$

Answer: C

Step-by-step solution

$\begin{gathered} \frac{\mathrm{R}_{1}}{\mathrm{R}_{2}}=\frac{40}{60}=\frac{2}{3} \end{gathered}$ $\begin{gathered} \frac{R_{1}}{\left(\frac{R_{2} \times 16}{R_{2}+16}\right)}=\frac{50}{50} \Rightarrow R_{1}=\frac{16 R_{2}}{16+R_{2}} \end{gathered}$ $\frac{2}{3} R_{2}=\frac{16 R_{2}}{16+R_{2}}$ $\frac{32}{3}+\frac{2 \mathrm{R}_{2}}{3}=16$ $\frac{2 \mathrm{R}_{2}}{3}=16-\frac{32}{3}=\frac{16}{3}$ $\mathrm{R}_{2}=8 \Omega$ By equation (1) $\mathrm{R}_{1}=\frac{2}{3} \mathrm{R}_{2}=\frac{16}{3} \Omega$

Answer key and solution verified before publishing.

Practise Current Electricity

Start with this question, then two more from the same chapter — with a tutor that explains every step. Free.

Solve a similar one free

Exam: JEE Main 2026
Paper: 22 January, Morning Shift
Subject: Physics
Chapter: Current Electricity
Topic: Combination of Resistors and cells, Wheatstone Bridge