Physics · Current Electricity

JEE Main 2026 — 24 January, Morning Shift — Question 35

Two resistors $2 \Omega$ and $3 \Omega$ are connected in the gaps of bridge as shown in figure. The null point is obtained with the contact of jockey at some point on wire XY . When an unknown resistor is connected in parallel with $3 \Omega$ resistor, the null point is shifted by 22.5 cm toward Y . The resistance of unknown resistor is $\Omega$ .

Option A:
3
Option B:
2
Correct
Option C:
4
Option D:
1

Answer: B

Step-by-step solution

Initially, $\frac{2}{3}=\frac{\mathrm{x}}{100-\mathrm{x}}$ $\Rightarrow \mathrm{x}=40 \mathrm{~cm}$ Now when ' $R$ ' connected in parallel $\frac{2}{\frac{3 R}{3+R}}=\frac{40+22.5}{60-22.5}=\frac{62.5}{37.5}$ $\therefore \mathrm{R}=2 \Omega$

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Exam: JEE Main 2026
Paper: 24 January, Morning Shift
Subject: Physics
Chapter: Current Electricity
Topic: Combination of Resistors and cells, Wheatstone Bridge

Answer: B

Step-by-step solution

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