In a meter bridge experiment to determine the value of unknown resistance, first the resistances 2 Ω and 3 Ω are connected in the left and right gaps of the bridge and the null point is obtained at a distance l cm from the left. Now when an unknown resistance x Ω is connected in parallel to 3 Ω resistance, the null point is shifted by 10 cm to the right of wire. The value of unknown resistance x is \ \ \ \ Ω .

Physics · Current Electricity

JEE Main 2026 — 24 January, Evening Shift — Question 46

In a meter bridge experiment to determine the value of unknown resistance, first the resistances $2 \Omega$ and $3 \Omega$ are connected in the left and right gaps of the bridge and the null point is obtained at a distance $l \mathrm{~cm}$ from the left. Now when an unknown resistance $\mathrm{x} \Omega$ is connected in parallel to $3 \Omega$ resistance, the null point is shifted by 10 cm to the right of wire. The value of unknown resistance x is $\_\_\_\_$ $\Omega$ .

Answer: 6

Numerical answer — enter this value.

Step-by-step solution

In case I

\frac{2}{3}=\frac{\ell}{(100-\ell)} \end{gathered}$$ $\ell=40 \mathrm{~cm}$ In case II $\frac{2}{\mathrm{R}}=\frac{\ell+10}{100-(\ell+10)}$ Put $\ell=40 \mathrm{~cm} \&$ solve $\mathrm{R}=2 \Omega$ $\therefore \frac{3 \mathrm{x}}{3+\mathrm{x}}=2$ $\mathrm{x}=6 \Omega$

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

Answer: 6

Step-by-step solution

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