Physics · Current Electricity

JEE Main 2024 — 4 April, Shift 2 — Question 57

Two wires A and B are made up of the same material and have the same mass. Wire A has radius of 2.0 mm and wire $B$ has radius of 4.0 mm . The resistance of wire B is $2 \Omega$ . The resistance of wire A is $\qquad$ $\Omega$ .

Answer: 32

Numerical answer — enter this value.

Step-by-step solution

$\because \mathrm{R}=\frac{\rho \ell}{\mathrm{A}}=\frac{\rho \mathrm{V}}{\mathrm{A}^{2}}$

\begin{aligned} & \therefore \frac{\mathrm{R}_{\mathrm{A}}}{\mathrm{R}_{\mathrm{B}}}=\frac{\mathrm{A}_{\mathrm{B}}^{2}}{\mathrm{~A}_{\mathrm{A}}^{2}}=\frac{\mathrm{r}_{\mathrm{B}}^{4}}{\mathrm{r}_{\mathrm{A}}^{4}} & \Rightarrow \frac{\mathrm{R}_{\mathrm{A}}}{2}=\left[\frac{4 \times 10^{-3}}{2 \times 10^{-3}}\right]^{4} & \Rightarrow \mathrm{R}_{\mathrm{A}}=32 \Omega \end{aligned}

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Exam: JEE Main 2024
Paper: 4 April, Shift 2
Subject: Physics
Chapter: Current Electricity
Topic: Ohm's Law and Calculation of Resistance