Current Electricity

JEE Main level, numerical answer, medium

50 of 380

Question 50 of 380

Two wires AA and BB are made up of the same material and have the same mass. Wire AA has radius of 2.0 mm2.0 \mathrm{~mm} and wire BB has radius of 4.0 mm4.0 \mathrm{~mm}. The resistance of wire BB is 2Ω2 \Omega. The resistance of wire AA is _____Ω\Omega.

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Answer: 32

∵R=ρℓA=ρVA2∴RARB=AB2 AA2=rB4rA4⇒RA2=[4×10−32×10−3]4⇒RA=32Ω.\begin{aligned} & \because \mathrm{R}=\frac{\rho \ell}{\mathrm{A}}=\frac{\rho \mathrm{V}}{\mathrm{A}^2} \\ & \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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