Question

A material B? has twice the specific resistance of ?A?. A circular wire made of B? has twice the diameter of a wire made of ?A?. Then for the two wires to have the same resistance, the ratio $l_{B} / l_{A} $ of their respective lengths must be :

• A. 1
• B. $\frac{1}{2}$
• C. $\frac{1}{4}$
• D. 2

Answer 1

The Correct Option is D

Let $\left(P_{A,} l_{A,} A_{A} \right) $ and $\left(P_{B,} l_{B,} A_{B}\right)$ are specific resistances, lengths, radii and areas of wires A and B respectively. Resistance of A $=R_{A}=\frac{\rho_{A}l_{A}}{A_{A}}=\frac{\rho_{A}l_{A}}{\pi r_{A}^{2}}$ Resistance of B $=R_{B}=|\frac{\rho_{B}l_{B}}{A_{B}}=\frac{\rho_{B}l_{B}}{\pi r_{B}^{2}}$ From given information $\rho_{B}=2\rho_{A}$ $r_{B}=2r_{A}$ and $\, \quad R_{A}=R_{B}$ $\therefore\, \quad\frac{\rho_{A}l_{A}}{\pi r_{A}^{2}}=\frac{\rho_{B}l_{B}}{\pi r_{B}^{2}}$ $\Rightarrow\, \quad\frac{\rho_{A}l_{A}}{\pi r_{A}^{2}}=\frac{2\rho_{A}\times I_{R}}{\pi\left(2r_{A}\right)^{2}}$ $\Rightarrow\, \quad\quad\frac{l_{B}}{l_{A}}=\frac{2}{1}=2:1$