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

Answer: (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 $\, \quad R_{A}=R_{B}$
$\therefore \, \quad\frac{\rho_{A}l_{A}}{\pi r_{A}^{2}}=\frac{\rho_{B}l_{B}}{\pi r_{B}^{2}}$
$\Rightarrow \, \quad\frac{\rho_{A}l_{A}}{\pi r_{A}^{2}}=\frac{2\rho_{A}\times I_{R}}{\pi\left(2r_{A}\right)^{2}}$
$\Rightarrow \, \quad\quad\frac{l_{B}}{l_{A}}=\frac{2}{1}=2:1$