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 $R_A=R_B$
$\therefore \frac{{\rho }_A{l}_A}{\pi r_A^2}=\frac{{\rho }_B{l}_B}{\pi r_B^2}$
$\Rightarrow \frac{{\rho }_A{l}_A}{\pi r_A^2}=\frac{2{\rho }_A\times I_R}{\pi {\left(2r_A\right)}^2}$
$\Rightarrow \frac{l_B}{l_A}=\frac{2}{1}=2:1$