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 $2$

Given

${\rho }_B=2{\rho }_A$ and $d_B=2d_A$
$R_B=R_A$

Resistance $R=\frac{\rho l}{A}$

$\Rightarrow \frac{{\rho }_B\times L_B}{\frac{\pi d_B^2}{4}}=\frac{{\rho }_A\times L_A}{\frac{\pi d_A^2}{4}}$

$\Rightarrow \frac{2{\rho }_A\times L_B}{4d_A^2}=\frac{{\rho }_A\times L_A}{d_A^2}$

$\Rightarrow \frac{L_B}{L_A}=\frac{2}{1}$