Question

A material 'B' has twice the specific resistance of 'A'. A circular wire made of 'B' has twice the diameter of wire made of 'A' then for the two wires to have the same resistance, the ratio, $\frac{I_B}{I_A}$ of their respective length must be

• A. 1
• B. 1/2
• C. 1/4
• D. 2

Answer 1

The correct option is D 2

$R=\rho \frac{L}{A}$

$⟹R=\rho \frac{4L}{D^2}$ where $D=$ diameter

Given that $\frac{{\rho }_B}{{\rho }_A}=2$

and, $\frac{D_A}{D_B}=\frac{1}{2}$

For same resistance-

${\rho }_A\frac{L_A}{D_A^2}={\rho }_B\frac{L_B}{D_B^2}$

$⟹\frac{L_B}{L_A}=\frac{{\rho }_A}{{\rho }_B}\frac{D_B^2}{D_A^2}$

$⟹\frac{L_B}{L_A}=\frac{1}{2}\times 2^2$

$⟹\frac{L_B}{L_A}=2$

Answer-(D)