A material ' $B$ ' has twice the specific resistance of ' $A$ '. A circular wire made of ' $B$ ' has twice…

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 $\ell_A / \ell_B$ of their respective lengths must be
  1. 2
  2. 1
  3. $\frac{1}{2}$
  4. $\frac{1}{4}$

Solution

$\mathrm{R}_1=\frac{\rho_{\mathrm{A}} \ell_{\mathrm{A}}}{\pi \mathrm{R}_{\mathrm{A}}^2} \quad \mathrm{R}_2=\frac{\rho_{\mathrm{B}} \ell_{\mathrm{B}}}{\pi \mathrm{R}_{\mathrm{B}}^2}$ $\frac{\ell_{\mathrm{A}}}{\ell_{\mathrm{B}}}=\frac{\rho_{\mathrm{B}} \mathrm{R}_{\mathrm{A}}^2}{\rho_{\mathrm{A}} \mathrm{R}_{\mathrm{B}}^2}=\frac{2 \rho_{\mathrm{A}} \mathrm{R}_{\mathrm{A}}^2}{\rho_{\mathrm{A}} \cdot 4 \mathrm{R}_{\mathrm{A}}^2} \Rightarrow \frac{\ell_{\mathrm{B}}}{\ell_{\mathrm{A}}}=2$

Asked in: JEE Main 2006

Practice more Current Electricity questions on Aicharya