Question

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 $\frac{l_A}{l_B}$ of their respective lengthaI,must be

• A. $\frac{1}{4}$
• B. $2$
• C. $1$
• D. $\frac{1}{2}$

Answer 1

The correct option is D $\frac{1}{2}$

Let the specific resistance of A $={\rho }_A$

Thus the specific resistance of B is ${\rho }_B=2{\rho }_A$ so $\frac{{\rho }_B}{{\rho }_A}=2$

If diameter of A is $D_A$ diameter of B is $D_B=2D_A$ so $\frac{D_B}{D_A}=2$

Here, resistance $R_A=R_B$

or $\frac{{\rho }_A{l}_A}{\pi (D_A/2{)}^2}=\frac{{\rho }_B{l}_B}{\pi (D_B/2{)}^2}$

or $\frac{l_A}{l_B}=\frac{{\rho }_B}{{\rho }_A}{\left[\frac{D_A}{D_B}\right]}^2=2\times \frac{1}{4}=\frac{1}{2}$