Question

A material $B$' has twice the specific resistance of $A^{\prime }$. A circular wire made of $B$' has twice the diameter of a wire made of $A^{\prime }$. Then for the two wires to have the same resistance, the ratio $P_A/P_B$ of their respective lengths must be

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

Answer 1

Answer: (C)

$\frac{1}{2}$

$R_1=\frac{{\rho }_A{l}_A}{\pi R_A^2}$

$R_2=\frac{{\rho }_B{l}_B}{\pi R_B^2}$

$R_1=R_2\Rightarrow \frac{l_A}{l_B}=\frac{{\rho }_B}{{\rho }_A}\frac{R_A^2}{R_B^2}=\frac{2{\rho }_A}{{\rho }_A}\frac{R_A^2}{4R_A^2}=0.5$