Question

Wires A and B have resistivities ${\rho }_A$ and ${\rho }_B$, $({\rho }_B=2{\rho }_A)$ and have lengths $l_A$ and $l_B$. If the diameter of the wire B is twice that of A and the two wires have same resistance, then $\frac{l_B}{l_A}$ is:

• A. $2$
• B. $1$
• C. $1/2$
• D. $1/4$

Answer 1

Answer: (A)

$2$

Given : ${\rho }_B=2{\rho }_A$ $D_B=2D_A$

Resistance $R=\rho \frac{l}{A}$ where $A=\frac{\pi D^2}{4}$

$⟹$ $R=\rho \frac{4l}{\pi D^2}$

According to the question, $R_A=R_B$

$\therefore$ ${\rho }_A\frac{4l_A}{\pi D_A^2}$ $={\rho }_B\frac{4l_B}{\pi D_B^2}$

${\rho }_A\frac{l_A}{D_A^2}$ $=2{\rho }_A\frac{l_B}{4D_A^2}$ $⟹\frac{l_B}{l_A}=2$