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 $\frac{l_B}{l_4}$ is

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

Answer 1

Answer: (A)

The given,
${\rho }_B=2{\rho }_A,R_A=R_B=R$
Let, $r_A=r$ $r_B=2r$
According to formula,
$R_A={\rho }_A\cdot \frac{l_A}{\pi r_A^2}...(i)$
$R_B={\rho }_B\cdot \frac{l_B}{\pi r_B^2}...(ii)$
$\because R_A=R_B$
${\rho }_A\cdot \frac{l_A}{\pi r_A^2}={\rho }_B\cdot \frac{l_B}{\pi r_B^2}$
$\frac{l_B}{l_A}=\frac{r_B^2}{r_A^2}\cdot \frac{{\rho }_A}{{\rho }_B}$
$=\frac{(2r{)}^2}{r^2}\cdot \frac{{\rho }_A}{2{\rho }_B}$
$=\frac{4r^2}{r^2}\cdot \frac{{\rho }_A}{2{\rho }_A}$
$=\frac{4}{2}=2$