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)

2

Answer 2

The given, $\rho_{B}=2 \rho_{A}, R_{A}=R_{B}=R$ Let, $r_{A}=r$ $r_{B}=2 r$ According to formula, $R_{A} =\rho_{A} \cdot \frac{l_{A}}{\pi r_{A}^{2}} \,\,\,...(i)$ $R_{B} =\rho_{B} \cdot \frac{l_{B}}{\pi r_{B}^{2}} \,\,\,...(ii)$ $\because R_{A} =R_{B}$ $\rho_{A} \cdot \frac{l_{A}}{\pi r_{A}^{2}}=\rho_{B} \cdot \frac{l_{B}}{\pi r_{B}^{2}}$ $\frac{l_{B}}{l_{A}}=\frac{r_{B}^{2}}{r_{A}^{2}} \cdot \frac{\rho_{A}}{\rho_{B}}$ $=\frac{(2 r)^{2}}{r^{2}} \cdot \frac{\rho_{A}}{2 \rho_{B}}$ $=\frac{4 r^{2}}{r^{2}} \cdot \frac{\rho_{A}}{2 \rho_{A}}$ $=\frac{4}{2}=2$