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, $r_{A}=r$ $r_{B}=2 r$
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 $