Question

Two copper wires are of same length one of the twice as thick as the other. Then the resistance of the two wires are in the ratio of:

• A. 1:16
• B. 1:8
• C. 1:4
• D. 1:2

Answer 1

Answer: (C)

Here : Thickness of first wire $r_{1}=2 r$
The thickness of second wire $r_{2}=r$
Using the selation for resistance of wire is,
$R=\frac{S \times l}{A}=S \times \frac{l}{\pi r^{2}} \propto \frac{l}{r^{2}}$
Hence $\frac{R_{1}}{R_{2}}=\left(\frac{r_{2}}{r_{1}}\right)^{2}$
$=\frac{(r)^{2}}{(2 r)^{2}}=\frac{1}{4}$