Question

A metal wire of \(1\) cm length and \(1\) mm radius has a resistance of \(3\times10^{-3}\ \Omega\). If a wire of the same metal of length \(3\) cm and radius \(0.5\) mm is drawn, the resistance of the wire is:

• A. \(0.036\ \Omega\)
• B. \(0.09\ \Omega\)
• C. \(1.2\ \Omega\)
• D. \(3.1\ \Omega\)

Hint:

Resistance of a wire varies directly with length and inversely with the square of the radius: \[ R\propto \frac{L}{r^2} \] A small decrease in radius causes a large increase in resistance.

Answer 1

The Correct Option is A

Step 1: Recall the formula for resistance.
The resistance of a wire is given by: \[ R = \rho \frac{L}{A} \] where $\rho$ is the resistivity, $L$ is the length, and $A = \pi r^2$ is the cross-sectional area. So: \[ R = \rho \frac{L}{\pi r^2} \implies R \propto \frac{L}{r^2} \] Since both wires are made of the same metal, resistivity $\rho$ is the same for both.
Step 2: Write the ratio of resistances.
For two wires of the same material: \[ \frac{R_2}{R_1} = \frac{L_2}{L_1} \times \frac{r_1^2}{r_2^2} \]
Step 3: Substitute the given values.
Wire 1: $L_1 = 1\,\text{cm}$, $r_1 = 1\,\text{mm}$, $R_1 = 3 \times 10^{-3}\,\Omega$. Wire 2: $L_2 = 3\,\text{cm}$, $r_2 = 0.5\,\text{mm}$. \[ \frac{R_2}{3 \times 10^{-3}} = \frac{3}{1} \times \frac{(1)^2}{(0.5)^2} \]
Step 4: Compute the ratio.
\[ \frac{R_2}{3 \times 10^{-3}} = 3 \times \frac{1}{0.25} = 3 \times 4 = 12 \] The radius halved (from 1 mm to 0.5 mm) reduces the area by a factor of 4, so resistance increases by 4 times from that alone. The length tripled increases resistance by 3 times. Combined factor: $4 \times 3 = 12$.
Step 5: Find $R_2$.
\[ R_2 = 12 \times 3 \times 10^{-3} = 36 \times 10^{-3} = 0.036\,\Omega \]
Step 6: State the final answer.
The resistance of the second wire is: \[ \boxed{0.036\,\Omega} \]