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 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:
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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.
Updated On: Jun 26, 2026
- \(0.036\ \Omega\)
- \(0.09\ \Omega\)
- \(1.2\ \Omega\)
- \(3.1\ \Omega\)
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The Correct Option is A
Solution and Explanation
Step 1: Use the formula for resistance.
The resistance of a wire is given by \[ R=\rho \frac{L}{A} \] where \[ A=\pi r^2 \] Thus, \[ R\propto \frac{L}{r^2} \] Since both wires are made of the same metal, resistivity \(\rho\) remains constant.
Step 2: Write the ratio of resistances.
Given: For the first wire, \[ R_1=3\times10^{-3}\ \Omega \] \[ L_1=1\ \text{cm} \] \[ r_1=1\ \text{mm} \] For the second wire, \[ L_2=3\ \text{cm} \] \[ r_2=0.5\ \text{mm} \] Using \[ \frac{R_2}{R_1}= \frac{L_2}{L_1} \times \frac{r_1^2}{r_2^2} \] Substituting the values, \[ \frac{R_2}{3\times10^{-3}} = \frac{3}{1} \times \frac{(1)^2}{(0.5)^2} \]
Step 3: Simplify the expression.
Since \[ (0.5)^2=0.25 \] Therefore, \[ \frac{R_2}{3\times10^{-3}} = 3\times\frac{1}{0.25} \] \[ = 3\times4 \] \[ =12 \] Hence, \[ R_2=12\times3\times10^{-3} \] \[ =36\times10^{-3} \] \[ =0.036\ \Omega \]
Step 4: Final conclusion.
Therefore, the resistance of the second wire is \[ \boxed{0.036\ \Omega} \]
The resistance of a wire is given by \[ R=\rho \frac{L}{A} \] where \[ A=\pi r^2 \] Thus, \[ R\propto \frac{L}{r^2} \] Since both wires are made of the same metal, resistivity \(\rho\) remains constant.
Step 2: Write the ratio of resistances.
Given: For the first wire, \[ R_1=3\times10^{-3}\ \Omega \] \[ L_1=1\ \text{cm} \] \[ r_1=1\ \text{mm} \] For the second wire, \[ L_2=3\ \text{cm} \] \[ r_2=0.5\ \text{mm} \] Using \[ \frac{R_2}{R_1}= \frac{L_2}{L_1} \times \frac{r_1^2}{r_2^2} \] Substituting the values, \[ \frac{R_2}{3\times10^{-3}} = \frac{3}{1} \times \frac{(1)^2}{(0.5)^2} \]
Step 3: Simplify the expression.
Since \[ (0.5)^2=0.25 \] Therefore, \[ \frac{R_2}{3\times10^{-3}} = 3\times\frac{1}{0.25} \] \[ = 3\times4 \] \[ =12 \] Hence, \[ R_2=12\times3\times10^{-3} \] \[ =36\times10^{-3} \] \[ =0.036\ \Omega \]
Step 4: Final conclusion.
Therefore, the resistance of the second wire is \[ \boxed{0.036\ \Omega} \]
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