Question:
A wire of resistance \(50 \, \Omega\) is elongated to get a resistance of \(60.5 \, \Omega\). The unelongated length of the wire is \(100 \, cm\). The wire is elongated by:
A wire of resistance \(50 \, \Omega\) is elongated to get a resistance of \(60.5 \, \Omega\). The unelongated length of the wire is \(100 \, cm\). The wire is elongated by:
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When wire is stretched with constant volume, resistance varies as \(R \propto L^2\).
Updated On: Jul 18, 2026
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The Correct Option is C
Solution and Explanation
Step 1: Resistance relation with length.
Resistance of a wire is: \[ R = \rho \frac{L}{A} \] When wire is stretched, volume remains constant: \[ AL = \text{constant} \]
Step 2: Effect of stretching on resistance.
If length increases by factor \(x\), then area decreases by same factor: \[ R \propto L^2 \] So: \[ \frac{R_2}{R_1} = \left(\frac{L_2}{L_1}\right)^2 \]
Step 3: Substituting values.
\[ \frac{60.5}{50} = \left(\frac{L_2}{L_1}\right)^2 \] \[ 1.21 = \left(\frac{L_2}{L_1}\right)^2 \]
Step 4: Taking square root.
\[ \frac{L_2}{L_1} = 1.1 \]
Step 5: Percentage increase.
\[ \% \text{increase} = (1.1 - 1)\times 100 = 10\% \]
Step 6: Final conclusion.
\[ \boxed{10\%} \]
Resistance of a wire is: \[ R = \rho \frac{L}{A} \] When wire is stretched, volume remains constant: \[ AL = \text{constant} \]
Step 2: Effect of stretching on resistance.
If length increases by factor \(x\), then area decreases by same factor: \[ R \propto L^2 \] So: \[ \frac{R_2}{R_1} = \left(\frac{L_2}{L_1}\right)^2 \]
Step 3: Substituting values.
\[ \frac{60.5}{50} = \left(\frac{L_2}{L_1}\right)^2 \] \[ 1.21 = \left(\frac{L_2}{L_1}\right)^2 \]
Step 4: Taking square root.
\[ \frac{L_2}{L_1} = 1.1 \]
Step 5: Percentage increase.
\[ \% \text{increase} = (1.1 - 1)\times 100 = 10\% \]
Step 6: Final conclusion.
\[ \boxed{10\%} \]
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