Question:

The resistance of a wire is 'R' ohm. If it is melted and stretched to 'n' times its original length, its new resistance will be:

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When a wire is stretched, its resistance increases by a factor of \( n^2 \), where \( n \) is the stretching factor.
Updated On: Jan 12, 2026
  • \(\dfrac{R}{n}\)
  • \(n^2 R\)
  • \(\dfrac{R}{n^2}\)
  • \(nR\)
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The Correct Option is C

Solution and Explanation

Step 1: The resistance of a wire is given by the formula \( R = \rho \dfrac{l}{A} \), where \( \rho \) is the resistivity, \( l \) is the length, and \( A \) is the cross-sectional area.
Step 2: If the wire is melted and stretched, its volume remains constant. The volume before stretching is \( A_1 l = A_2 (n l) \), where \( A_1 \) and \( A_2 \) are the cross-sectional areas before and after stretching. This gives \( A_2 = \dfrac{A_1}{n^2} \).
Step 3: The new resistance \( R' = \rho \dfrac{n l}{A_2} = \dfrac{R}{n^2} \).

Final Answer: \[ \boxed{\dfrac{R}{n^2}} \]
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