Question

A copper wire is stretched to increase its length by 1%. Then the change in its resistance is close to:

• A. 1%
• B. 4%
• C. $-4$%
• D. 2%

Hint:

The volume of the stretched wire is constant, so R is proportional to L squared. Double the percentage change in length.

Answer 1

The Correct Option is D

Step 1: Understanding the Concept:
Resistance of a wire is \(R = \dfrac{\rho L}{A}\).
When we stretch a wire, its volume stays the same. So a longer wire must also become thinner.
Both the length and the area change, and both change \(R\).

Step 2: Key Formula or Approach:
Volume is constant, so \(V = AL\) gives \(A = V/L\).
Put this into the resistance formula:
\[ R = \frac{\rho L}{V/L} = \frac{\rho L^2}{V} \]
So \(R \propto L^2\) for a wire of fixed volume.

Step 3: Detailed Explanation:
Take the logarithm and differentiate: \(\dfrac{\Delta R}{R} = 2\,\dfrac{\Delta L}{L}\).
The length rises by 1%, so \(\dfrac{\Delta L}{L} = 0.01\).
\[ \frac{\Delta R}{R} = 2 \times 1\% = 2\% \]
The exact value is \((1.01)^2 - 1 = 0.0201\), which is 2.01%. This is very close to 2%.

Step 4: Check the options:
Option (A) 1% would be true only if the area stayed the same. But the wire gets thinner, so the area drops.
Option (B) 4% double counts the effect, as if \(R \propto L^4\).
Option (C) \(-4\)% has the wrong sign. A longer, thinner wire always has more resistance.
Option (D) 2% matches our result.

Final Answer:
The resistance increases by about 2%, which is option (D). \[ \boxed{2\%} \]