Question

A wire of resistance R is stretched uniformly to double its original length. The new resistance will be:

• A. \( 2R \)
• B. \( 4R \)
• C. \( \frac{R}{2} \)
• D. \( \frac{R}{4} \)

Hint:

If length becomes \(nL\), then: \[ R \propto n^2 \]

Answer 1

The Correct Option is B

Step 1: Recall the resistance formula.
The resistance of a wire is $R = \rho\dfrac{L}{A}$, where $\rho$ is the material constant, $L$ is the length, and $A$ is the cross-section area.

Step 2: Use the volume idea.
Stretching does not add or remove metal, so the volume stays the same. Volume is length times area, so $L \times A$ is constant.

Step 3: Find the new length.
The wire is stretched to twice its length, so the new length is $L_2 = 2L$.

Step 4: Find the new area.
Since volume is fixed, doubling the length must halve the area: $A_2 = \dfrac{A}{2}$.

Step 5: Compute the new resistance.
Put the new values in: \[ R_2 = \rho\,\frac{2L}{A/2} = \rho\,\frac{2L \times 2}{A} = 4\,\rho\frac{L}{A} = 4R. \]

Step 6: State the answer.
Because resistance depends on length squared when volume is fixed, the new resistance is four times the old. \[ \boxed{4R} \]