A copper wire is stretched to make it 0.5% longer. The percentage change in its electrical resistance if its volume remains unchanged is:

We have a copper wire whose original length is $$L$$ and original cross-sectional area is $$A$$. The original resistance is given by the basic relation

$$R = \rho \dfrac{L}{A},$$

where $$\rho$$ is the resistivity of copper. Resistivity does not change because only the dimensions are altered, not the material or temperature.

The wire is stretched so that its length becomes $$L' = L + \Delta L$$, and the question tells us that the percentage elongation is $$0.5\%.$$ Therefore, in fractional form,

$$\dfrac{\Delta L}{L} = 0.5\% = 0.005.$$

The stretching is done in such a way that the volume of the wire remains unchanged. Volume before stretching is

$$V = A L,$$

and after stretching it must still be the same:

$$V = A' L'.$$

Hence

$$A L = A' L' \quad\Longrightarrow\quad A' = \dfrac{A L}{L'}.$$

Substituting $$L' = L + \Delta L = L(1 + \dfrac{\Delta L}{L})$$ we get

$$A' = \dfrac{A L}{L(1 + \dfrac{\Delta L}{L})} = \dfrac{A}{1 + \dfrac{\Delta L}{L}}.$$

For small changes we use the first-order binomial approximation

$$\dfrac{1}{1 + x} \approx 1 - x \quad\text{when } x \text{ is small}.$$

Putting $$x = \dfrac{\Delta L}{L},$$

$$A' \approx A\!\left(1 - \dfrac{\Delta L}{L}\right).$$

Therefore the fractional change in area is

$$\dfrac{\Delta A}{A} = \dfrac{A' - A}{A} \approx \left(1 - \dfrac{\Delta L}{L}\right) - 1 = -\,\dfrac{\Delta L}{L}.$$

So the area decreases by exactly the same fraction by which the length increases.

Now we calculate the new resistance $$R'$$:

$$R' = \rho \dfrac{L'}{A'} = \rho \dfrac{L(1 + \dfrac{\Delta L}{L})}{A\!\left(1 - \dfrac{\Delta L}{L}\right)}.$$

Combining the two first-order factors and again neglecting higher-order terms $$\bigl((\Delta L/L)^2\bigr),$$ we have

$$\dfrac{1 + \dfrac{\Delta L}{L}}{1 - \dfrac{\Delta L}{L}} \approx 1 + \dfrac{\Delta L}{L} + \dfrac{\Delta L}{L} = 1 + 2\,\dfrac{\Delta L}{L}.$$

Thus

$$R' \approx \rho \dfrac{L}{A}\!\left(1 + 2\,\dfrac{\Delta L}{L}\right) = R \left(1 + 2\,\dfrac{\Delta L}{L}\right).$$

The fractional change in resistance is therefore

$$\dfrac{\Delta R}{R} = 2\,\dfrac{\Delta L}{L}.$$

Putting $$\dfrac{\Delta L}{L} = 0.005$$ gives

$$\dfrac{\Delta R}{R} = 2 \times 0.005 = 0.01.$$

Converting to percentage,

$$0.01 \times 100\% = 1.0\%.$$

So the resistance increases by $$1.0\%.$$ Hence, the correct answer is Option B.