1. Use the resistance formula

For a wire, $R = \rho \frac{L}{A}$ where $\rho$ is resistivity, $L$ is length, and $A$ is cross-sectional area.

2. Interpret “length increased by $100\%$”

If the original length is $L$, then the new length is $L' = 2L$

3. Wire is stretched, so volume remains constant

For the same wire, $A L = A' L'$ Since $L' = 2L$, $A' = \frac{AL}{L'} = \frac{AL}{2L} = \frac{A}{2}$

So the new area becomes half.

4. Find the new resistance

$R' = \rho \frac{L'}{A'} = \rho \frac{2L}{A/2} = \rho \frac{4L}{A} = 4R$

Thus resistance becomes four times the original.

5. Calculate percentage change

Increase in resistance: $\Delta R = R' - R = 4R - R = 3R$

Percentage increase: $\frac{\Delta R}{R} \times 100 = \frac{3R}{R}\times 100 = 300\%$

6. Check options

• A: $200\%$ ❌
• B: $100\%$ ❌
• C: $50\%$ ❌
• D: $300\%$ ✅

Therefore, the correct answer is D.