Let the original diameter of the wire be \(2r\) units, its length be \(h\) units, and its volume be \(V\) cubic units. \[ \therefore V = \pi r^2 h \quad \text{(since radius = \(r\))} \] If the diameter is reduced by 50%, the new diameter becomes: \[ 2r - 2r \times \frac{50}{100} = 2r - r = r \] So, the new radius is: \[ \frac{r}{2} \] To keep the volume unchanged, let the new length be \(H\) units. Then the new volume is: \[ V = \pi \left(\frac{r}{2}\right)^2 \cdot H = \frac{\pi r^2 H}{4} \] According to the question: \[ \frac{\pi r^2 H}{4} = \pi r^2 h \Rightarrow H = 4h \] So, the increase in length is: \[ H - h = 4h - h = 3h \] Percentage increase in length: \[ \frac{3h}{h} \times 100\% = 300\% \] Therefore, the length of the wire must be increased by 300%