If the original radius of the sphere is \(r\) units, then after a 50% increase, the new radius becomes: \[ r + \frac{r \times 50}{100} = r + \frac{r}{2} = \frac{3r}{2} \] The original curved surface area of the sphere was: \[ 4πr^2 \text{ square units} \] The new curved surface area becomes: \[ 4π\left(\frac{3r}{2}\right)^2 = 4π \times \frac{9r^2}{4} = 9πr^2 \text{ square units} \] So, the increase in surface area is: \[ 9πr^2 - 4πr^2 = 5πr^2 \text{ square units} \] Therefore, the percentage increase in the curved surface area is: \[ \frac{5πr^2}{4πr^2} \times 100 = 125\% \] (Answer)