Let the radius of the sphere be r unit.
`therefore` the of the curved surface of the sphere=`4pir^(2)` sq-unit.
If the radius of the sphere be increased by 50% then the new radius will be `(r+rxx(50)/(100)) “unit”=(3r)/(2)` unit.
Then the curved surface area of the sphere will be `4pi((3r)/(2))^(2)”sq.units”=4pi.(9r^(2))/(4)` sq.units
`=9pir^(2)` sq.units.
`therefore` increase of corved surface area `=(9pir^(2)-4pir^(2)) “sq.unit”=5pir^(2)` sq.unit
`therefore` the percent of increment of the curved surface area of the sphere =`(5pir^(2))/(4pir^(2))xx100%=125%`
Hence the required percent=125%