The correct option is B 10 cm2/min
Let V be the volume and r the radius of the balloon at any time.
Then, V=(43)πr3
∴dVdt=4πr2drdt
⇒40=4πr2drdt (Given)
⇒drdt=10πr2 ...(1)
Now, let S be the surface area of the balloon when its radius is r. Then
S=4πr2
∴dSdt=8πrdrdt
From (1) and (2),
dSdt=8πr10πr2=80r
When r=8, the rate of increase of surface area is, S=808=10 cm2/min.