The correct option is A 0.025 cm
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.
Therefore, increase of S in 12 min=10×(12)=5 cm2/min.
If r1 is the radius of the balloon after 12 min, then 4πr21=4π(8)2+5
⇒r21−82=54π=0.397 nearly.
∴ Required increase in the radius =r1=8.025−8
=0.025 cm