Given that, surface area of balloon is increasing in constant rate. ∴
ds
dt
= constant =k (Assume) ⇒k=8πr⋅
dr
dt
⇒∫kdt=∫8πrdr ⇒kt=8π×
r2
2
+c ⇒kt=4πr2+c.....(1) Given at t=0, radius r=3 So, 0=4π(32)+c ⇒c=−36π ∴ Equation (1) becomes kt=4πr2−36π Also given, at t=5, radius r=7 ∴k(5)=4π(7)2−36 ⇒k=32π ∴ Equation (1) is 32πt=4πr2−36π Now at t=9 ⇒32π(9)=4πr2−36π ⇒8×9=r2−9 ⇒r2=81⇒r=9