Saved Bookmarks
| 1. |
The surface area of a balloon of spherical shape being inflated, increases at a constant rate. If initially, the radius of balloon is 3 units and after 5 seconds, it becomes 7 units, then its radius after 9 seconds is :(A) 9(B) 10(C) 11(D) 12 |
|
Answer» Correct option is (A) 9 Let r be the radius of spherical balloon S = Surface area S = 4πr2 \(\frac{dS}{dt}=8πr \times\frac{dr}{dt}=k(constant)\) 4πr2 = kt + C (C is constant of integration) For t = 0, r = 3 ⇒ 36π = C For t = 5, r = 7 ⇒ K = 32π 4πr2 = 32πt + 36π r2 = 8t + 9 for t = 9 r2 = 81 r = 9 |
|