A spherical balloon expands at a rate proportional to its surface area. Initially its radius is 2 cm and 5 minutes later it increases upto 7 cm, then surface area of spherical balloon after 12 minutes will be

Step-by-step solution

Let V be the volume and S be the surface area of the spherical balloon. Given that the balloon expands at a rate proportional to its surface area: dV dt S dV dt = kS Using V = 4 3 r^3 and S = 4 r^2 , differentiating V with respect to t gives: dV dt = 4 r^2 dr dt = S dr dt Equating the two expressions for dV dt : S dr dt = kS dr dt = k Integrating both sides with respect to t : r = kt + C Initially, at t = 0 , r = 2 cm: 2 = k(0) + C C = 2 At t = 5 minutes, r = 7 cm: 7 = 5k + 2 5k = 5 k = 1 Thus, the radius at any ti