JEE Main2013MathematicsApplication of DerivativesActual
A spherical balloon is being inflated at the rate of 35 cc / min . The rate of increase in the surface area (in cm ^2 / min .) of the balloon when its diameter is 14 ~cm , is :
Options
- A10
- B10
- C100
- D10 10
Correct answer
A. 10
Step-by-step solution
Volume of sphere V = 4 3 r^3 d ~V d t = 4 3 3 r^2 d r d t 35=4 r^2 d r d t or d r d t = 35 4 r^2 Surface area of sphere = S =4 r^2 d ~S d t =4 2 r d r d t =8 r d r d t d ~S d t = 70 r Now, diameter =14 ~cm , r=7 d ~S d t =10