Q. A spherical balloon is being inflated at the rate of 30$cc^{\ \ }/\min$. The rate of increase of the surface area of the balloon when its diameter is 14cm, is
A) 7 sq cm/min
B) 10 sq cm/min
C) 17.5 sq cm/min
D) 28 sq cm/min
Explanation
(b)
Let volume of sphere($V$)=$\frac{4}{3}\pi r^{3}$
$\Rightarrow \frac{dV}{dt}4\pi r^{2}\frac{dr}{dt}\ \ \Rightarrow \frac{dr}{dt} = \frac{35}{4\pi r^{2}}$
Let surface area of sphere(S)$= 4\pi r^{2}$
$\Rightarrow \frac{dS}{dt} = 2 \times a\pi r\frac{dr}{dt}$
$\ \ = 2 \times 4\pi r \times \frac{35}{4\pi r^{2}}$
$\therefore\ \ \ \ \frac{dS}{dt} = 10sq\ cm/\min$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.