JEE Main 2014 (Online) 19th AprilQuestion 1
If the volume of a spherical ball is increasing at the rate of $4\pi$ cc/sec then the rate of increase of its radius (in cm/sec), when the volume is $288\pi$ cc is
Differentiation and its Application
Solution
Concept. Differentiate $V=\frac43\pi r^3$ with respect to time. Formulas used. (1) $\dfrac{dV}{dt}=4\pi r^2\dfrac{dr}{dt}$Step 1 - Radius. $\frac43\pi r^3=288\pi$ gives $r^3=216$, $r=6$. Step 2 - Rate. By (1), $4\pi=4\pi\cdot36\cdot\frac{dr}{dt}$, so $\dfrac{dr}{dt}=\dfrac{1}{36}$. Answer: (D)