A balloon remains spherical while inflated by $900$ cm$^3$/s of gas. When the radius is $15$ cm, the radius is increasing at the rate of

1. $V=\dfrac{4}{3}\pi r^3$. 2. $\dfrac{dV}{dt}=4\pi r^2\dfrac{dr}{dt}$. 3. Set $=900$, $r=15$: $900=4\pi(225)\dfrac{dr}{dt}=900\pi\dfrac{dr}{dt}$. 4. $\dfrac{dr}{dt}=\dfrac{900}{900\pi}=\dfrac{1}{\pi}$. _Source: NCERT Class 12 Mathematics Ch 6 "Application of Derivatives", p.4_