Sand pours from a pipe at $12$ cm$^3$/s, forming a cone whose height is always one-sixth of the base radius. When the height is $4$ cm, the height of the cone is increasing at the rate of

1. Height $h=\dfrac{r}{6}\Rightarrow r=6h$. 2. Volume $V=\dfrac{1}{3}\pi r^2 h=\dfrac{1}{3}\pi(36h^2)h=12\pi h^3$. 3. $\dfrac{dV}{dt}=36\pi h^2\dfrac{dh}{dt}$. 4. Set $=12$: $12=36\pi h^2\dfrac{dh}{dt}$. 5. At $h=4$: $12=36\pi(16)\dfrac{dh}{dt}=576\pi\dfrac{dh}{dt}$. 6. $\dfrac{dh}{dt}=\dfrac{12}{576\pi}=\dfrac{1}{48\pi}$. _Source: NCERT Class 12 Mathematics Ch 6 "Application of Derivatives", p.4_