A water tank is an inverted cone with semi-vertical angle $\tan^{-1}(0.5)$. Water is poured at $5$ m$^3$/h. When the depth is $4$ m, the water level rises at the rate of ($\pi=\tfrac{22}{7}$)

1. $\tan\alpha=\dfrac{r}{h}=0.5 \Rightarrow r=\dfrac{h}{2}$. 2. $V=\dfrac{1}{3}\pi r^2 h=\dfrac{1}{3}\pi\left(\dfrac{h}{2}\right)^2 h=\dfrac{\pi h^3}{12}$. 3. $\dfrac{dV}{dt}=\dfrac{\pi}{4}h^2\dfrac{dh}{dt}$. 4. At $h=4$, $\dfrac{dV}{dt}=5$: $5=\dfrac{\pi}{4}(16)\dfrac{dh}{dt}=4\pi\dfrac{dh}{dt}$. 5. $\dfrac{dh}{dt}=\dfrac{5}{4\pi}=\dfrac{5}{4}\cdot\dfrac{7}{22}=\dfrac{35}{88}$ m/h. _Source: NCERT Class 12 Mathematics Ch 6 "Application of Derivatives", p.31_