If water is poured into an inverted hollow cone whose semi-vertical angle is ${30}^0,$ its depth (measured along the axis) increases at the rate of 1 cm/s. If the rate at which  the volume of water increases when the depth is 24 cm is $k\pi c{m}^3/\sec$ then $k$ is equal to

Let A be the vertex and AO axis of the cone. Let O’A=h be the depth of water in the  cone. In $\Delta AO\text{'}C, \tan {30}^0=\frac{O\text{'}C}{h}orO\text{'}C=\frac{h}{\sqrt{3}}=\text{radius}$

V=Volume of water in the cone $=\frac{1}{3}\pi {(O\text{'}C)}^2\times AO\text{'}=\frac{1}{3}\pi \left(\frac{h^2}{3}\right)\times h$

$V=\frac{\pi }{9}{h}^3$Or $\frac{dV}{dt}=\frac{\pi }{3}{h}^2\frac{dh}{dt}\left(1\right)$

But given that depth of water increases at the rate of 1 cm/s. So, $\frac{dh}{dt}=1cm/s\left(2\right)$

From  (1) and (2), $\frac{dV}{dt}=\frac{\pi h^2}{3}$

When $h=24$ cm, the rate of increases of volume is $\frac{dV}{dt}=\frac{\pi {\left(24\right)}^2}{3}=192 \mathrm{\pi }c{m}^3/s$