VERTEX and AO the axis of the cone. Let \[O'A=h\] be the DEPTH of water in the cone. In \[\Delta AO'C,\] \[\tan 30{}^\circ =\frac{O'C}{h}\] or \[O'C=\frac{h}{\SQRT{3}}=radius\] \[V=\] Volume of water in the cone \[=\frac{1}{3}\pi {{(O'C)}^{2}}\times AO'\] \[=\frac{1}{3}\pi \left( \frac{{{h}^{2}}}{3} \right)\times h=\frac{\pi }{9}{{h}^{3}}\] or \[\frac{dV}{dt}=\frac{\pi }{3}{{h}^{2}}\frac{dh}{dt}\] ? (1) But given that depth of water increases at the rate of 1 cm/s \[\frac{dh}{dt}=1\,\,cm/s\] ?. (2) From (1) and (2), \[\frac{dV}{dt}=\frac{\pi {{h}^{2}}}{3}\] When \[h=24\,\,cm,\] the rate of increase of volume is \[\frac{dV}{dt}=\frac{\pi {{(24)}^{2}}}{3}=192\,\,c{{m}^{3}}/s\].