• # question_answer An inverted cone is 10 cm in diameter and $10\,\,cm$ deep. Water is poured into it at the rate of $4c{{m}^{3}}/\min$. When the depth of water level is $6\,\,cm$, then it is rising at the rate A) $\frac{9}{4\pi }c{{m}^{3}}/\min$          B) $\frac{1}{4\pi }c{{m}^{3}}/\min$C) $\frac{1}{9\pi }c{{m}^{3}}/\min$          D) $\frac{4}{9\pi }c{{m}^{3}}/\min$

Let $y$ be the level of water at time $t$ and $x$ the radius of the surface and $V$, the volume of water. We know that the volume of cone                 $=\frac{1}{3}\pi {{(radius)}^{2}}\times height$ $\therefore$  $V=\frac{1}{3}\pi {{x}^{2}}y.$ Let$\angle BAD=\alpha$ $\Rightarrow$               $\tan \alpha =\frac{BD}{AD}=\frac{5}{10}=\frac{1}{2}$ Again, from right angled $\Delta AMR$, we have                 $\tan \alpha =\frac{MR}{AR}=\frac{x}{y};\,\,\Rightarrow \frac{1}{2}=\frac{x}{y};\,\,\,\,x=\frac{y}{2}$. $\therefore$$V=\frac{1}{3}\pi {{x}^{2}}y=\frac{1}{3}\pi {{\left( \frac{y}{2} \right)}^{2}}\cdot y=\frac{\pi }{12}{{y}^{2}}$ ? (1) By question, the rate of change of volume                 $=\frac{dV}{dt}=4\,\,cub.cm/\min$ We have to find out the rate of increase of water-level$i.e.,$$\frac{dy}{dt}.$ Differentiating (1) with respect to$t$, we get $\frac{dV}{dt}=\frac{\pi }{12}.3{{y}^{2}}\cdot \frac{dy}{dt};\,\,\therefore 4=\frac{\pi }{4}{{y}^{2}}\cdot \frac{dy}{dt};\,\,\therefore \frac{dy}{dt}=\frac{16}{\pi {{y}^{2}}}$When$y=6\,\,cm,\,\,\frac{dy}{dt}=\frac{16}{\pi {{6}^{2}}}=\frac{4}{9\pi }cub.cm/\min$