A water tank has the shape of an inverted right circular cone, whose semi-vertical angle is tan^-1( 1 2). Water is poured into it at a constant rate…

The given water tank is of the shape shown by the diagram.

The semi-vertical angle $\theta ={\mathrm{tan}}^{-1}(\frac{1}{2}),\Rightarrow \mathrm{tan}\theta =\frac{1}{2}$θ=tan−1(21​),⇒tanθ=21​

Let at any time $tmin,$tmin, height of water level is $hm$hm and radius of cone filled with water be $rm.$rm.

Also, we have $\mathrm{tan}\theta =\frac{r}{h}$tanθ=hr​

$$\Rightarrow \frac{r}{h}=\frac{1}{2}$$⇒hr​=21​

$$\Rightarrow r=\frac{h}{2}...(i)$$⇒r=2h​...(i)

Now, the volume of the water at time $tmin$tmin in the cone is $V=\frac{1}{3}\pi {r}^{2}h$V=31​πr2h

On putting the value of $r$r from the equation $(i),$(i), we get

$$V=\frac{\pi }{3}(\frac{{h}^{3}}{4})=\frac{\pi }{12}({h}^{3})$$V=3π​(4h3​)=12π​(h3)

Now, differentiating with respect to $t,$t, we get

$$\frac{dV}{dt}=\frac{\pi }{12}\cdot (3{h}^{2}\frac{dh}{dt})$$dtdV​=12π​⋅(3h2dtdh​)

Put the given value of \frac{dV}{dt}&h

$$\Rightarrow 5=\frac{\pi }{4}\cdot (100\frac{dh}{dt})$$⇒5=4π​⋅(100dtdh​)

$$\Rightarrow \frac{dh}{dt}=\frac{1}{5\pi }m/min$$⇒dtdh​=5π1​m/min