A water tank has the shape of an inverted right circular cone, whose semi-vertical angle is $$\tan^{-1}\left(\frac{1}{2}\right)$$. Water is poured into it at a constant rate of 5 cubic m/min. Then the rate (in m/min), at which the level of water is rising at the instant when the depth of water in the tank is 10 m; is:

We consider the inverted right-circular cone that forms the tank. Let the semi-vertical angle be $$\alpha$$. It is given that $$\tan\alpha=\dfrac{1}{2}$$.

In any right-circular cone, the radius $$r$$ and the height $$h$$ are related by the tangent of this angle:

$$\tan\alpha=\dfrac{r}{h}.$$

Substituting $$\tan\alpha=\dfrac{1}{2}\;,$$ we get

$$\dfrac{r}{h}=\dfrac{1}{2}\quad\Longrightarrow\quad r=\dfrac{h}{2}.$$

Now we express the volume $$V$$ of water in the cone at time $$t$$. For a cone, the formula for volume is

$$V=\dfrac{1}{3}\pi r^{2}h.$$

Replacing $$r$$ by $$\dfrac{h}{2}$$ gives

$$V=\dfrac{1}{3}\pi\left(\dfrac{h}{2}\right)^{2}h =\dfrac{1}{3}\pi\left(\dfrac{h^{2}}{4}\right)h =\dfrac{\pi}{12}\,h^{3}.$$

The rate at which water is poured is constant: $$\dfrac{dV}{dt}=5\ \text{m}^{3}\!/\text{min}.$$

We differentiate $$V=\dfrac{\pi}{12}h^{3}$$ with respect to time $$t$$:

$$\dfrac{dV}{dt}=\dfrac{\pi}{12}\cdot 3h^{2}\dfrac{dh}{dt} =\dfrac{\pi}{4}\,h^{2}\dfrac{dh}{dt}.$$

Equating this to the given inflow rate, we have

$$5=\dfrac{\pi}{4}\,h^{2}\dfrac{dh}{dt}.$$

Solving for $$\dfrac{dh}{dt}$$ yields

$$\dfrac{dh}{dt}=\dfrac{5\cdot 4}{\pi h^{2}} =\dfrac{20}{\pi h^{2}}.$$

We require the value when the water depth is $$h=10\ \text{m}$$. Substituting $$h=10$$ gives

$$\dfrac{dh}{dt}=\dfrac{20}{\pi(10)^{2}} =\dfrac{20}{100\pi} =\dfrac{1}{5\pi}\ \text{m/min}.$$

Hence, the correct answer is Option C.