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Semi-vertical angle $$=\tan ^{-1}\left(\frac{1}{2}\right)$$

Let $$\alpha=\tan ^{-1}\left(\frac{1}{2}\right)$$

$$\begin{aligned} & \tan \alpha=\frac{1}{2} \\ & \therefore \quad \frac{\mathrm{r}}{\mathrm{h}}=\frac{1}{2} \\ & \mathrm{r}=\frac{\mathrm{h}}{2} \end{aligned}$$

Given, $$\frac{\mathrm{dV}}{\mathrm{dt}}=5 \mathrm{~m}^3 / \mathrm{min}$$.

$$\mathrm{V}=$$ Volume of cone

Volume of cone $$=\frac{1}{3} \pi \mathrm{r}^2 \mathrm{~h}$$

$$\begin{aligned} & \mathrm{V}=\frac{1}{3} \pi\left(\frac{\mathrm{h}}{2}\right)^2 \times \mathrm{h} \\ & \mathrm{V}=\frac{1}{12} \pi \mathrm{h}^3 \end{aligned}$$

Differentiating w. r. t. t, we get $$\frac{\mathrm{dV}}{\mathrm{dt}}=\frac{1}{12} \times \pi \times 3 \mathrm{~h}^2 \times \frac{\mathrm{dh}}{\mathrm{dt}}$$

$$5=\frac{1}{4} \pi \mathrm{h}^2 \frac{\mathrm{dh}}{\mathrm{dt}}$$

$$\frac{\mathrm{dh}}{\mathrm{dt}}=\frac{20}{\pi \mathrm{h}^2}$$

Now, $$\mathrm{h}=10\quad$$ .... [Given]

$$\begin{aligned} \therefore \quad \frac{\mathrm{dh}}{\mathrm{dt}} & =\frac{20}{\pi \times(10)^2} \\ \frac{\mathrm{dh}}{\mathrm{dt}} & =\frac{1}{5 \pi} \end{aligned}$$

$$\therefore \quad$$ Rate of change of water level is $$\frac{1}{5 \pi} \mathrm{m} / \mathrm{min}$$.

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