A water tank has a shape of inverted right circular cone whose semi-vertical angle is $\tan…

A water tank has a shape of inverted right circular cone whose semi-vertical angle is $\tan ^{-1}\left(\frac{1}{2}\right)$. Water is poured into it at constant rate of 5 cubic meter/minute. The rate in meter/minute at which level of water is rising. at the instant when depth of water in the tank is. $10 \mathrm{~m}$ is
  1. $\frac{1}{5 \pi}$
  2. $\frac{1}{15 \pi}$
  3. $\frac{2}{\pi}$
  4. $\frac{1}{10 \pi}$

Solution

Semi-vertical angle $=\tan ^{-1}\left(\frac{1}{2}\right)$ Let $\alpha=\tan ^{-1}\left(\frac{1}{2}\right)$ $\tan \alpha=\frac{1}{2}$ $\frac{\mathrm{r}}{\mathrm{h}}=\frac{1}{2}$ $\mathrm{r}=\frac{\mathrm{h}}{2}$ Given, $\frac{d \mathrm{~V}}{\mathrm{dt}}=5 \mathrm{~m}^3 / \mathrm{min}$. $\mathrm{V}=$ Volume of cone Volume of cone $=\frac{1}{3} \pi r^2 \mathrm{~h}$ $\mathrm{V}=\frac{1}{3} \pi\left(\frac{\mathrm{h}}{2}\right)^2 \times \mathrm{h}$ $\mathrm{V}=\frac{1}{12} \pi \mathrm{h}^3$ Differentiating w, r.t. t, we get $\frac{\mathrm{dV}}{\mathrm{dt}}=\frac{1}{12} \times \pi \times 3 \mathrm{~h}^2 \times \frac{\mathrm{dh}}{\mathrm{dt}}$ $5=\frac{1}{4} \pi h^2 \frac{\mathrm{dh}}{\mathrm{dt}}$ $\frac{\mathrm{dh}}{\mathrm{dt}}=\frac{20}{\pi \mathrm{h}^2}$ Now, $\mathrm{h}=10$... [Given] $\therefore \quad \frac{\mathrm{dh}}{\mathrm{dt}}=\frac{20}{\pi \times(10)^2}$ $\frac{\mathrm{dh}}{\mathrm{dt}}=\frac{1}{5 \pi}$ $\therefore \quad$ Rate of change of water level is $\frac{1}{5 \pi} \mathrm{m} / \mathrm{min}$.

Asked in: MHT CET 2023 (09 May Shift 2)

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