Question

In a hospital, used water is collected in a cylindrical tank of diameter $2\mathrm{m}$ and height $5\mathrm{m}$. After recycling, this water is used to irrigate a park of hospital whose length is $25\mathrm{m}$ and breadth is $20\mathrm{m}$. If tank is filled completely then what will be the height of standing water used for irrigating the park.

Answer 1

Step 1: Find the volume of the tank.

Given: Diameter of cylindrical tank, $\mathrm{d}=2\mathrm{m}$

Height of cylindrical tank, $\mathrm{h}=5\mathrm{m}$

So radius, $\mathrm{r}=\frac{2}{2}\left[\because \mathrm{r}=\frac{\mathrm{d}}{2}\right]$

$=1\mathrm{m}$

We know that the volume of the cylinder$\mathbf{=}{\mathbf{\pi r}}^{\mathbf{2}}\mathbf{h}$, where $\mathrm{r}$ is the radius and $\mathrm{h}$is the height of the cylinder

So, the volume of the cylindrical tank$={\mathrm{\pi r}}^2\mathrm{h}$

$$\begin{gathered} =\mathrm{\pi }\times 1^2\times 5\left[\because \mathrm{r}=1\mathrm{m}\mathrm{and}\mathrm{h}=5\mathrm{m}\right] \\ =5\mathrm{\pi }{\mathrm{m}}^3 \end{gathered}$$

Step 2: Find the required height.

Given : Length of park, $\mathrm{l}=25\mathrm{m}$

Breadth of park, $\mathrm{b}=20\mathrm{m}$

We know that the volume of the cuboid $\mathbf{=}\mathbf{}\mathbf{lbh}$, where $\mathrm{l}$ is the length, $\mathrm{b}$ is the breadth and $\mathrm{h}$is the height of the cuboid.

So, the volume of water park $=\mathrm{lbh}$, here $\mathrm{h}$is the height of standing water.

$=\left(25\times 20\times \mathrm{h}\right){\mathrm{m}}^3$

As, Water from the tank is used to irrigate the park.

So, Volume of water in the park $=$ the volume of the cylindrical tank

$$\begin{gathered} \Rightarrow 25\times 20\times \mathrm{h}=5\mathrm{\pi } \\ \Rightarrow 500\mathrm{h}=5\mathrm{\pi } \\ \Rightarrow \mathrm{h}=\frac{5\mathrm{\pi }}{500} \\ \Rightarrow \mathrm{h}=\frac{5(3.14)}{500}\left[\because \mathrm{\pi }=3.14\right] \\ \Rightarrow \mathrm{h}=\frac{3.14}{100} \\ \Rightarrow \mathrm{h}=0.0314\mathrm{m} \end{gathered}$$

Hence, the height of standing water used for irrigating the park is $0.0314\mathrm{m}$.