Question

Water is flowing through a cylindrical pipe, with an internal diameter of $2\mathrm{cm}$, into a cylindrical tank with a base radius of $40\mathrm{cm}$, at the rate of$0.4\mathrm{m}/\mathrm{s}$.

Determine the rise in the level of water In the tank in half hours.

Answer 1

Step 1. Find the speed of water.

The radius of a cylindrical pipe, $r=1\mathrm{cm}$

The radius of the cylindrical tank, $R=40\mathbf{}\mathrm{cm}$

It is known that,

Length of water through a pipe, $\mathit{l}\mathbf{=}\mathbf{rate}\mathbf{\times }\mathbf{time}\mathbf{}\mathbf{taken}$.

Diameter of cylindrical pipe having circular end $=2\mathrm{cm}$
Radius ${\text{r}}_1$ of circular end of pipe is,

$$\begin{gathered} r_1=\frac{2}{2}\left[\because \mathrm{radius}=\frac{\mathrm{diameter}}{2}\right] \\ =1\mathrm{cm} \\ =0.01\mathrm{m}\left[\because 1\mathrm{m}=100\mathrm{cm}\right] \end{gathered}$$
Area of cross-section is,

$$\begin{gathered} A=\mathrm{\pi }\times {r_1}^2 \\ =\mathrm{\pi }\times {\left(0.01\right)}^2 \\ =0.0001\mathrm{\pi }{\mathrm{m}}^2 \end{gathered}$$
Speed of water is,

$$\begin{gathered} s=0.4\mathrm{m}/\mathrm{s} \\ =0.4\times 60\mathrm{m}/\min \left[\because 1\mathrm{second}=\frac{1}{60}\min \right] \end{gathered}$$

Step 2. Find the radius of base of cylindrical tank.

The volume of water that flows in one minute from the pipe will be equal to the area of a cross-section of the cylindrical pipe multiplied by the speed of flow of water through it.
So, the Volume of water that flows in one minute from the pipe is,

$$\begin{gathered} =24\left(\text{m}/\text{minute}\right)\times 0.0001\mathrm{\pi }{\text{m}}^2 \\ =0.0024\mathrm{\pi }{\text{m}}^3 \end{gathered}$$
So using the unitary method,
The volume of water that flows in $30$ minutes from the pipe $=30\times 0.0024\pi {\text{m}}^3=0.072\pi {\text{m}}^3$
Radius of base of cylindrical tank, $r_2=40\mathrm{cm}\mathrm{or}0.4\mathrm{m}$ $\left(\text{using}1\text{cm =}\text{.01 m}\text{.}\right)$

Step 3. Find the rise in the level of water.
Let the cylindrical tank be filled up to him in $30$ minutes.
We know that,
The volume of water filled in a cylindrical tank in $30$ minutes is equal to the volume of water flowing out in 30 minutes from the pipe.

$$\begin{gathered} \mathrm{\pi }{r_2}^2\times h=0.072\mathrm{\pi } \\ \Rightarrow \mathrm{\pi }{\left(0.4\right)}^2\times h=0.072\mathrm{\pi } \\ \Rightarrow \mathrm{\pi }\times 0.16\times h=0.072\mathrm{\pi } \\ \Rightarrow h=\frac{0.072}{0.16} \\ \therefore h=0.45\mathrm{m}\mathrm{or}45\mathrm{cm} \end{gathered}$$
Hence, the rise in the level of water in the tank in half an hour (or in $30$ minutes) is $45\mathrm{cm}$.