A farmer connects a pipe of internal diameter 20 cm from a canal into a cylindrical tank which is 10 m in diameter and 2 m deep.

Question:

A farmer connects a pipe of internal diameter 20 cm from a canal into a cylindrical tank which is 10 m in diameter and 2 m deep. If the water flows through the pipe at the rate of 4 km/hr, then in how much time will the tank be filled completely?

Solution:

We have,

Internal radius of the pipe, $r=\frac{20}{2}=10 \mathrm{~cm}=0.1 \mathrm{~m}$,

Radius of the cylindrical tank, $R=\frac{10}{2}=5 \mathrm{~m}$ and

Height of the cylindrical tank, $H=2 \mathrm{~m}$

Also, the speed of the water flow in the pipe, $h=4 \mathrm{~km} / \mathrm{hr}=\frac{4 \times 1000 \mathrm{~m}}{1 \mathrm{hr}}=4000 \mathrm{~m} / \mathrm{hr}$

Now,

The volume of the water flowing out of the pipe in a hour $=\pi r^{2} h$

$=\frac{22}{7} \times 0.1 \times 0.1 \times 4000$

$=\frac{880}{7} \mathrm{~m}^{3}$

And,

The volume of the cylindrical $\operatorname{tank}=\pi R^{2} H$

$=\frac{22}{7} \times 5 \times 5 \times 2$

$=\frac{1100}{7} \mathrm{~m}^{3}$

So,

The time taken to fill the tank $=\frac{\text { Volume of the cylindrical tank }}{\text { Vo } 2}$

$=\frac{\left(\frac{1100}{7}\right)}{\left(\frac{880}{7}\right)}$

$=\frac{1100}{880}$

$=\frac{5}{4} \mathrm{hr}$

$=1 \frac{1}{4} \mathrm{hr}$

$=1 \mathrm{hr}$ and $\frac{1}{4} \times 60 \mathrm{~min}$

$=1 \mathrm{hr} 15 \mathrm{~min}$

So, the tank will be completely filled in 1 hour 15 minutes.

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