A farmer runs a pipe of internal diameter $ 20 \mathrm{~cm} $ from the canal into a cylindrical tank in his field which is $ 10 \mathrm{~m} $ in diameter and $ 2 \mathrm{~m} $ deep. If water flows through the pipe at the rate of $ 3 \mathrm{~km} / \mathrm{h} $, in how much time will the tank be filled?

Given:

A farmer runs a pipe of internal diameter \( 20 \mathrm{~cm} \) from the canal into a cylindrical tank in his field which is \( 10 \mathrm{~m} \) in diameter and \( 2 \mathrm{~m} \) deep.

The water flows through the pipe at the rate of \( 3 \mathrm{~km} / \mathrm{h} \).

To do:

We have to find the time it takes to fill the tank.

Solution:

Diameter of the pipe $=20 \mathrm{~cm}$

This implies,

Radius of the pipe $r=\frac{20}{2}$

$=10 \mathrm{~cm}$

$=\frac{10}{100}=\frac{1}{10} \mathrm{m}$

Diameter of the cylindrical tank $=10 \mathrm{~m}$

This implies,

Radius of the tank $\mathrm{R})=\frac{10}{2}$

$=5 \mathrm{~m}$

Depth of the tank $H=2 \mathrm{~m}$
Therefore,

Volume of water filled in the tank $=\pi R^{2} H$

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

$=\frac{1100}{7} \mathrm{~m}^{3}$
Let the length of flow of water from the pipe $=\mathrm{h}$
This implies,

$\pi r^{2} \mathrm{h}=\frac{1100}{7}$

$\frac{22}{7} \times \frac{1}{10} \times \frac{1}{10} \times \mathrm{H}=\frac{1100}{7}$

$\mathrm{H}=\frac{1100}{7} \times \frac{700}{22}$

$=5000$

$=5 \mathrm{~km}$
Rate of flow of water $=3 \mathrm{~km} / \mathrm{hr}$

Therefore,

Time taken to fill the tank $=\frac{5}{3} \mathrm{hr}$

$=1 \frac{2}{3}$ hours

$=1\ hr\ 40\ minutes$

The time it takes to fill the tank is 1 hour 40 minutes.

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Updated on: 10-Oct-2022

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