Water is flowing at the rate of $ 2.52 \mathrm{~km} / \mathrm{h} $ through a cylindrical pipe into a cylindrical tank, the radius of the base is $ 40 \mathrm{~cm} $. If the increase in the level of water in the tank, in half an hour is $ 3.15 \mathrm{~m} $, find the internal diameter of the pipe.

Given:

Water is flowing at the rate of \( 2.52 \mathrm{~km} / \mathrm{h} \) through a cylindrical pipe into a cylindrical tank, the radius of the base is \( 40 \mathrm{~cm} \).

The increase in the level of water in the tank, in half an hour is \( 3.15 \mathrm{~m} \).

To do:

We have to find the internal diameter of the pipe.

Solution:

Rate of water flow $=2.52 \mathrm{~km} / \mathrm{hr}$

Radius of the base of the cylindrical tank $=40 \mathrm{~cm}$

Increase in the water level per half an hour $=3.15 \mathrm{~m}$

$=\frac{315}{100} \mathrm{~m}$

Therefore,

Volume of water in the tank $=\pi r^{2} h$

$=\frac{22}{7} \times \frac{40}{100} \times \frac{40}{100} \times \frac{315}{100}$

$=\frac{198}{125} \mathrm{~m}^{3}$

Le the inner radius of the pipe be $R$.

This implies,

Length of column of water flown in half an hour $=\frac{2.52}{2} \mathrm{~km}$

$=1.26 \mathrm{~km}$

$=1260 \mathrm{~m}$

Therefore,

Volume of water flown from the pipe $=$ Volume of water filled in the tank

$=\frac{198}{125} \mathrm{~m}^{3}$

$\Rightarrow \frac{22}{7} \mathrm{R}^{2} \times 1260=\frac{198}{125}$

$\Rightarrow \mathrm{R}^{2}=\frac{198 \times 7}{125 \times 22 \times 1260}$

$\Rightarrow \mathrm{R}^{2}=\frac{1}{2500}$

$\Rightarrow \mathrm{R}^{2}=(\frac{1}{50})^{2}$

$\Rightarrow \mathrm{R}=\frac{1}{50} \mathrm{~m}$

$=\frac{1}{50} \times 100 \mathrm{~cm}$

$=2 \mathrm{~cm}$

Diameter of the pipe $=2 \mathrm{R}$

$=2 \times 2$

$=4 \mathrm{~cm}$

The internal diameter of the pipe is $4\ cm$.

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Updated on: 10-Oct-2022

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