A rectangular tank is $80\ m$ long and $25\ m$ broad. Water-flows into it through a pipe whose cross-section is $25\ cm^2$, at the rate of $16\ km$ per hour. How much the level of the water rises in the tank in $45$ minutes.

Given:

A rectangular tank is $80\ m$ long and $25\ m$ broad.

Water flows into it through a pipe whose cross-section is $25\ cm^2$, at the rate of $16\ km$ per hour. 

To do:

We have to find the level of the water rise in the tank in $45$ minutes.

Solution:

Length of the tank $(l) = 80\ m$

Breadth of the tank $(b) = 25\ m$

Area of cross section of the mouth of the pipe $= 25\ cm^2$

Speed of water flow $=16\ km/h$

Therefore,

Volume of water flow in 45 minutes $=\frac{16 \times 3}{4} \times 1000 \times \frac{25}{10000}$

$=\frac{12 \times 25}{10}$

$=\frac{300}{10}$

$=30 \mathrm{~m}^{3}$

This implies,

Level of water in the tank $=\frac{\text { Volume of water }}{\text { Area of tank }}$

$=\frac{30}{80 \times 25}$

$=\frac{3}{200}$

$=\frac{3}{200} \times 100$

$=1.5 \mathrm{~cm}$

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Updated on: 10-Oct-2022

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