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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