500 persons have to dip in a rectangular tank which is $ 80 \mathrm{~m} $ long and $ 50 \mathrm{~m} $ broad. What is the rise in the level of water in the tank, if the average displacement of water by a person is $ 0.04 \mathrm{~m}^{3} $?
Given:
500 persons have to dip in a rectangular tank which is \( 80 \mathrm{~m} \) long and \( 50 \mathrm{~m} \) broad.
The average displacement of water by a person is \( 0.04 \mathrm{~m}^{3} \).
To do:
We have to find the rise in the level of water in the tank.
Solution:
Length of the rectangular tank $l=80 \mathrm{~m}$
Breadth of the rectangular tank $b=50 \mathrm{~m}$
Average displacement of water by 1 person $=0.04 \mathrm{~m}^{3}$
This implies,
Displacement of water by 500 persons $=0.04 \times 500 \mathrm{~m}^{3}$
$=20 \mathrm{~m}^{3}$
Let the rise of water level in the tank be $H$.
Therefore,
Rise of water level in the tank $H=\frac{\text { Volume of water }}{l \times b}$
$=\frac{20}{80 \times 50}$
$=\frac{1}{200}$
$=0.5 \mathrm{~cm}$
The rise in the level of water in the tank is $0.5\ cm$.
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