500 men took dip in a tank which is 80 m long, 50 m broad. What is the rise in water level if the average displacement of water by a man is 4 $m^{3}$

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Solution

Volume of water displaced by $500$ men = $(4 \times 500) = 2000 m^{3}$

Area of the tank = $80 \times 50 = 4000 m^{2}$ .
We know that, Volume of cuboid $= Area \times height$
Let the rise in the level of water $= h$

Then $4000 \times h = 2000$
$h = 0.5 m$

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500 men took dip in a tank which is 80 m long, 50 m broad. What is the rise in water level if the average displacement of water by a man is 4 m3

Volume of water displaced by 500 men = (4×500)=2000 m3 Area of the tank = 80×50=4000 m2.We know that, Volume of cuboid =Area×height Let the rise in the level of water =h Then 4000×h=2000 h=0.5 m

500 men took dip in a tank which is 80 m long, 50 m broad. What is the rise in water level if the average displacement of water by a man is 4 $m^{3}$

Volume of water displaced by $500$ men = $(4 \times 500) = 2000 m^{3}$

Area of the tank = $80 \times 50 = 4000 m^{2}$ .
We know that, Volume of cuboid $= Area \times height$
Let the rise in the level of water $= h$

Then $4000 \times h = 2000$
$h = 0.5 m$