In a swimming pool measuring 90, m by 40, m,1 | vedclass

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(D) The average displacement of water by one man is $8\, m^3$.The total displacement of water by $150$ men is $150 	imes 8\, m^3 = 1200\, m^3$.Let th

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$33.33$

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(D) The average displacement of water by one man is $8\, m^3$.The total displacement of water by $150$ men is $150 	imes 8\, m^3 = 1200\, m^3$.Let the rise in the water level be $h$ meters.The volume of the displaced water is equal to the area of the pool multiplied by the rise in height: $90\, m 	imes 40\, m 	imes h = 1200\, m^3$.$3600 	imes h = 1200$.$h = \frac{1200}{3600} = \frac{1}{3}\, m$.To convert the height into centimeters,multiply by $100$: $h = \frac{1}{3} 	imes 100 = 33.33\, cm$.

In a swimming pool measuring $90\, m$ by $40\, m$,$150$ men take a dip. If the average displacement of water by a man is $8\, m^3$,what will be the rise in water level? (in $cm$)

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