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(B) Let the rise of water level in the pond be $h \, m$,when $500$ persons are taking a dip into a cuboidal pond.Given that,Length of the cuboidal pon

Vedclass · Accepted Answer

$0.5$

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(B) Let the rise of water level in the pond be $h \, m$,when $500$ persons are taking a dip into a cuboidal pond.Given that,Length of the cuboidal pond $= 80 \, m$Breadth of the cuboidal pond $= 50 \, m$Now,the volume of the water rise in the pond is given by:Volume $= 	ext{Length} 	imes 	ext{Breadth} 	imes 	ext{Height}$Volume $= 80 	imes 50 	imes h = 4000h \, m^3$The average displacement of water by one person $= 0.04 \, m^3$So,the total displacement of water by $500$ persons $= 500 	imes 0.04 \, m^3 = 20 \, m^3$According to the problem,the volume of the rise in water level equals the total displacement of water by $500$ persons:$4000h = 20$$h = \frac{20}{4000} \, m$$h = \frac{1}{200} \, m = 0.005 \, m$To convert the height into centimeters $(cm)$:$h = 0.005 	imes 100 \, cm = 0.5 \, cm$Hence,the required rise of water level in the pond is $0.5 \, cm$.

$500$ persons are taking a dip into a cuboidal pond which is $80 \, m$ long and $50 \, m$ broad. What is the rise of water level in the pond,if the average displacement of the water by a person is $0.04 \, m^3$? (in $cm$)

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