Question

500 persons are taking a dip into a cylinder pond which is 80metres long and 50 metres broad. what is the raise of water level in the pond if the average displacement of water by a person is 0.04metre^3?

Answer 1

$\textbf{\huge\underline{\underline\red{solution} : - }} \\ \\ \bold{\underline\purple{here \: we \: know}} \\ \\ \red{1.} \bold \blue{ \: length \: (l)\: of \: cuboidal \: \: tank \: is \: 80 \: m .} \\ \red{2.} \bold \blue{ \: breadth \:( b)\: of \: cuboidal \: tank \: is \: 50 \: m.} \\ \red{3.} \bold \blue{ \: displacement \: of \: water \: by \: a \: one \: } \\ \bold \blue{person \: in \: a \: tank \: is \:0.04 \: {m}^{3} .} \\ \\ \underline \bold\red{so \: \:displacement \: of \: water \: in \: a \: tank \: by \:} \\ \underline \bold\red{ 500 \: persons } = \blue{500 \times0.04 = 20\: }\bold\blue{ {m}^{3} \: \: \: \: ..(1)} \\ \\ \underline \bold \purple{let \: height \: of \: cuboidal \: tank \: is \: h} \\ \\ \underline\bold\blue{to \: find \: volume \: of \: water \: in \: tank} \\ \\ \underline \bold \red{volume\: of \: cuboidal \: tank \: } = \bold \purple{ l\times b\times h} \\ \\ \bold \red{v.o.c.t} = \bold \purple{80 \times 50 \times h} \\ \\ \bold \red{v.o.c.t} = \bold \purple{(4000 \times h \: \: ) {m}^{3} \: \: \: ..(2) } \\ \\ \underline\bold\blue{as \: we\: know} \\ \\ \underline\bold\purple{volume \: of \: raised \: water \: in \: cuboidal } \\ \underline\bold\purple{tank \: is \: equal \: to \: displacement \: of \: water \: in } \\ \underline\bold\purple{ tank\: by \:500 \: person .} \\ \\ \underline\bold\red{hence} \\ \\ \bold\red{volume \: of \: raised \: water \: in \: cuboidal} \\\bold\red{tank \:} = \bold\purple{ displacement \: of \: water \: in } \\ \bold\purple{ tank\: by \:500 \: person .} \\ \\ = > \bold \blue{4000 \times h} = \bold \blue{20} \\ \\ = > \bold \blue{h} = \bold \blue{ \frac{20}{4000}} = \bold \blue{ \frac{1}{200} } = \bold \blue{0.005 \: m} \\ \\ \underline \bold\red{Hence \: \purple{0.005 \: m} \: level \: of \: water \: rise \: in \: tank.}$