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Q. 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 m³ ?

Answer 1

Dear Student,
Here is the solution of your asked query:
$$\begin{gathered} \mathrm{Let}\mathrm{initial}\mathrm{height}\mathrm{of}\mathrm{the}\mathrm{pond}\mathrm{be}\mathrm{h}\mathrm{m}. \\ \mathrm{Length}\mathrm{of}\mathrm{pond}=80\mathrm{m} \\ \mathrm{And}\mathrm{breadth}\mathrm{of}\mathrm{pond}=50\mathrm{m} \\ \mathrm{So},\mathrm{initial}\mathrm{volume}\mathrm{of}\mathrm{tank}=80\times 50\times \mathrm{h}=4000\mathrm{h}{\mathrm{m}}^3 \\ \mathrm{Let}\mathrm{x}\mathrm{be}\mathrm{rise}\mathrm{in}\mathrm{water}\mathrm{level}\mathrm{of}\mathrm{the}\mathrm{pond}. \\ \mathrm{Now}\mathrm{volume}\mathrm{of}\mathrm{tank}=80\times 50\times \left(\mathrm{h}+\mathrm{x}\right)=4000\left(\mathrm{x}+\mathrm{h}\right) \\ \mathrm{Final}\mathrm{volume}\mathrm{of}\mathrm{tank}=\mathrm{initial}\mathrm{volume}\mathrm{of}\mathrm{tank}+\mathrm{volume}\mathrm{of}500\mathrm{persons} \\ \Rightarrow 4000\left(\mathrm{x}+\mathrm{h}\right)=4000\mathrm{h}+500\times 0.04 \\ \Rightarrow 4000\mathrm{x}+4000\mathrm{h}=4000\mathrm{h}+20 \\ \Rightarrow \mathrm{x}=\frac{20}{4000}=0.005 \\ \mathrm{Therefore}\mathrm{rise}\mathrm{in}\mathrm{water}\mathrm{level}\mathrm{of}\mathrm{the}\mathrm{pond}\mathrm{is}0.005\mathrm{m} \end{gathered}$$
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