A) 2.5 m/s
B) 0.25 m/s
C) 1.5 m/s
D) 3.5 m/s
Correct Answer: C
Solution :
| Length of tank = 225 m, breadth = 162 m |
| The volume of water accumulated in tank in 5 h |
| \[=225\times 162\times \frac{20}{100}\,\,{{\text{m}}^{3}}\] |
| Volume of water which flows in 1 h through the aperture |
| \[=\frac{1}{5}\times 225\times 162\times \frac{20}{100}\] |
| \[=1458\,\,{{\text{m}}^{3}}\] |
| The area of the cross-section of aperture |
| \[=\frac{60}{100}\times \frac{45}{100}=\frac{27}{100}\,\,{{\text{m}}^{3}}\] |
| \[\therefore \] Velocity of flow of water per hour |
| \[=\frac{\text{The}\,\,\text{volume}\,\,\text{of}\,\,\text{water}\,\,\text{which}\,\,\text{flows}\,\,\text{in}\,\,\text{1}\,\,\text{h}}{\text{The}\,\,\text{area}\,\,\text{of}\,\,\text{the}\,\,\text{cross-section}}\] |
| \[=\frac{1458}{27/100}\text{m/h}\]\[=\frac{1458\times 100}{27}\text{m/h}=5400\,\text{m/h}\] |
| \[\therefore \]The velocity per second \[=\frac{5400}{60\times 60}=1.5\,\,\text{m/s}\] |
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