InterviewSolution
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InterviewSolution
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Figure shows a large closed cylindrical tank containing water. Initially, the air trapped above the water surface has a height `h_(0)` and pressure `2p_(0)` where `rh_(0)` is the atmospheric pressure. There is a hole in the wall of the tank at a depth `h_(1)` below the top from which water comes out. A long vertical tube is connected as shown. Find the speed with which water comes out of the holeA. `(1)/(rho)[p_(0)-rhog(h_(1)-2h_(0))]^(1//2)`B. `[(2)/(rho)[p_(0)+rhog(h_(1)-h_(0))]]^(1//2)`C. `[(3)/(rho)[p_(0)+rhog(h_(1)+h_(0))]]^(1//2)`D. `[(4)/(rho)[p_(0)-rhog(h_(1)-h_(0))]]^(1//2)` |
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Answer» Correct Answer - B `2P_(0)=(h_(2)+h_(0))rhog+p_(0)` (since liquids at the same level have the same pressure) `P_(0)=h_(2)rhog+h_(0)rhog,h_(2)rhog=P_(0)-h_(0)rhog` `h_(2)=(P_(0))/(rhog)+(h_(0)rhog)/(rhog)=(P_(0))/(rhog)-h_(0)` KE to the water `=` pressure energy of the water at that layer `(1)/(2)mV^(2)=mxx(P)/(rho)` `V^(2)=(2P)/(rho)=(2)/(rho)[P_(0)+rhog(h_(1)+h_(0))]` `V=[(2)/(rho){P_(0)+rhog(h_(1)-h_(0))}]^(1//2)` We know `2P_(0)+rhog(h_(1)-h_(0))=P_(0)+rhogX` `impliesX=(P_(0))/(rhog)+(h_(1)-h_(0))=h_(2)+h_(1)` i.e., X is `h_(1)` meter below the top or X is `h_(2)` above the top. |
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