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 height of the water in the long tube above the top when the water stops coming out of the hole.A. `-2h_(0)`B. `h_(0)`C. `h_(2)`D. `-h_(1)` |
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Answer» Correct Answer - D `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` of the water `=` Pressure energy of the water at that layer `1/2mV^(2)=mxxP/(rho)` `V^(2)=(2P)/rho=2/(rho)[P_(0)+rhog(h_(1)-h_(2))]` `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)` metre below the top or `X` is `-h_(1)` above the top. |
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