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.

Options

A. `-2h_(0)`
B. `h_(0)`
C. `h_(2)`
D. `-h_(1)`

Answer

D

Explanation

`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.