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 hole

Options

A. `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_(2))]]^(1/2)`

Answer

B

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` 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.