Correct Answer - A
The water is filled up to height H and then the top is closed . The pressure of air column now is atmospheric pressure ,`P_(0)`
when the tap is opened and no more water is flowing, the pressure inside P+ pressure due to water column is atmospheric pressure.
`P+hgrho=P_(0)`
`P+(200)xx10^(-3)xx10xx1000=P_(0)" "...(i)`
`therefore P=P_(0)-hgrho`
`implies P=1.0xx10^(5)-200xx10^(-3)xx10xx1000`
`P=0.98xx10^(5) N//m^(2)" "...(ii)`
`PV`= constant
Let area of cross section of the vessel be A.
`P_(0)(500-H)xx10^(-3)xxA=P(300xx10^(-3))xxA`
`implies P_(0)(500-H)=300 P`
From equation `(i) P_(0)(5-H)/(300)+2000=P_(0)`
`therefore H=206` mm
The final height of air column =200 mm Fall in the height =6 mm.