In the figure shown the velocity and pressure of the liquid at the cross section (2) are given by (If `P_(0)` is the atmospheric pressure).

Options

A. `sqrt(2hg),P_(0)+(rhohg)/2`
B. `sqrt(hg), P_(0)+(rhohg)/2`
C. `sqrt((hg)/2)P_(0)+(3rhohg)/4`
D. `sqrt((hg)/2,P_(0)+(3rhohg)/4`

Answer

C

Explanation

Velocity of the liquid just after coming out of section `3` is
`v_(3)=sqrt(2gh)`…..i
Using continuity equatioin at equation `2` and section `3`
`A/2 v_(2)=A/4v_(3)`
`v_(2)=1/2v_(3)=sqrt((hg)/2)`
Now using Beroulli's theorem at section `2` and at the opening end of pipe `3`
`P_(0)+1/2rhov_(3)^(2)=P_(2)+1/2rhov_(2)^(2)`
`impliesP_(2)-P_(0)+1/2rho(v_(3)^(2)-v_(2)^(2))=P_(0)+1/2rho(2gh-(gh)/2)`
`P_(2)=P_(0)+(3rhogh)/4`