As shown in figure, a liquid of density `rho` is standing in sealed container to a height h. The container contains compressed air at a gauge pressure of p.The horizontal outlet pipe has a cross-sectional area A at C and D. The cross-sectional area is `A//2` at E. Find correct options:

Options

A. The velocity of liquid at C will be `[((P+rhogh))/(4rho)]^(1//2)`
B. The velocity of liquid at C will be `[(2(P+rhogh))/(rho)]^(1//2)`
C. the discharge rate is given by `(A)/(2rho)(p+rhogh)^(1//2)`
D. The discharge rate is given by `(A)/(2sqrt(rho))(p+rhogh)^(1//2)`

Answer

A, D

Explanation

`p_(A)=p+p_(0)+rhogh`. (1)
Apply bernoulli's theorem at A and C and E
`(p+p_(0))+rhogh=p_(c)+rho(V_(c)^(2))/(2)`
`=p_(0)+rho(V_(E)^(2))/(2)` ….(2)
`=p_(c)+p_(0)+rhogh_(1)` .(3)
`A_(C)V_(C)=A_(E)V_(E),A_(C)V_(C)=((A_(C))/(2))V_(E)` .(4)
from eq. (2) `(p_(0)+rhogh_(1))+rho(V_(C)^(2))/(2)=p_(0)+rho(2V_(C))^(2)`
or `h_(1)=(7V_(C)^(2))/(2g)`
for eq. (2), `(p+p_(0))+rhogh=(p_(0)+rhogh_(1))+rho(V_(C)^(2))/(2)`
`p+rhogh=rhog((7V_(C)^(2))/(2g))+(rhoV_(C)^(2))/(2),V_(C)^(2)=((p+rhogh))/(4rho)`
Discharge rate `=A_(C)V_(C)=A(((p+rhogh))/(4rho))^(1//2)`