Water flows through a horizontal tube as shown in figure. If the difference of heights of water column in the vertical tubes is 2 cm and the area of cross section at A and B are `4cm^2 and 2cm^2` respectively, find the rate of flow of water across any section.

Options

A. `130xx10^(-6)m^(3)//s`
B. `146xx10^(-6)m^(3)//s`
C. `160xx10^(-6)m^(3)//s`
D. `170xx10^(-6)m^(3)//s`

Answer

B

Explanation

`v_(A)a_(A)=v_(B)xxa_(B)=v_(A)xx4=v_(B)xx2v_(B)=2v_(A)`………i
Again `1/2, rhov_(A)^(2)+rhogh_(A)+p_(A)=1/2rhov_(B)^(2)+rhogh_(B)+p_(B)`
`implies1/2rhog_(A)^(2)+p_(A)=1/2rhov_(B)^(2)+p_(B)` (as `h_(A)=h_(B)`)
`impliesp_(A)-p_(B)=1/2rho(v_(B)^(2)-v_(A)^(2))=1/2xx1xx(4v_(A)^(2)-v_(A)^(2))`
`implies2x1xx1000=1/2xx1xx3v_(A)^(2)`
`p_(A)=p_(B)=2cm` of water column `=2xx1xx1000dyn//cm^(2)`
`:.v_(A)=sqrt(4000/3)=36.51cm//s`
So, rate of flow `=V_(a)a_(A)=36.51xx4=146cm^(3)//s`.