Water flows through a horizontal tube as shown in figure. If the difference of height of water column in the vertical tubes in `2cm` and the areas of corss-section at A and B are `4 cm^(2)` and `2 cm^(2)` respectively. Find the rate of flow of water across any section.
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Answer

A, C, D

Explanation

Applying Bernoulli's equation at `A` and `B`
`p_(A)+(1)/(2)rho v_(A)^(2)+rhogh_(A)=p_(B)+(1)/(2)rhov_(B)^(2)+rho gh_(B) (h_(A)=h_(B))`
`or, (1)/(2)rho v_(B)^(2)-(1)/(2)rho v_(A)^(2)=rhog (h_(A)=h_(B))`
or `v_(B)^(2)-v_(A)^(2)=2g (h_(A)=h_(B))`
`=2xx10xx0.02`
or `v_(B)^(2) -v_(A)^(2) = 0.4 m^(2)//s^(2)`....(i)
`v_(A)A_(A)=v_(B)A_(B)`
or, `4v_(A)=2v_(B)`
`:. v_(B)=2v_(A)` ...(ii)
Solving Eqs, (i) and (ii), we get
`v_(A)=0.363 m//s`
Volume Flow rate`=v_(A)A_(A)`
`=(0.365)(4xx10^(-4))`
`=1.46xx10^(-4) m^(3)//s=146cm^(3)//s`.