A non-viscous liquid of constant density `1000kg//m^3` flows in a streamline motion along a tube of variable cross section. The tube is kept inclined in the vertical plane as shown in Figure. The area of cross section of the tube two point P and Q at heights of 2 metres and 5 metres are respectively `4xx10^-3m^2` and `8xx10^-3m^2`. The velocity of the liquid at point P is `1m//s`. Find the work done per unit volume by the pressure and the gravity forces as the fluid flows from point P to Q.

Answer

A, B, C, D

Explanation

Given that
`rho=1000kg//m^3, h_1=2m, h_2=5m`
`A_1=4xx10^-3m^2, A_2=8xx10^-3m^2, v_1=1m//s`
Equation of continuity
`A_1v_1=A_2v_2` `:.v_2=(A_1v_1)/(A_2)=0.5m//s`
According to Bernoulli's theorem,
`(p_1-p_2)=rhog(h_2-h_1)-1/2rho(v_2^2-v_1^2)`
where `(p_1=p_2)` =work done/vol. [by the pressure]
`rhog(h_2-h_1)`=work done/vol. [by gravity forces]
Now, work done/vol. by gravity forces
`rhog(h_2-h_1)=10^3xx9.8xx3=29.4xx10^3J//m^3`.
And `1/2rho(v_2^2-v_1^2)=1/2xx10^3[1/4-1]=-3/8xx10^3J//m^3`
`=-0.375xx10^3J//m^3`
`:.` Work done/vol. by pressure
`=29.4xx10^3-0.375xx10^3J//m^3=29.025xx10^3J//m^3`.