State and prove Bernoulli's theorem.

Explanation

1st part: N/A
2nd part: Let an incompressible, non-viscous liquid entering into the cross -sectional area `A_1` at point A with velocity `V_1` and coming out a height `h_2` at point B with velocity `v_2` [Fig. 3.52].
The value of the potential energy (PE) and kinetic energy (KE) are increased , since `h_2 gt h_1 and v_2 gt v_1` . This is done by the pressure at the time of work done on the liquid. If `p_1 and p_2` are the pressure at A and B, for a small displacement at A and B, the work done on the liquid at A.
`Deltax_1=p_1 A_1v_1 Delta t`
So, net work done by pressure
`=(p_1-p_2)A_0v_0 Delta t` [ where `A_1 v_1=A_2v_2=A_0v_0]`
From conservation of energy,
`(p_1-p_2)A_0 v_0 Delta t =` change in KE + change in PE
or, `(p_1 -p_2) A_0v_0 Delta t =1/2A_0v_0 Delta t rho (v_2^2-v_1^2)+A_0 v_0 rho Delta tg (h_2-h_1)`
or, `p_1-p_2=rhog(h_2-h_1)+(rho)/(2)(v_2^2-v_1^2)`
or, `p_1+rho gh_1+(rho)/(2)v_1^2=p_2+rho gh_2+(rho)/(2)v_2^2`
`therefore (p)/(rhog)+h+(v^2)/(2g) =` constant