State and prove bernoullis theorm.

Bernoulli's principle states that an increase in the speed of a fluid occurs simultaneously with a decrease in static pressure or a decrease in the fluid's potential energy.
To prove Bernoulli's theorem, consider a fluid of negligible viscosity moving with laminar flow. as shown in Figure.
Let the velocity, pressure and area of the fluid cloumn be p₁, v₁ and A₁ at Q and p₂, v₂ and A₂ at R. Let the  volume bounded by Q and R move to S and T where QS = L₁, and RT = L₂

If the fluid is incompressible:
The work done by the pressure difference per unit volume = gain in kinetic energy per unit volume + gain in potential energy per unit volume. Now:
A₁L₁ = A₂L₂
Work done is given by:
W = F {tex}\times{/tex} d = p {tex}\times{/tex} volume
{tex}\Rightarrow{/tex} W_net = p₁ - p₂
{tex}\Rightarrow{/tex} K.E = {tex}\frac{1}{2}{/tex}mv² = {tex}\frac{1}{2}{/tex}V {tex}\rho{/tex}v² = {tex}\frac{1}{2}{/tex}{tex}\rho{/tex}v² ({tex}\because{/tex} V = 1)
{tex}\Rightarrow{/tex} K.E_gained = {tex}\frac{1}{2} \rho\left(v_{2}^{2}-v_{1}^{2}\right){/tex}
P₁ + {tex}\frac{1}{2} \rho v_{1}^{2}{/tex} + {tex}\rho{/tex}gh₁ = P₂ + {tex}\frac{1}{2} \rho v_{2}^{2}{/tex} + {tex}\rho{/tex}gh₂
{tex}\therefore{/tex} P + {tex}\frac{1}{2} \rho v^{2}{/tex} + {tex}\rho{/tex}gh = const.
For a horizontal tube
{tex}\because{/tex} h₁ = h₂
{tex}\therefore{/tex} P + {tex}\frac{1}{2} \rho v^{2}{/tex} = const.
Therefore, this proves Bernoulli's theorem. Here we can see that if there is an increase in velocity there must be a decrease in pressure and vice versa.