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 $\times$ d = p $\times$ volume
$\Rightarrow$ W_net = p₁ - p₂
$\Rightarrow$ K.E = $\frac{1}{2}$mv² = $\frac{1}{2}$V $\rho$v² = $\frac{1}{2}$$\rho$v² ($\because$ V = 1)
$\Rightarrow$ K.E_gained = $\frac{1}{2} \rho\left(v_{2}^{2}-v_{1}^{2}\right)$
P₁ + $\frac{1}{2} \rho v_{1}^{2}$ + $\rho$gh₁ = P₂ + $\frac{1}{2} \rho v_{2}^{2}$ + $\rho$gh₂
$\therefore$ P + $\frac{1}{2} \rho v^{2}$ + $\rho$gh = const.
For a horizontal tube
$\because$ h₁ = h₂
$\therefore$ P + $\frac{1}{2} \rho v^{2}$ = 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.