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**Hint:**Here, we first know what Bernoulli's principle states, and then further we can prove the theorem by deriving the expression which relates the pressure and potential energy of the fluid. At last we will also note down the limitations of this theorem.

**Formula used:**

$\eqalign{

& K.{E_{gained}} = \dfrac{1}{2}\rho ({v_2}^2 - {v_1}^2) \cr

& P.{E_{gained}} = \rho g({h_2} - {h_1}) \cr} $

**Complete answer:**

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 column be ${p_1}$, ${v_1}$ and ${A_1}$ at Q and ${p_2}$, ${v_2}$ and ${A_2}$ at R. Let the volume bounded by Q and R move to S and T where QS =${L_1}$, and RT = ${L_2}$.

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:

${A_1}{L_1} = {A_2}{L_2}$

Work done is given by:

$\eqalign{& W = F \times d = p \times volume \cr

& \Rightarrow {W_{net}} = {p_1} - {p_2} \cr} $

$\eqalign{& \Rightarrow K.E = \dfrac{1}{2}m{v^2} = \dfrac{1}{2}V\rho {v^2} = \dfrac{1}{2}\rho {v^2}(\because V = 1) \cr

& \Rightarrow K.{E_{gained}} = \dfrac{1}{2}\rho ({v_2}^2 - {v_1}^2) \cr} $

$\eqalign{& {P_1} + \dfrac{1}{2}\rho {v_1}^2 + \rho g{h_1} = {P_2} + \dfrac{1}{2}\rho {v_2}^2 + \rho g{h_2} \cr

& \therefore P + \dfrac{1}{2}\rho {v^2} + \rho gh = const. \cr} $

For a horizontal tube

$\eqalign{& \because {h_1} = {h_2} \cr

& \therefore P + \dfrac{1}{2}\rho {v^2} = const. \cr} $

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.

**Additional information:**

Bernoulli's principle is named after Daniel Bernoulli who published this in his book Hydrodynamica in 1738.

Pressure applied on an object is given by the force exerted on the object per unit area. The S.I unit of pressure is Pascal.

Work done on an object is defined as the force applied on the object for a certain displacement. Further, if we define a volume, it is the quantity of three-dimensional space enclosed by a closed surface.

**Note:**

We should remember that no fluid is totally incompressible whereas in practice the general qualitative assumptions still hold for real fluids. One should also notice that in Bernoulli's theorem, it is given that the velocity of every particle of liquid across any cross-section is uniform which is not correct, because the velocity of the particles is different in different layers.

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