Bernaullis theorem derivation 

$Ans . Proof : Let the velocity, pressure and area of a fluid column at a point X be v_1, p_{1}and A_1$

$and at another point Y be v_2, p_2 and A_2. Let the volume that is bounded by X and Y$

$be moved to M and N. let XM = L_1 and YN = L_2. Now if we can compress the fluid then we have,$

$A_1 \times L_1= A_2 \times L_2$

$We know that that the work done by the pressure difference per volume of the$

$unit is equal to the sum of the gain in kinetic energy and$

$gain in potential energy per volume of the unit$.

$This implies Work done = force \times dis\tan ce \Rightarrow Work done = p \times volume$

$Therefore, net work done per volume = p_1 – p_2$

$Also, kinetic energy per unit volume = \frac{1}{2} m{v}^2 = \frac{1}{2} \rho v^2$

$Therefore, we have,$

$Kinetic energy gained per volume of unit = \frac{1}{2} \rho \left({\left(v_2\right)}^2 - {\left(v_1\right)}^2\right)$

$And potential energy gained per volume of unit = pg (h_2 – h_1)$

$Here, h_1 and h_2 are heights of X and Y above the reference level taken in common.$

$Finally we have p_1 – p_2 = \frac{1}{2}\rho \left({\left(v_2\right)}^2-{\left(v_1\right)}^2\right) + \rho g (h_2 – h_1)$

$\Rightarrow p_1 + \frac{1}{2} \rho (v_1{)}^2 + \rho g{h}_1 = p_2 + \frac{1}{2}\rho (v_2{)}^2 + \rho g{h}_2$

$\Rightarrow p + \frac{1}{2}\rho v^2 + \rho gh is a cons\tan t When we have h_1 = h_2$

$Then we have, p + \frac{1}{2} \rho v^{2}is a cons\tan t. This proves the Bernoulli’s Theorem$

$Ans. According to Bernoulli’s theorem in physics, whenever there is an increase in the speed of the liquid, there is a simul\tan eous decrease in the potential energy of the fluid or we can say that there is a decrease in the pressure of the fluid. Basically, it is a principle of conservation of energy in the case of ideal fluids. If the fluid flows horizontally such that there is no change in the gravitational potential energy of the fluid then increase in velocity of the fluid results in a decrease in pressure of the fluid and vice versa.$