Solution

Bernoulli's Theorem:
According to Bernoulli's theorem, the sum of the energies possessed by a flowing ideal liquid at a point is constant provided that the liquid is incompressible and non-viseous and flow in streamline.
Potential energy + Kinetic energy + Pressure energy = Constant
$$P + \frac{1} {2}pv^{2} + pgh = Constant$$
$$gh + \frac{1} {2}v^{2} + \frac{P} {p} = C$$ ............(11.11)
Where C is a constant.
This relation is called Bernoulli's theorem.
Dividing eqn. (11.11) by g, we get
$$h + \frac{P} {pg} + \frac{1} {2} \frac{v^{2}} {g} = C'$$ ............(11.12)
Where C is another constant.
For horizontal flow, h remains same throughout.
So,
$$ \frac{P} {pg} + \frac{v^{2}} {2g} = Constant$$
or; $$ P + \frac{1} {2} pv^{2} = Constant$$
P is static pressure of the liquid and $$\frac{1} {2} pv^{2}$$ is its dynamic and velocity pressure.
Thus, for horizontal motion, the sum of static and dynamic pressure is constant. If $$p_1v_1$$ and $$p_2v_2$$ represent pressure and velocities at two points. Then
$$ P_1 + \frac{1} {2} pv^{2}_1 = P_2 + \frac{1} {2} pv^{2}_2 $$
Concepts : 1. In Bernoulli's theorem $$ P + pgh + \frac{1} {2} ev^{2}$$
= constant. The term $$\left(p + pgh \right)$$ is called static pressure and the term $$\frac{1} {2} pv^{2} $$ is the dynamic pressure of the fluid.
2. Bernoulli theorem is fundamental principle of the energy.
3. The equation $$\frac{P} {pg} + \frac{1} {2} \frac{v^{2}} {g} + h = constant$$
the term $$\frac{P} {pg}$$ = pressure head
the term $$\frac{v^{2}} {2g}$$ = velocity head
h = gravitational head.
Derivation of Bernoulli's Theorem :
The energies possessed by a flowing liquid are mutually convertible. When one type of energy increases, the other type of energy decreases and vice-versa. Now, we will derive the Bernoulli's theorem using the work-energy theorem.
Consider the flow of liquid. Let at any time, the liquid lies between two areas of flowing liquid $$A_1$$ and $$A_2$$. In time interval $$\Delta t$$, the liquid displaces from $$A_1$$ by $$\Delta x_1 = v_1\Delta t$$ and displaces from $$A_2$$ by $$\Delta x_2 = v_2\Delta t$$. Here $$v_1$$ and $$v_2$$ are the velocities of the liquid at $$A_1$$ and $$A_2$$.

The work done on the liquid is $$P_1 A_1 \Delta x_1$$ by the force and $$P_2 A_2 \Delta x_2$$ against the force respectively.
Net work done,
$$W = P_1 A_1 \Delta x_1 - P_2 A_2 \Delta x_2$$
$$\Rightarrow W = P_1 A_1 v_1 \Delta t - P_2 A_2 v_2 \Delta t$$
$$= \left (P_1 - P_2 \right) \Delta V$$ .........(11.13)
Here, $$\Delta V \rightarrow $$ the volume of liquid that flows through a cross-section is same (from equation of continuity).
But, the work done is equal to net change in energy (K.E. + P.E.) of the liquid, and
$$\Delta K = \frac{1} {2} p \Delta V \left (v^{2}_1 - v^{2}_2 \right)$$ ........ (11.14)
and $$\Delta U = p \Delta V \left (h_2 - h_1 \right)$$ ........ (11.15)
$$\therefore \left (P_1 - P_2 \right) \Delta V = \frac{1} {2} p \Delta V \left (v^{2}_1 - v^{2}_2 \right) + pg \Delta V \left (h_2 - h_1 \right)$$
$$P_1 + \frac{1} {2} v^{2}_1 + pgh_1 = P_2 + \frac{1} {2} v^{2}_2 + pgh_2$$
or $$ P + \frac{1} {2} pv^{2} + pgh = constant$$ .......... (11.16)
This is the required relation for Bernoulli's theorem.
$$\therefore A_1 v_1 = A_2 v_2$$
So, more the cross-sectional area, lesser is the velocity and vice-versa.
So, Bernoulli's theorem,
$$ P_1 + \frac{1} {2} p_1 v^{2}_1 = P_2 + \frac{1} {2} p_2 v^{2}_2$$