Bernoulli's Theorem NEET MCQ - NEET Practice Questions with Answers

 For a point in a fluid flow, Bernoulli's Theorem relates its pressure, its velocity, and its height from a reference point.

Bernoulli's Theorem states that the total energy (Pressure energy, Potential energy, and Kinetic energy ) per unit volume or mass of an incompressible and non-viscous fluid in steady flow through a pipe remains constant throughout the flow. (Provided that there is no source or sink of the fluid along the length of the pipe).

Mathematically for a liquid flowing through a pipe.

We can write Bernoulli's  equation as 

$$
P+\rho g h+\frac{1}{2} \rho v^2=\text { constant }
$$

$P \rightarrow$ Pressure energy per unit volume
$\rho g h \rightarrow$ Potential Energy per unit volume $\frac{1}{2} \rho v^2 \rightarrow$ Kinetic Energy per unit volume

Bernoulli's Theorem can be proved with the help of work-energy theorem.

Bernoulli's equation also represents the conservation of mechanical energy in the case of moving fluids.

• Bernoulli's theorem for the unit mass of liquid flowing through the pipe is given by

$$
\frac{P}{\rho}+g h+\frac{1}{2} v^2=\text { constant }
$$

If we divide the above equation by $g$ we get

$$
\frac{P}{\rho g}+h+\frac{v^2}{2 g}=\text { constant }
$$

Where
$\mathrm{h}=$ gravitational head
$\frac{P}{\rho g} \rightarrow$ Pressure head
$\frac{v^2}{2 g} \rightarrow$ velocity head

• The below figure

 With the help of Bernoulli's  equation 

We can write

$P_1+\rho g h_1+\frac{1}{2} \rho v_1^2=P_2+\rho g h_2+\frac{1}{2} \rho v_2^2=$ constant

The velocity of Efflux or Torricelli's Theorem-

If a liquid is filled in a vessel up to height H and a hole is made at a depth h below the free surface of the liquid as shown in fig.

Now take the level of the hole as the reference level (i.e., zero point of potential energy)

By applying Bernoulli's  equation we get

$v=\sqrt{2 g h}$

This v is called the Velocity of Efflux.

This formula is only valid when (Area of Hole) <<< (Area of the vessel)

Thus Torricelli's Theorem relates the speed of fluid flowing out of an orifice.

(Note-  The speed that an object would acquire in falling

from rest through a distance h is equal to $v=\sqrt{2 g h}$

And this is the same as that of Velocity of Efflux.)

• The velocity of efflux is independent of the nature of the liquid $\left(\rho_{)}\right.$, the quantity of liquid in the vessel, and the area of the orifice/hole.

• The velocity of efflux depends on h (i.e depth below the free surface)

            I.e Means the Greater the distance of the hole from the free surface of the liquid, the  greater will be the velocity of efflux

• The distance of the hole from the ground is (H-h) and its initial vertical velocity at the hole is zero.

So Time taken by liquid  to reach the Ground =t is given by

$T=\sqrt{\frac{2(H-h)}{g}}$

Where

H - the height of the vessel

And  h= depth below the free surface

• Range (x)-

During time t liquid is moving horizontally with constant velocity v,

And it will hit the base level at a horizontal distance x as shown in the above figure.

This horizontal distance x is also called a Horizontal range.

Using x=vt

We get Range as

$$
x=R=2 \sqrt{h(H-h)}
$$

This range will be maximum when $h=\frac{H}{2}$
And Maximum value of the range is H
Means $x_{\max }=R_{\max }=H$

Venturi meter-

• It is a device used for measuring the rate of flow of liquid through pipes.

• This device is based on the application of Bernoulli's theorem.

• The image of the Venturi Meter device is given below

For the above figure

a₁ and a₂ are an area of cross-section of tubes A and B respectively

And $P_1$ and $P_2$ are the Liquid pressure at A and B respectively
Then $P_1-P_2=\rho g h$
Where $\rho=$ density of flowing liquid
And $h=$ difference of fluid level between the vertical tube $D$ and $E$
Now applying Bernoulli's equation for the horizontal flow of liquid we get

$$
P_1+\frac{1}{2} \rho v_1^2=P_2+\frac{1}{2} \rho v_2^2 \cdots .
$$

If $\vee=$ rate of the flow of liquid through a pipe
Then from the continuity equation, we can write

$$
V=a_1 v_1=a_2 v_2 \ldots \ldots(3)
$$
 

From equation (1) and (2) and (3)

We get

V=r ate of the flow of liquid through a pipe

 As

$V=a_1 a_2 \sqrt{\frac{2 g h}{a_1^2-a_2^2}}$

• Aspirator pumps-

This works on the principle of Bernoulli's Theorem.

Example of Aspirator pumps is paint-gun, scent-spray or insect-sprayer, etc.

In such devices, high-speed air is passed over a tube T with the help of motion of a piston P in a cylinder C and this helps to spray the liquid L as shown in the above figure.

 The high-speed air creates low pressure in the tube and because of the low-pressure liquid rise in it. And thus liquid gets sprayed with expelled air.

• Change of plane of motion of spinning ball-

This can be with the help of  the principle of Bernoulli's Theorem

Magnus effect- When a spinning ball is thrown it deviates from its usual path in flight. This effect is called the Magnus effect. 

This effect plays a very important role in sports like cricket, tennis, and football, etc.

• Working of an aeroplane-

This is also based on  Bernoulli's principle.

• During  a tornado or hurricane, blowing off roofs by wind storms can be explained 

with the help of  the principle of Bernoulli's Theorem