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Bernoulli's Theorem MCQ - Practice Questions with Answers

Edited By admin | Updated on Sep 25, 2023 25:23 PM | #NEET

Quick Facts

  • Applications of Bernoulli's Theorem(I) is considered one of the most asked concept.

  • 57 Questions around this concept.

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Water flows steadily through a horizontal pipe of a variable cross-section. If the pressure of water is p at a point where the velocity of flow is v, what is the pressure at another point where the velocity of flow is \mathrm{2 v ; \rho} the density of water?

A liquid is kept in a cylindrical vessel. When the vessel is rotated about its axis, the liquid rises at its sides. If the radius of the vessel is 0.05 \mathrm{~m} and the frequency of rotation is 2 revolutions per second, the difference in the heights of the liquid at the centre and at the sides of the vessel will be \mathrm{(take \, \, g=10 \mathrm{~ms}^{-2}\, \, and \, \, \pi^2=10 )}

Water flows through a frictionless duct with a cross-section varying as shown in figure. Pressure p at points along the axis is represented by

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A cyclindrical vessel of cross-sectional area 1000 \, \, \mathrm{cm}^2, is fitted with a frictionless piston of mass 10 \mathrm{~kg}, and filled with water completely. A small hole of crosssectional area 10 \mathrm{~mm}^2 is opened at a point 50 \mathrm{~cm} deep from the lower surface of the piston. The velocity of efflux from the hole will be

A horizontal pipeline carries water in a streamlined flow. At a point along the tube where the cross-sectional area is 10^{-2} \mathrm{~m}^2, the water velocity is 2 \mathrm{~ms}^{-1} and the pressure is 8000 \mathrm{~Pa}. The pressure of water at another point where the cross-sectional area is 0.5 \times 10^{-2} \mathrm{~m}^2  is :

A liquid of density \rho is coming out of a hose pipe of radius a with horizontal speed v and hits a mesh. 50% of the liquid passes through the mesh unaffected. 25% looses all of its momentum and 25% comes back with the same speed. The resultant pressure (in \rhov2) on the mesh will be:

Fountains usually seen in gardens are generated by a wide pipe with an enclosure at one end having many small holes. Consider one such fountain which is produced by a pipe of internal diameter \mathrm{2 \mathrm{~cm}} in which water flows at a rate 3 \mathrm{~ms}^{-1}.  The enclosure has 100 holes each of diameter \mathrm{0.05 \mathrm{~cm}}. The velocity of water coming out of the holes is \mathrm{(in \, \, \mathrm{ms}^{-1} ):}

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A water barrel stands on a table of height h. If a small hole is punched in the side of the barrel at its base, it is found that the resultant stream of water strikes the ground at a horizontal distance R from the barrel. The depth of water in the barrel is

A non-viscous liquid of constant density $10^3 \mathrm{~kg} / \mathrm{m}^3$ flows in streamlined motion along a vertical tube PQ of variable cross-section. The heights of P and Q are 2 m and 2.5 m respectively. The area of the tube at Q is equal to 3 times the area of the tube at P. Find the work done per unit volume by pressure as the liquid flows from $P$ to $Q$. Speed of the liquid at $P$ is $3 \mathrm{~m} / \mathrm{s}\left(\mathrm{g}=10 \mathrm{~m} / \mathrm{s}^2\right)$

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There are two identical small holes of area of cross-section a on the opposite sides of a tank containing a liquid of density p. The difference in height between the holes is h. Tank is resting on a smooth horizontal surface. Horizontal force which will have to be applied on the tank to keep it in equilibrium is:

Concepts Covered - 4

Bernoulli's Theorem

 For a point in a fluid flow, Bernoulli's Theorem relates between 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 gh+\frac{1}{2}\rho v^{2}=constant

P\rightarrow Pressure \: energy \ per \ unit \ volume

\rho gh\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 case of moving fluids.

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

\frac{P}{\rho }+gh+\frac{1}{2}v^{2}= constant

If we divide the above equation by g we get

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

Where

  h=gravitational head

\frac{P}{\rho g}\rightarrow Pressure\: head

\frac{v^{2}}{2 g}\rightarrow velocity \: head


 

  • For the below figure

 

 With the help of Bernoulli's  equation 

We can write

P_1+\rho gh_1+\frac{1}{2}\rho v_1^{2}=P_2+\rho gh_2+\frac{1}{2}\rho v_2^{2}=constant

Applications of Bernoulli's Theorem(I)

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 reference level (i.e., zero point of potential energy)

And by applying Bernoulli's  equation we get

v=\sqrt{2gh}

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{2gh}

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

  • The velocity of efflux is independent of the nature of liquid (\rho), 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 Greater is the distance of the hole from the free surface of the liquid, greater will be the velocity of efflux

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

Applications of Bernoulli's Theorem(II)

Venturimeter-

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

  • This device based on application 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 tube A and B respectively

And v1 and v2 are the Velocities of the flow of liquid through 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 gh ….(1)

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 .....(2)

If V=rate of the flow of liquid through a pipe

Then from the continuity equation, we can write

V=a_1v_1=a_2v_2...... (3)

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

We get

V=rate of the flow of liquid through pipe

 As

V= a_1a_2\sqrt{\frac{2gh}{a_1^2-a_2^2}}

Applications of Bernoulli's Theorem(III)
  • 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

 

Study it with Videos

Bernoulli's Theorem
Applications of Bernoulli's Theorem(I)
Applications of Bernoulli's Theorem(II)

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Books

Reference Books

Bernoulli's Theorem

Physics Part II Textbook for Class XI

Page No. : 258

Line : 58

Applications of Bernoulli's Theorem(I)

Physics Part II Textbook for Class XI

Page No. : 259

Line : 50

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