I E Irodov Physical Fundamentals Of Mechanics (Hydrodynamics) Q. 1.335 Solution

I E Irodov Physical Fundamentals of Mechanical Engineering is a very popular advanced book for Physics for students preparing for JEE Advanced, Physics Olympiads and other competitive exams.

The Hydrodynamics section includes fluid pressure, viscosity, Bernoulli’s equation, pressure-head loss and flow of liquids through pipes. These problems provide students with the opportunity to relate fluid-energy concepts to viscosity effects.

Here in this page, you will get a step by step solution to the I E Irodov Physical Fundamentals of Mechanics (Hydrodynamics) Question 1.335 along with the concepts and formulae used.

I E Irodov Hydrodynamics Question 1.335

Question:

In the arrangement shown in the figure, a viscous liquid of density

[
\rho=1.0\text{ g/cm}^3
]

flows through a tube out of a wide tank (A).

Find the velocity of the liquid flow if:

[
h_1=10\text{ cm}
]

[
h_2=20\text{ cm}
]

[
h_3=35\text{ cm}
]

All the distances (l) shown in the arrangement are equal.

Concepts Used in I E Irodov Q. 1.335

This problem is based on the flow of a viscous liquid through a horizontal tube.

Important concepts include:

• Bernoulli’s equation
• Pressure head
• Viscous energy loss
• Steady liquid flow
• Equal pipe-length sections
• Conversion of pressure energy into kinetic energy
• Atmospheric pressure at the outlet

In a real viscous liquid mechanical energy is lost when the liquid flows through the tube. This is a loss which we obtain from the pressure-head readings.

Easy Solution of I E Irodov Hydrodynamics Q. 1.335

Given:

[
h_1=10\text{ cm}
]

[
h_2=20\text{ cm}
]

[
h_3=35\text{ cm}
]

Density of liquid:

[
\rho=1.0\text{ g/cm}^3
]

All the marked distances along the tube are equal to (l).

We need to find the velocity (v) of the liquid leaving the tube.

Step 1: Understand the Pressure-Head Readings

The vertical liquid columns represent the pressure head at various points of the arrangement.

The section of the horizontal tube is constant and consequently the velocity of the liquid is constant in different points of the tube.

But the liquid is viscous, so its pressure drops as it flows along the pipe.

The pressure-head loss over one length (l) is found from:

[
h_2-h_1
]

Substituting the given values:

[
h_2-h_1=20-10
]

[
h_2-h_1=10\text{ cm}
]

Therefore, the viscous loss of pressure head over each equal length (l) is:

[
10\text{ cm}
]

Step 2: Calculate the Available Pressure Head

The total pressure-head difference represented by (h_3) and (h_2) is:

[
h_3-h_2
]

Substituting:

[
h_3-h_2=35-20
]

[
h_3-h_2=15\text{ cm}
]

Out of this (15\text{ cm}), a pressure head of (10\text{ cm}) is used to overcome viscous resistance over the equal pipe length.

Therefore, the remaining pressure head is:

[
\Delta h=(h_3-h_2)-(h_2-h_1)
]

Substitute the values:

[
\Delta h=(35-20)-(20-10)
]

[
\Delta h=15-10
]

[
\Delta h=5\text{ cm}
]

Convert into metres:

[
\Delta h=0.05\text{ m}
]

Step 3: Convert the Remaining Head into Kinetic Energy

The remaining pressure head is converted into the kinetic energy of the liquid at the outlet.

Using Bernoulli’s energy relation:

[
\rho g\Delta h=\frac12\rho v^2
]

The density (\rho) appears on both sides and cancels:

[
g\Delta h=\frac12v^2
]

Therefore:

[
v=\sqrt{2g\Delta h}
]

Step 4: Substitute the Values

Using:

[
g=9.8\text{ m/s}^2
]

and:

[
\Delta h=0.05\text{ m}
]

we get:

[
v=\sqrt{2(9.8)(0.05)}
]

[
v=\sqrt{0.98}
]

[
v\approx0.99\text{ m/s}
]

Therefore:

[
v\approx1.0\text{ m/s}
]

Final Answer

The velocity of the liquid flowing out of the tube is:

[
\boxed{v\approx1.0\text{ m/s}}
]

Simple Explanation of the Answer

The frictional loss of some of the pressure energy of the liquid flowing through a viscous tube.

The pressure-head readings indicate that the liquid loses:

[
10\text{ cm}
]

of head over each equal tube length.

The total available head difference is:

[
15\text{ cm}
]

After subtracting the (10\text{ cm}) viscous loss, only:

[
5\text{ cm}
]

of pressure head remains.

This remaining energy becomes the kinetic energy of the liquid at the outlet, producing a flow velocity of approximately:

[
1.0\text{ m/s}
]

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Key Takeaways

Common Mistakes Students Should Avoid

Students commonly make these mistakes:

• Applying the ideal Bernoulli equation without considering viscosity
• Using the entire (15\text{ cm}) head as kinetic head
• Forgetting to subtract the pressure loss due to viscosity
• Confusing pressure-head readings with actual tube heights
• Forgetting to convert centimetres into metres
• Keeping density in the final equation even though it cancels
• Using (h_3-h_1) directly without accounting for the equal pipe sections

Always identify the viscous head loss before calculating the outlet velocity.

Why Practice I E Irodov Hydrodynamics Problems?

I E Irodov Hydrodynamics problems help students understand real fluid flow, where viscosity causes energy loss.

Benefits include:

• Better understanding of Bernoulli’s equation
• Clear knowledge of pressure head
• Improved understanding of viscous losses
• Better application of energy conservation
• Stronger analytical thinking
• Preparation for advanced Fluid Mechanics questions

Students should solve selected I E Irodov questions after completing standard JEE-level Fluid Mechanics concepts.

Is I E Irodov Q. 1.335 a JEE Main Question?

I E Irodov Hydrodynamics Q. 1.335 is more advanced than a typical JEE Main problem.

The question requires students to interpret a figure and combine:

• Pressure-head readings
• Viscous energy loss
• Equal pipe lengths
• Bernoulli’s equation
• Outlet kinetic energy

It is more suitable for:

• JEE Advanced aspirants
• Physics Olympiad students
• Advanced Fluid Mechanics practice
• Students familiar with Bernoulli’s equation

For JEE Main, students should first focus on NCERT concepts, previous year questions and standard numerical problems.

Is Hydrodynamics Important for JEE?

Fluid Mechanics constitutes an important chapter in JEE Physics but students should not assume that there will be a fixed number of questions every year.

Important topics in fluid mechanics include:

• Fluid pressure
• Archimedes’ principle
• Buoyant force
• Continuity equation
• Bernoulli’s theorem
• Viscosity
• Pressure-head loss
• Reynolds number
• Stokes’ law
• Terminal velocity
• Surface tension

The number and difficulty of questions may vary between exam sessions.

Important Topics in I E Irodov Physical Fundamentals of Mechanics

How Should Students Use I E Irodov for JEE Advanced?

Recommended preparation sequence:

I E Irodov should be used as an advanced practice resource after students have developed strong fundamentals.

I E Irodov for NRI JEE Students

Selected I E Irodov Hydrodynamics Problems for Advanced Physics Practice for NRI Students in the USA, UAE, Saudi Arabia, Qatar, Oman, Kuwait and Bahrain Preparing for JEE

Students studying AP, IB, British, American or any other international curriculum should first cross-check their school syllabus with the JEE Physics syllabus.

A useful preparation sequence is:

• Complete JEE Fluid Mechanics concepts
• Solve chapter-wise JEE questions
• Practice JEE previous year papers
• Attempt JEE Advanced numerical problems
• Solve selected I E Irodov questions
• Review difficult concepts with a teacher or mentor

This method helps NRI students use advanced reference books efficiently.

Related I E Irodov Hydrodynamics Questions

After completing Question 1.335, students can also practise:

• I E Irodov Hydrodynamics Q. 1.324
• I E Irodov Hydrodynamics Q. 1.328
• I E Irodov Hydrodynamics Q. 1.336
• I E Irodov Hydrodynamics Q. 1.339

Each question should link to its corresponding TestPrepKart solution page.

Practice Next

After solving this question, students can continue with:

• Bernoulli’s equation problems
• Viscous pressure-loss questions
• Continuity equation questions
• Reynolds number numerical problems
• JEE Advanced Fluid Mechanics previous year papers
• Physics revision notes
• JEE Advanced mock tests

Useful JEE, DASA & CIWG Resources for I E Irodov Physics

FAQs on I E Irodov Hydrodynamics Q. 1.335

1. What is the answer to I E Irodov Q. 1.335?

The velocity of the liquid flow is approximately:

[
\boxed{1.0\text{ m/s}}
]

2. Which concepts are used in I E Irodov Q. 1.335?

The problem uses pressure head, viscous energy loss and Bernoulli’s equation.

3. What is the viscous head loss over one pipe length?

The head loss over one equal pipe length is:

[
h_2-h_1=20-10=10\text{ cm}
]

4. What pressure head is converted into kinetic energy?

The available kinetic head is:

[
(h_3-h_2)-(h_2-h_1)
]

[
=15-10=5\text{ cm}
]

5. Why does the liquid density cancel?

Both pressure energy and kinetic energy are proportional to the liquid density:

[
\rho g\Delta h=\frac12\rho v^2
]

Therefore, (\rho) cancels from both sides.

6. Why cannot the entire pressure head be converted into velocity?

Part of the pressure energy is lost in overcoming viscous resistance while the liquid flows through the tube.

7. Is I E Irodov Q. 1.335 useful for JEE Main?

It is more advanced than a typical JEE Main question and is better suited to JEE Advanced Fluid Mechanics practice.