I E Irodov Physical Fundamentals Of Mechanics (Hydrodynamics) Q. 1.328 Solution

I E Irodov Physical Fundamentals of Mechanics, is a popular advanced Physics book for students preparing for JEE Advanced, Physics Olympiads and other competitive examinations.

Hydrodynamics is the chapter on fluid flow, momentum transfer, reaction forces, pressure, viscosity, Bernoulli’s equation and the movement of liquids through pipes. These problems assist learners in comprehending the forces and turning effects fluids exert on pipes and other surfaces.

This page contains a simple step by step solution of I E Irodov Physical Fundamentals of Mechanics (Hydrodynamics) Question 1.328 with concepts and formulas.

I E Irodov Hydrodynamics Question 1.328

Question:

Water flows out of a large tank through a tube bent at right angles.

The inside radius of the tube is:

[
r=0.50\text{ cm}
]

The length of the horizontal section of the tube is:

[
l=22\text{ cm}
]

The volume flow rate of water is:

[
Q=0.50\text{ litres per second}
]

Find the moment of the reaction forces exerted by the flowing water on the walls of the tube about point (O).

Concepts Used in I E Irodov Q. 1.328

This problem is based on momentum transfer when water changes its direction inside a bent tube.

Important concepts include:

• Volume flow rate
• Fluid velocity
• Mass flow rate
• Momentum flux
• Impulse–momentum theorem
• Reaction force on a pipe
• Moment of force
• Newton’s third law

When the water reaches the bend, it turns (90^\circ). The tube must exert a force to change the momentum. Newton’s third law states that the flowing water exerts an equal and opposite reaction force on the tube.

Easy Solution of I E Irodov Hydrodynamics Q. 1.328

Given:

Inside radius of the tube:

[
r=0.50\text{ cm}
]

Convert into metres:

[
r=0.0050\text{ m}
]

Length of the horizontal section:

[
l=22\text{ cm}
]

[
l=0.22\text{ m}
]

Water flow rate:

[
Q=0.50\text{ L/s}
]

Since:

[
1\text{ L}=10^{-3}\text{ m}^3
]

therefore:

[
Q=0.50\times10^{-3}\text{ m}^3/\text{s}
]

[
Q=5.0\times10^{-4}\text{ m}^3/\text{s}
]

Density of water:

[
\rho=1000\text{ kg/m}^3
]

We need to calculate the moment of the reaction force about point (O).

Step 1: Calculate the Cross-Sectional Area of the Tube

The cross-sectional area of the tube is:

[
A=\pi r^2
]

Substitute the radius:

[
A=\pi(0.0050)^2
]

[
A=\pi(2.5\times10^{-5})
]

[
A\approx7.85\times10^{-5}\text{ m}^2
]

Step 2: Calculate the Velocity of Water

The volume flow rate is related to the cross-sectional area and flow velocity by:

[
Q=Au
]

where (u) is the velocity of water inside the tube.

Therefore:

[
u=\frac{Q}{A}
]

Since:

[
A=\pi r^2
]

we can write:

[
u=\frac{Q}{\pi r^2}
]

Substituting the values:

[
u=
\frac{5.0\times10^{-4}}
{\pi(0.0050)^2}
]

[
u\approx6.37\text{ m/s}
]

Therefore, the water flows through the tube with a speed of approximately:

[
\boxed{u\approx6.37\text{ m/s}}
]

Step 3: Calculate the Mass Flow Rate

The mass of water flowing through the tube per second is:

[
\dot{m}=\rho Q
]

Substituting:

[
\dot{m}=1000(5.0\times10^{-4})
]

[
\dot{m}=0.50\text{ kg/s}
]

Step 4: Determine the Force Caused by the Change in Momentum

At the right-angle bend, the direction of the water velocity changes.

The magnitude of each momentum-force component is:

[
F=\dot{m}u
]

Since:

[
\dot{m}=\rho Q
]

we get:

[
F=\rho Qu
]

Substituting the expression for (u):

[
F=
\rho Q\left(\frac{Q}{\pi r^2}\right)
]

Therefore:

[
F=\frac{\rho Q^2}{\pi r^2}
]

Substitute the values:

[
F=
\frac{
1000(5.0\times10^{-4})^2
}{
\pi(0.0050)^2
}
]

[
F\approx3.18\text{ N}
]

This is the magnitude of the relevant force component exerted because of the change in water momentum.

Step 5: Calculate the Moment About Point O

The moment of force is:

[
N=F\times l
]

The force component producing the turning effect acts at a perpendicular distance (l) from point (O).

Therefore:

[
N=
\frac{\rho Q^2}{\pi r^2}l
]

or:

[
N=
\frac{\rho Q^2l}{\pi r^2}
]

Substituting the values:

[
N=
\frac{
1000(5.0\times10^{-4})^2(0.22)
}{
\pi(0.0050)^2
}
]

[
N\approx0.70\text{ N·m}
]

Final Answer

The moment of the reaction forces of the flowing water on the tube walls about point (O) is:

[
\boxed{
N=\frac{\rho Q^2l}{\pi r^2}
}
]

For the given values:

[
\boxed{N\approx0.70\text{ N·m}}
]

Therefore, the required moment of force is:

0.70 N·m

Simple Explanation of the Answer

Water enters the horizontal part of the tube . The water direction changes at the bend (90^\\circ).

Changing the direction of the water changes its momentum. The tube must exert a force on the water to effect this change.

Newton’s third law states that the water pushes back on the tube with an equal and opposite force.

This reaction force is at a distance (22\text{ cm}) from (O) and will cause a turning moment about (O).

The moment of result is:

[
0.70\text{ N·m}
]

Direction of the Moment

The precise clockwise or anti-clockwise direction is dependent on the orientation of the original figure.

The size of the moment is:

[
\boxed{0.70\text{ N·m}}
]

Students should use the direction of the outlet flow shown in the diagram to assign the correct sign to the moment.

Download Free JEE E-Book for DASA Students

Planning DASA counselling through JEE Main for NIT, IIIT, SPA or engineering admission in India? Download the free TestPrepKart JEE E-Book and understand the right preparation strategy, syllabus planning, mock test approach and admission roadmap for NRI students.

Download JEE E-BookExplore JEE Courses

Key Takeaways

Common Mistakes Students Should Avoid

Students commonly make these mistakes:

• Using the diameter instead of the radius
• Forgetting to convert centimetres into metres
• Forgetting to convert litres per second into cubic metres per second
• Using (Q) directly as velocity
• Ignoring the change in momentum at the bend
• Forgetting that the water exerts a reaction force opposite to the force exerted by the tube
• Calculating force but not multiplying by the perpendicular distance
• Using the resultant force magnitude instead of the force component that produces the moment about (O)
• Forgetting that moment is measured in newton-metres

Always draw the inlet and outlet velocity directions before applying the momentum equation.

Why Practice I E Irodov Hydrodynamics Problems?

I E Irodov Hydrodynamics problems assist learners in comprehending the principles of mechanics applied to flowing liquids.

Benefits are:

• Better understanding of momentum flux
• Improved application of Newton’s laws to fluids
• Stronger knowledge of reaction forces
• Better understanding of moments and torque
• Improved analytical thinking
• Stronger preparation for advanced Fluid Mechanics questions

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

Is I E Irodov Q. 1.328 a JEE Main Question?

I E Irodov Hydrodynamics Q. 1.328 is more advanced than a typical JEE Main question.

The problem combines:

• Volume flow rate
• Fluid velocity
• Mass flow rate
• Momentum change
• Reaction force
• Moment of force

It is more suitable for:

• JEE Advanced aspirants
• Physics Olympiad students
• Advanced Fluid Mechanics practice
• Students comfortable with momentum-flux problems

For JEE main, students should first concentrate on NCERT concepts, standard numerical problems, previous year questions and mock tests.

Is Hydrodynamics Important for JEE?

Fluid Mechanics constitutes a big topic for Physics preparation. However, students should not assume that the same number of questions will be asked each year.

Important topics in Fluid Mechanics are:Fluid pressure

• Buoyant force
• Archimedes’ principle
• Continuity equation
• Bernoulli’s theorem
• Momentum transfer in fluids
• Viscosity
• Reynolds number
• Stokes’ law
• Terminal velocity
• Surface tension

The number and difficulty of questions may vary between examination 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 problem-solving resource after students have developed strong Physics fundamentals.

I E Irodov for NRI JEE Students

NRI students preparing for JEE from USA, UAE, Saudi Arabia, Qatar, Oman, Kuwait and Bahrain can use Selected I E Irodov Hydrodynamics problems for advanced Physics practice.

Students studying in AP, IB, British, American or any international curriculum should check their school syllabus with 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 without spending excessive time on topics before building the required fundamentals.

Related I E Irodov Hydrodynamics Questions

After completing Question 1.328, students can also practise:

• I E Irodov Hydrodynamics Q. 1.324
• I E Irodov Hydrodynamics Q. 1.335
• 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:

• Fluid momentum numerical problems
• Reaction force on pipe bends
• Continuity equation questions
• Bernoulli’s theorem questions
• Viscosity and Reynolds number 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. 1.328

1. What is the answer to I E Irodov Q. 1.328?

The moment of the reaction forces about point (O) is:

[
\boxed{0.70\text{ N·m}}
]

2. Which concepts are used in I E Irodov Q. 1.328?

The problem uses volume flow rate, momentum flux, reaction force, Newton’s third law and moment of force.

3. How is the velocity of water calculated?

The water velocity is calculated using:

[
Q=Au
]

Therefore:

[
u=\frac{Q}{\pi r^2}
]

For the given values:

[
u\approx6.37\text{ m/s}
]

4. Why does the bent tube experience a reaction force?

The water changes direction at the bend. A force is required to change its momentum, and the water exerts an equal and opposite reaction force on the tube.

5. What is the force component used to calculate the moment?

The relevant force component has magnitude:

[
F=\rho Qu
]

or:

[
F=\frac{\rho Q^2}{\pi r^2}
]

For the given values:

[
F\approx3.18\text{ N}
]

6. Why is the length (l) used in the moment calculation?

The force acts at a perpendicular distance (l) from point (O). Therefore:

[
N=Fl
]

7. Is I E Irodov Q. 1.328 useful for JEE Main?

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