I E Irodov Physical Fundamentals Of Mechanics (Hydrodynamics) Q. 1.339

I E Irodov Physical Fundamentals of Mechanics has become a very popular advanced Physics book for the students preparing for JEE Advanced, Physics Olympiads and other competitive exams.

In the Hydrodynamics chapter we study the motion of fluids, viscosity, Stokes’ law, terminal velocity, buoyancy and the motion of objects through liquids. These problems will help students to build up their understanding of fluid mechanics and differential equations.

This page contains a simple step by step solution of I E Irodov Physical Fundamentals of Mechanics (Hydrodynamics) Problem 1.339. Along with the concepts and formulas used in the solution.

I E Irodov Hydrodynamics Question 1.339

Question:

A steel ball of diameter

[
d=3.0\text{ mm}
]

starts sinking from rest in olive oil whose viscosity is

[
\eta=0.90\text{ P}
]

How soon after the beginning of motion will the velocity of the ball differ from its steady-state velocity by

[
1.0%
]

?

Concepts Used in I E Irodov Q. 1.339

This problem is based on the motion of a sphere through a viscous liquid.

Important concepts include:

• Stokes’ law
• Viscous drag
• Buoyant force
• Terminal velocity
• Exponential approach to terminal velocity
• Relaxation time
• Differential equation of motion

The ball starts from rest and accelerates to its terminal or steady-state velocity. The difference between the instantaneous and terminal velocities decreases exponentially with time.

Easy Solution of I E Irodov Hydrodynamics Q. 1.339

Given:

Diameter of the steel ball:

[
d=3.0\text{ mm}
]

Radius:

[
r=\frac d2=1.5\text{ mm}
]

[
r=1.5\times10^{-3}\text{ m}
]

Viscosity of olive oil:

[
\eta=0.90\text{ P}
]

Since

[
1\text{ P}=0.1\text{ Pa·s}
]

therefore,

[
\eta=0.09\text{ Pa·s}
]

For steel, take the density approximately as:

[
\rho=7.8\times10^3\text{ kg/m}^3
]

Required difference from steady velocity:

[
1%=0.01
]

Step 1: Write the Viscous Drag Force

According to Stokes’ law, the viscous force acting on a sphere moving through a liquid is:

[
F_v=6\pi\eta rv
]

where:

• (\eta) is the viscosity of the liquid
• (r) is the radius of the ball
• (v) is its instantaneous velocity

The viscous force acts opposite to the direction of motion.

Step 2: Write the Equation of Motion

The ball is acted upon by:

• Gravitational force
• Buoyant force
• Viscous drag force

The equation of motion can be written as:

[
m\frac{dv}{dt}=F_0-6\pi\eta rv
]

Here, (F_0) represents the constant effective downward force after accounting for buoyancy.

At terminal velocity (v_s), acceleration becomes zero.

Therefore:

[
F_0=6\pi\eta rv_s
]

Substituting this into the equation of motion:

6\pi\eta r(v_s-v)
]

Step 3: Obtain the Velocity-Time Relation

Rearranging:

\frac{6\pi\eta r}{m}dt
]

After integration and using the initial condition (v=0) at (t=0):

[
v=v_s\left(1-e^{-t/\tau}\right)
]

where (\tau) is the relaxation time:

[
\tau=\frac{m}{6\pi\eta r}
]

The difference between steady velocity and instantaneous velocity is:

[
v_s-v=v_se^{-t/\tau}
]

Therefore:

[
\frac{v_s-v}{v_s}=e^{-t/\tau}
]

Step 4: Apply the 1% Condition

The velocity differs from the steady-state velocity by 1%.

Therefore:

[
\frac{v_s-v}{v_s}=0.01
]

Hence:

[
e^{-t/\tau}=0.01
]

Taking the natural logarithm:

[
-\frac{t}{\tau}=\ln(0.01)
]

Since:

[
\ln(0.01)=-\ln(100)
]

we obtain:

[
t=\tau\ln(100)
]

[
t\approx4.605\tau
]

Step 5: Calculate the Relaxation Time

The mass of the steel ball is:

[
m=\frac43\pi r^3\rho
]

Therefore:

\frac{\frac43\pi r^3\rho}
{6\pi\eta r}
]

After simplifying:

[
\tau=\frac{2\rho r^2}{9\eta}
]

Substitute the values:

\frac{
2(7.8\times10^3)(1.5\times10^{-3})^2
}
{9(0.09)}
]

[
\tau\approx0.0433\text{ s}
]

Step 6: Calculate the Required Time

[
t=4.605\tau
]

[
t=4.605(0.0433)
]

[
t\approx0.20\text{ s}
]

Final Answer

The velocity of the steel ball will differ from its steady-state velocity by 1% after approximately:

[
\boxed{t\approx0.20\text{ s}}
]

Therefore, the required time is approximately:

0.20 seconds

Simple Explanation of the Answer

The steel ball starts from rest and accelerates downwards through the oil.

It is viscous, and the faster it moves the greater the force against it. Eventually the viscous force and buoyancy balance the effective weight of the ball and the ball reaches its steady-state or terminal velocity.

The ball doesn’t reach terminal velocity immediately. Instead it approaches terminal velocity asymptotically.

After about 0.20 seconds the difference between the actual speed of the ball and its terminal speed is only 1%.

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Key Takeaways

Common Mistakes Students Should Avoid

Students commonly make these mistakes:

• Using the diameter instead of the radius in the formula
• Forgetting to convert poise into SI units
• Treating terminal velocity as being reached instantly
• Using a linear velocity-time equation
• Forgetting that the difference from terminal velocity decreases exponentially
• Using (0.1) instead of (0.01) for a 1% difference
• Ignoring the density of the ball while calculating relaxation time

Always convert all quantities into SI units before substituting values.

Why Practice I E Irodov Hydrodynamics Problems?

I E Irodov Hydrodynamics problems help students understand how mechanics principles apply to liquids and moving bodies.

Benefits include:

• Strong understanding of viscosity
• Better application of Stokes’ law
• Improved differential-equation skills
• Clear understanding of terminal velocity
• Better analytical thinking
• Preparation for advanced Fluid Mechanics questions

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

Is I E Irodov Q. 1.339 a JEE Main Question?

I E Irodov Hydrodynamics Q. 1.339 is above the usual JEE Main level.

The problem combines Stokes’ law, terminal velocity, exponential motion and a differential equation. It is more suitable for:

• JEE Advanced aspirants
• Physics Olympiad students
• Advanced Fluid Mechanics practice
• Students comfortable with differential equations

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

Is Hydrodynamics Important for JEE?

Fluid Mechanics is relevant to JEE preparation, but students should not claim that a fixed number of questions is asked every year.

Important areas include:

• Fluid pressure
• Buoyant force
• Archimedes’ principle
• Continuity equation
• Bernoulli’s theorem
• Viscosity
• Stokes’ law
• Terminal velocity
• Surface tension

The number and difficulty of questions can vary from one examination session to another.

Important Topics in I E Irodov Physical Fundamentals of Mechanics

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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.

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How Should Students Use I E Irodov for JEE Advanced?

Recommended preparation sequence:

Students should use I E Irodov as an advanced problem-solving resource rather than their first Physics textbook.

I E Irodov for NRI JEE Students

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

Students following AP, IB, British or American curricula should first compare 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 problems
• Solve selected I E Irodov questions
• Review mistakes with a teacher or mentor

This approach helps NRI students use advanced books without spending excessive time on questions outside their immediate preparation needs.

Related I E Irodov Hydrodynamics Questions

After completing Question 1.339, students can also practise:

• I E Irodov Hydrodynamics Q. 1.324
• I E Irodov Hydrodynamics Q. 1.328
• I E Irodov Hydrodynamics Q. 1.335
• I E Irodov Hydrodynamics Q. 1.336

Each question should link to its corresponding TestPrepKart solution page.

Practice Next

After solving this question, students can continue with:

• Stokes’ law numerical problems
• Terminal velocity questions
• Viscosity problems
• Bernoulli’s theorem questions
• JEE Advanced Fluid Mechanics 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. 1.339

1. What is the answer to I E Irodov Q. 1.339?

The steel ball’s velocity differs from its terminal velocity by 1% after approximately:

[
\boxed{0.20\text{ s}}
]

2. Which concept is used in I E Irodov Q. 1.339?

The problem uses Stokes’ law, viscous drag, terminal velocity and exponential relaxation.

3. Why does the ball approach terminal velocity gradually?

The viscous resistance increases with velocity. As the ball speeds up, the net force and acceleration decrease until the ball approaches terminal velocity.

4. What is the relaxation time in this problem?

The relaxation time is:

[
\tau=\frac{2\rho r^2}{9\eta}
]

For the given values:

[
\tau\approx0.0433\text{ s}
]

5. Why is (\ln(100)) used?

A 1% difference means:

[
e^{-t/\tau}=0.01=\frac1{100}
]

Therefore:

[
t=\tau\ln(100)
]

6. Is I E Irodov Q. 1.339 useful for JEE Main?

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

7. Is the density of olive oil required?

The liquid density affects terminal velocity through buoyancy, but the relaxation-time expression used here depends on the ball’s mass, radius and the liquid’s viscosity.