NCERT Solutions for Class 11 Physics Chapter 9 Mechanical Properties Of Fluids

NCERT Solutions for Class 11 Physics Chapter 9 Mechanical Properties Of Fluids

Vishal kumarUpdated on 19 Sep 2025, 01:01 AM IST

Did you ever wonder why the water runs out, or why the nozzle of a garden hose runs faster when you squeeze it, or how big aeroplanes are able to remain floating in the air? These normal miracles can be easily explained with the assistance of NCERT Solutions to Class 11 Physics Chapter 9 - Mechanical Properties of Fluids, where students can observe the interesting nature of liquids and gases.

This Story also Contains

  1. Mechanical Properties of Fluids NCERT Solutions: Download PDF
  2. Mechanical Properties of Fluids NCERT Solutions: Exercise Questions
  3. Mechanical Properties of Fluids NCERT Solutions: Higher Order Thinking Skills (HOTS) Questions
  4. Class 11 Physics Chapter 9 - Mechanical Properties of Fluids: Topics
  5. Approach to Solve the Questions of Class 11 Physics Chapter 9 - Mechanical Properties of Fluids
  6. What Extra Should Students Study Beyond NCERT for JEE/NEET?
  7. NCERT Solutions for Class 11 Physics Chapter-Wise
NCERT Solutions for Class 11 Physics Chapter 9 Mechanical Properties Of Fluids
NCERT SOlution for class 11 chapter 9

The key ideas discussed in this chapter are fluid pressure, Pascal's Law, viscosity, buoyant force, Archimedes' Principle, and surface tension and establish a good base in fluid mechanics. The NCERT Solutions for Class 11 Physics Chapter 9 - Mechanical Properties of Fluids are compiled by highly qualified teachers and are developed in accordance with the recent CBSE syllabus, thus they are a trustworthy source of board examination and also in competitive examinations like JEE, NEET and Olympiad. Students also receive valuable formulas, solved examples and smart preparation tips in addition to step-by-step solutions, so that they can reinforce their problem-solving skills. Also, the NCERT Solutions for Class 11 Physics Chapter 9 - Mechanical Properties of Fluids can be downloaded freely in a PDF format, and hence, students can always revise the solutions anywhere, anytime. These NCERT solutions can be used to simplify the preparation for exams, but they also increase the clarity of the concepts, which allows students to excel academically with confidence by relating the concepts studied in physics with the real world.

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Mechanical Properties of Fluids NCERT Solutions: Download PDF

Class 11 Physics Chapter 9 - Mechanical Properties of Fluids question answers present detailed knowledge behind each and every textbook problem, thus helping in understanding concepts such as pressure, buoyancy, viscosity, and surface tension. Each solution is prepared based on the current syllabus of CBSE and can be downloaded easily because they are available in PDF format only.

Download PDF

Mechanical Properties of Fluids NCERT Solutions: Exercise Questions

The Mechanical Properties of Fluids class 11 question answers are clear and definite solutions to the exercises in the textbook. These will assist students in reinforcing their knowledge of principles such as Pascal's law, Archimedes' principle and Bernoulli's principle, facilitating sufficient preparation to write exams.

Q 9.1 (a) Explain why

The blood pressure in humans is greater at the feet than at the brain

Answer:

The pressure in a fluid column increases with the height of the column, as the height of the blood column is more than that for the brain the blood pressure in feet is more than the blood pressure in the brain.

Q 9.1 (b) Explain why

Atmospheric pressure at a height of about 6 km decreases to nearly half of its value at the sea level, though the height of the atmosphere is more than 100 km

Answer:

This is because the air density does not remain the same in the atmosphere. It decreases exponentially as height increases.

Q 9.1 (c) Explain why

Hydrostatic pressure is a scalar quantity even though pressure is force divided by area.

Answer:

When a force is applied on fluid the pressure which gets generated gets uniformly transmitted to all directions and therefore has no particular direction and is a scalar quantity. We talk of division of force with area only while considering the magnitudes. The actual vector form of the relation is F=pA

Q 9.2 (a) Explain why

The angle of contact of mercury with glass is obtuse, while that of water with glass is acute.

Answer:

1650445487018

TSL = Surface tension corresponding to the solid-liquid layer

TLA = Surface tension corresponding to liquid-air layer

TSA = Surface tension corresponding to solid-air layer

The angle of contact is θ

Since the liquid is not flowing over the solid surface the components of TSL , TLA and TSA along the solid surface must cancel out each other.

TSL+TLAcosθ=TSA
cosθ=TSATSLTLA

In case of mercury TSA < TSL and therefore cosθ<0 and therefore θ>π2 i.e the angle of contact of mercury with glass is obtuse.

Q 9.2 (b) Explain why

Water on a clean glass surface tends to spread out while mercury on the same surface tends to form drops. (Put differently, water wets glass while mercury does not.)

Answer:

Cohesive forces between water molecules is much lesser than adhesive forces between water and glass molecules and that's why water tends to spread out on glass whereas cohesive forces within mercury is comparable to adhesive forces between mercury and glass and that's why mercury tends to form drops.

Q 9.2 (c) Explain why

Surface tension of a liquid is independent of the area of the surface

Answer:

Surface tension is the force acting per unit length at the interface of a liquid and another surface. Since this force itself is independent of area, the surface tension is also independent of area.

Q 9.2 (d) Explain why

Water with detergent dissolved in it should have small angles of contact.

Answer:

As we know, detergent with water rises very fast in the capillaries of clothes, which is only possible when the cosine of the angle of contact is large, i.e. angle of contact must be small.

Q 9.2 (e) Explain why

A drop of liquid under no external forces is always spherical in shape

Answer:

While in a spherical shape, the surface area of the drop of liquid will be minimum, and thus the surface energy would be minimum. A system always tends to be in a state of minimum energy, and that's why in the absence of external forces, a drop of liquid is always spherical in shape.

Q 9.3 Fill in the blanks using the word(s) from the list appended with each statement:

(a) Surface tension of liquids generally ... with temperatures (increases/decreases)

(b) Viscosity of gases ... with temperature, whereas viscosity of liquids ... with temperature (increases/decreases)

(c) For solids with elastic modulus of rigidity, the shearing force is proportional to ..., while for fluids it is proportional to ... (shear strain/rate of shear strain)

(d) For a fluid in a steady flow, the increase in flow speed at a constriction follows (conservation of mass / Bernoulli’s principle)

(e) For the model of a plane in a wind tunnel, turbulence occurs at a ... speed for turbulence for an actual plane (greater / smaller)

Answer:

(a) The surface tension of liquids generally decreases with temperature.

(b) The viscosity of gases increases with temperature, whereas the viscosity of liquids decreases with temperature.

(c) For solids with elastic modulus of rigidity, the shearing force is proportional to shear strain, while for fluids, it is proportional to the rate of shear strain.

(d) For a fluid in a steady flow, the increase in flow speed at a constriction follows from conservation of mass, while the decrease in pressure there follows from Bernoulli’s principle.

(e) For the model of a plane in a wind tunnel, turbulence occurs at a greater speed for turbulence for an actual plane.

Q 9.4 (a) Explain why

To keep a piece of paper horizontal, you should blow over, not under it

Answer:

As per Bernoulli's principle, when we blow over a piece of paper, the pressure there decreases while the pressure under the piece of paper remains the same, and that's why it remains horizontal.

Q 9.4 (b) Explain why

When we try to close a water tap with our fingers, fast jets of water gush through the openings between our fingers

Answer:

This is because when we cover the tap, there are very small gaps remaining for the water to escape, and it comes out at very high velocity in accordance with the equation of continuity.

Q 9.4 (c) Explain why

The size of the needle of a syringe controls flow rate better than the thumb pressure exerted by a doctor while administering an injection

Answer:

Because of the extremely small size of the opening of a needle, its size can control the flow with more precision than the thumb of a doctor.

According to the equation of continuity, area * velocity constant. If the area is very small, the velocity must be large. Thus, if the area is small flow becomes smooth

Q 9.4 (d) Explain why

A fluid flowing out of a small hole in a vessel results in a backward thrust on the vessel

Answer:

Through a small area, the velocity will be large. A fluid flowing out of a small hole in a vessel results in a backward thrust on the vessel in accordance with the law of conservation of linear momentum.

Q 9.4 (e) Explain why

A spinning cricket ball in air does not follow a parabolic trajectory

Answer:

The ball, while travelling, rotates about its axis as well, causing a difference in air velocities at different points around it, thus creating a pressure difference which results in external forces. In the absence of air, a ball would have travelled along the expected parabolic path.

Q 9.5 A 50 kg girl wearing high-heeled shoes balances on a single heel. The heel is circular with a diameter 1.0 cm. What is the pressure exerted by the heel on the horizontal floor?

Answer:

Mass of the girl m = 50 kg.

Gravitational acceleration g = 9.8 m s-2

Weight of the girl (W), mg = 490 N

Pressure=ForceArea=490π(1×1022)2=6.24×106Pa

Q 9.6 Torricelli's barometer used mercury. Pascal duplicated it using French wine of density 984 kg m3 . Determine the height of the wine column for normal atmospheric pressure.

Answer:

Atmospheric pressure is P=1.01×105Pa

The density of French wine ρw=984 kg m3

Height of the wine column h w would be

hw=Pρwg


hw=1.01×105984×9.8


hw=10.474m

Q 9.7 A vertical off-shore structure is built to withstand a maximum stress of 109 Pa. Is the structure suitable for putting up on top of an oil well in the ocean? Take the depth of the ocean to be roughly 3 km, and ignore ocean currents.

Answer:

The density of water is ρw=1000 kg m3

The depth of the ocean is 3 km

The pressure at the bottom of the ocean would be

P=ρwgh
P=1000×9.8×3×1000
P=2.94×107Pa

The above value is much less than the maximum stress the structure can withstand, and therefore, it is suitable for putting up on top of an oil well in the ocean.

Q 9.8 A hydraulic automobile lift is designed to lift cars with a maximum mass of 3000 kg. The area of cross-section of the piston carrying the load is 425 cm2. What maximum pressure would the smaller piston have to bear?

Answer:

The Maximum Pressure that the piston would have to bear is

Pmax=Maximum weight of a carArea of the piston=3000×9.8425×104=6.917×105Pa

Q 9.9 A U-tube contains water and methylated spirit separated by mercury. The mercury columns in the two arms are in level with 10.0 cm of water in one arm and 12.5 cm of spirit in the other. What is the specific gravity of spirit?

Answer:

Since the mercury columns in the two arms are equal, the pressure exerted by the water and the spirit column must be the same.

hwρwg=hsρsg


ρsρw=hwhs


ρsρw=1012.5


ρsρw=0.8

Therefore, the specific gravity of spirit is 0.8.

Q 9.10 In the previous problem, if 15.0 cm of water and spirit each are further poured into the respective arms of the tube, what is the difference in the levels of mercury in the two arms? (Specific gravity of mercury = 13.6)

Answer:

Let the difference in the levels of mercury in the two arms be hHg

(ρwρs)gh=ρHgghHg(1ρsρw)h=ρHgρwhHg(10.8)×15=13.68hHghHg=0.2×1513.6hHg=0.22 cm

Q 9.11 Can Bernoulli’s equation be used to describe the flow of water through a rapid in a river? Explain.

Answer:

No. Bernoulli's equation can be used only to describe streamline flow, and the flow of water in a river is turbulent.

Q 9.12 Does it matter if one uses gauge instead of absolute pressures in applying Bernoulli’s equation? Explain

Answer:

No, unless the atmospheric pressures at the two points where Bernoulli’s equation is applied are significantly different.

Q 9.13 Glycerine flows steadily through a horizontal tube of length 1.5 m and radius 1.0 cm. If the amount of glycerine collected per second at one end is 4.0 × 103 kg s1 , what is the pressure difference between the two ends of the tube ? (Density of glycerine = 1.3 × 103 kg m3 and viscosity of glycerine = 0.83 Pa s). [You may also like to check if the assumption of laminar flow in the tube is correct].

Answer:

The volumetric flow rate of glycerine would be given by

V=Amount of glycerine flow per second Density of glycerine
V=Mρ
V=4×1031.3×103
V=3.08×106ms1

The viscosity of glycerine is η=0.83 Pa s

Assuming Laminar flow for a tube of radius r, length l, having a pressure difference P across its ends, a fluid with viscosity η would flow through it with a volumetric rate of

V=πPr48ηl


P=8Vηlπr4


P=8×3.08×106×0.83π×(102)4


P=9.75×102 Pa

Reynolds' number is given by

R=4ρVπdη


R=2ρVπrη


R=2×1.3×103×3.08×106π×0.01×0.83R0.3

Since the Reynolds Number is coming out to be 0.3, our Assumption of laminar flow was correct.

Q 9.14 In a test experiment on a model aeroplane in a wind tunnel, the flow speeds on the upper and lower surfaces of the wing are 70 m s1 and 63 ms1 respectively. What is the lift on the wing if its area is 2.5 m2 ? Take the density of air to be 1.3 kg m3

Answer:

The speeds of air above and below the wings are given to be v1 = 70 m s1 and v2 = 63 m s1, respectively.

Let the pressure above and below the wings be P1 and P2, and let the model aeroplane be flying at a height h from the ground.

Applying Bernoulli's Principle on two points above and below the wings, we get

P1+ρgh+12ρv12=P2+ρgh+12ρv22P2P1=12ρ(v12v22)ΔP=12×1.3×(702632)ΔP=605.15 Pa

The pressure difference between the regions below and above the wing is 605.15 Pa

The lift on the wing is F

F=ΔP× Area of wing F=605.15×2.5F=1512.875 N

Q 9.15 Figures (a) and (b) refer to the steady flow of a (non-viscous) liquid. Which of the two figures is incorrect? Why?

Answer:

By the continuity equation, the velocity of the non-viscous liquid will be larger at the kink than at the rest of the tube, and therefore, pressure would be lower here by Bernoulli's principle, and the air column above it, therefore, should be of lesser height. Figure (a) is therefore incorrect.

Q 9.16 The cylindrical tube of a spray pump has a cross-section of 8.0 cm2 one end of which has 40 fine holes each of diameter 1.0 mm. If the liquid flow inside the tube is 1.5 m min1 , what is the speed of ejection of the liquid through the holes?

Answer:

Cross-sectional area of cylindrical tube is a1=8.0 cm2
The total area of the 40 fine holes is a2

a2=40π(d24)a2=40×π×(103)24a2=3.143×105 m2

Speed of liquid inside the tube is v1=1.5 m min1
Let the speed of ejection of fluid through the holes be v2
Using the continuity equation

a1v1=a2v2v2=a1v1a2v2=8×104×1.560×3.143×105v2=0.636 m s1

Q 9.17 A U-shaped wire is dipped in a soap solution, and removed. The thin soap film formed between the wire and the light slider supports a weight of 1.5 × 102 N (which includes the small weight of the slider). The length of the slider is 30 cm. What is the surface tension of the film?

Answer:

Total weight supported by the film W=1.5×102N
Since a soap film has two surfaces, the total length of the liquid film is 60 cm.
Surface Tension is T

T=WlT=1.5×10260×102T=2.5×102Nm1

Q 9.18 Figure (a) shows a thin liquid film supporting a small weight = 4.5 × 102 N. What is the weight supported by a film of the same liquid at the same temperature in Fig. (b) and (c) ? Explain your answer physically.

Answer:

As the liquid and the temperature are the same in all three, the surface tension will also be the same. Since the length is also given to be equal (40 cm) in all three cases, the weight being supported is also the same and equal to 4.5×102 N.

Q 9.19 What is the pressure inside the drop of mercury of radius 3.00 mm at room temperature? Surface tension of mercury at that temperature (20 °C) is 4.65 × 101 N m1. The atmospheric pressure is 1.01 × 105 Pa. Also give the excess pressure inside the drop.

Answer:

Surface Tension of Mercury is T=4.65×101 N m1
The radius of the drop of Mercury is r=3.00 mm
Excess pressure inside the Mercury drop is given by

ΔP=2TrΔP=2×4.65×1013×103ΔP=310 Pa

Atmospheric Pressure is P0=1.01×105 Pa
Total Pressure inside the Mercury drop is given by

PT=ΔP+P0PT=310+1.01×105PT=1.0131×105

Q 9.20 What is the excess pressure inside a bubble of soap solution of radius 5.00 mm, given that the surface tension of soap solution at the temperature (20C) is 2.50×102 N m1? If an air bubble of the same dimension were formed at depth of 40.0 cm inside a container containing the soap solution of relative density 1.20 ), what would be the pressure inside the bubble? ( 1 atmospheric pressure is 1.01× 105 Pa ).

Answer:

Excess Pressure inside a bubble is given by

ΔP=4Tr (It's double the usual value because of the presence of 2 layers in the case of a soap bubble)

where T is surface tension and r is the radius of the bubble

ΔP=4×2.5×1025×103ΔP=20Pa

Atmospheric Pressure is Pa=1.01×105 Pa
The density of the soap solution is ρs=1.2×103 kg m3
The pressure at a depth of 40 cm (h) in the soap solution is

Ps=Pa+ρsghPs=1.01×105+1.2×103×9.8×40×102Ps=1.01×105+4704Ps=1.057×105 Pa

Total Pressure inside an air bubble at that depth

Pt=Ps+2TrPt=1.057×105+2×2.5×1025×103Pt=1.057×105+10Pt=1.0571×105 Pa

Mechanical Properties of Fluids NCERT Solutions: Higher Order Thinking Skills (HOTS) Questions

When learning Physics, it is not only necessary to study the formulas but also to apply the concepts to real life. Class 11 Physics Chapter 9 - Mechanical Properties of Fluids question answers contain Higher Order Thinking Skills (HOTS) questions, which require the student to think outside the box. Such enhanced problems are useful in building analytical skills, enhancing numerical skills, and preparing successfully for JEE, NEET, and board exams.

Q1: A particle of mass m is dropped inside a liquid medium which applies a bouncy force v and a drag force k.v². If acceleration due to gravity g is constant. What is the terminal velocity attained by the particle?


Answer:

From the figure,

mg=B+kvt2mgB=kvt2vt2=mgBkvt=mgBk

Q2: For the arrangement shown in the figure, what is the density of oil?

Answer :

ρ0+ρwgl=ρ0+ρoil (l+d)gρoil =ρωll+d=1000×135(135+12.3)ρoil =916 kg/m3

Q:3 In the given figure, the velocity V3 will be

Answer:

From the continuity equation

A1V1=A2V2+A3V3V3=A1V1A2V2A3V3=(0.2×4)(0.2×2)0.4V3=0.80.40.4=0.40.4V3=1 m/s

Q4: A tank filled with water (density ρw = 1000 kg/m3) and oil (density ρo = 900 kg/m3). The height of the water is 1.0m, and that of the oil is 4.0m. Find the velocity of efflux through a hole at the bottom of the tank.

Answer :

The height of water which exerts the same pressure on the interface as whatever oil exerts:

hρwg=4ρoilg

or

h=49001000=3.6 m

Effective height of water over the hole:

H=1+3.6=4.6m

ve=2gh=2104.6=246=96ve=9.5 m/s

Q5: The velocity of liquid coming out of a small hole of a vessel containing two different liquids of densities 2ρ and ρ, as shown in the figure, is-

Answer: With the help of Bernoulli's equation, we get-
P0+ρg(2h)+(2ρ)gh=12(2ρ)v2+P0P0+4ρgh=ρv2+P04ρgh=ρv2v=2gh

Class 11 Physics Chapter 9 - Mechanical Properties of Fluids: Topics

Mechanical Properties of Fluids class 11 question answers are based on concepts such as Pascal’s law, Archimedes’ principle, viscosity, surface tension, Bernoulli’s theorem, etc. These topics lay the foundation to understand real-world applications like fluid pressure in a dam, the flying of aircraft and blood circulation in the living system.

9.1 Introduction
9.2 Pressure
9.2.1 Pascal’s law
9.2.2 Variation of pressure with depth
9.2.3 Atmospheric pressure and gauge pressure
9.2.4 Hydraulic machines
9.3 Streamline flow
9.4 Bernoulli’s principle
9.4.1 Speed of efflux: Torricelli’s law
9.4.2 Dynamic lift
9.5 Viscosity
9.5.1 Stokes’ law
9.6 Surface tension
9.6.1 Surface energy
9.6.2 Surface energy and surface tension
9.6.3 Angle of contact
9.6.4 Drops and bubbles
9.6.5 Capillary rise

Approach to Solve the Questions of Class 11 Physics Chapter 9 - Mechanical Properties of Fluids

Efficient problem solving in Physics is not only a question of memorizing the formulas but it also involves the construction of a systematic way of thinking. A step-by-step technique is used to prevent confusion during fluid mechanics, particularly when several principles interact, such as the principle of pressure, the principle of buoyancy, and the law of Bernoulli. Students are able to arrive at correct solutions with great confidence when they can break down the problem into smaller and logical steps.

  • Visualizing a Physical Situation: Take time to read through the problem and draw a simple diagram (such as a column of liquid, a submerged object or flowing fluid). The visualization aids in determining forces acting on the system, including buoyant force, pressure or viscous drag forces.

  • Determine the Applicable Concept: Decide which law applies - The law of pressure transmission by Pascal, the principle of buoyancy problems according to Archimedes, the principle of fluid motion of Bernoulli, Viscosity and Stoke law of motion through fluids.

  • Write Governing Equations Clearly: Start with the standard equations, such as

$P=h \rho g$ for pressure in fluids,
$F_b=V \rho g$ for buoyant force,
$P+\frac{1}{2} \rho v^2+\rho g h=$ constant for Bernoulli's equation.

  • Apply Boundary Conditions: Care should be taken when writing values (such as fluid density, velocity, depth, or height). Make sure you use the right conditions (e.g. velocity at one end compared to another one, or forces on an immersed body).

  • Check for Units and Dimensions: It is always better to ensure that the final answer is written in the correct unit (N, Pa, m/s, etc.). Formulas can be cross-checked before calculation with the help of dimensional analysis.

  • Relate to Real-Life Examples: Explain the physical meaning of the result (e.g. why the object in question floats or sinks, why pressure rises with depth or why fluid flow at higher velocity produces low pressure). This measure reinforces conceptualization and eliminates formula applications blindly.

What Extra Should Students Study Beyond NCERT for JEE/NEET?

NCERT textbook gives basic knowledge of the Class 11 Physics Chapter 9 - Mechanical Properties of Fluids in JEE/NEET preparation, but to perform better, students require advanced knowledge of the subject. The competitive exams require more knowledge on practical impact, higher order problem solving and mathematics on its application to concepts such as Bernoulli theorem, viscosity and surface tension.

NCERT Solutions for Class 11 Physics Chapter-Wise

NCERT Solutions for Class 11 Physics are designed to make learning simple and effective by providing clear, step-by-step answers to every chapter. These solutions not only help in preparing for school exams but also strengthen the base for competitive exams like JEE and NEET. Here you can access chapter-wise links for easy navigation and focused study.

NCERT Solutions for Class 11 Subject-wise

Also Check NCERT Books and NCERT Syllabus here

Subject wise NCERT Exemplar solutions

Frequently Asked Questions (FAQs)

Q: Where can I practice numericals on Mechanical Properties of Fluids Class 11 PDF format?
A:

You can find Numericals on Mechanical Properties of Fluids Class 11 PDF from the above given link. These PDFs include step-by-step solutions to numerical problems based on pressure, viscosity, buoyancy, and other related concepts.

Q: Is there a free Mechanical Properties of Fluids Class 11 NCERT PDF available?
A:

Yes, many websites offer the Mechanical Properties of Fluids Class 11 NCERT PDF for free. It usually includes theory, solved questions, and summary notes to help students understand the chapter better.

Q: What is Pascal’s Law and how is it applied?
A:

Pascal’s Law states that any change in pressure applied to an enclosed fluid is transmitted equally in all directions. It is used in hydraulic brakes and lifts.

Q: Why does surface tension occur in liquids?
A:

Surface tension arises due to cohesive forces between liquid molecules at the surface, causing the liquid to behave like a stretched elastic sheet, as seen in water droplets forming spherical shapes.

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