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.
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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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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.
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
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
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:
The angle of contact is
Since the liquid is not flowing over the solid surface the components of
In case of mercury
Q 9.2 (b) Explain why
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
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
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.
Answer:
Mass of the girl m = 50 kg.
Gravitational acceleration g = 9.8 m s-2
Weight of the girl (W), mg = 490 N
Answer:
Atmospheric pressure is
The density of French wine
Height of the wine column h w would be
Answer:
The density of water is
The depth of the ocean is 3 km
The pressure at the bottom of the ocean would be
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.
Answer:
The Maximum Pressure that the piston would have to bear is
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.
Therefore, the specific gravity of spirit is 0.8.
Answer:
Let the difference in the levels of mercury in the two arms be hHg
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.
Answer:
No, unless the atmospheric pressures at the two points where Bernoulli’s equation is applied are significantly different.
Answer:
The volumetric flow rate of glycerine would be given by
The viscosity of glycerine is
Assuming Laminar flow for a tube of radius r, length l, having a pressure difference P across its ends, a fluid with viscosity
Reynolds' number is given by
Since the Reynolds Number is coming out to be 0.3, our Assumption of laminar flow was correct.
Answer:
The speeds of air above and below the wings are given to be
Let the pressure above and below the wings be
Applying Bernoulli's Principle on two points above and below the wings, we get
The pressure difference between the regions below and above the wing is 605.15 Pa
The lift on the wing is F
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.
Answer:
Cross-sectional area of cylindrical tube is
The total area of the 40 fine holes is
Speed of liquid inside the tube is
Let the speed of ejection of fluid through the holes be
Using the continuity equation
Answer:
Total weight supported by the film
Since a soap film has two surfaces, the total length of the liquid film is 60 cm.
Surface Tension is T
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
Answer:
Surface Tension of Mercury is
The radius of the drop of Mercury is
Excess pressure inside the Mercury drop is given by
Atmospheric Pressure is
Total Pressure inside the Mercury drop is given by
Answer:
Excess Pressure inside a bubble is given by
where
Atmospheric Pressure is
The density of the soap solution is
The pressure at a depth of 40 cm (h) in the soap solution is
Total Pressure inside an air bubble at that depth
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,
Q2: For the arrangement shown in the figure, what is the density of oil?
Answer :
Q:3 In the given figure, the velocity
Answer:
From the continuity equation
Q4: A tank filled with water (density
Answer :
The height of water which exerts the same pressure on the interface as whatever oil exerts:
or
Effective height of water over the hole:
Q5: The velocity of liquid coming out of a small hole of a vessel containing two different liquids of densities
Answer: With the help of Bernoulli's equation, we get-
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
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.
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 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.
Frequently Asked Questions (FAQs)
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.
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.
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.
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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