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The NCERT Solutions for Class 11 Physics Chapter 9 Mechanical Properties of Fluids are provided here to help students prepare effectively for their Class 11 final exams and competitive exams like JEE and NEET. These solutions include detailed answers to all exercise questions, along with important problems, important topics, important formulas, and what you need to learn extra for the competitive exam. These NCERT solutions are prepared by subject experts as per the latest CBSE Syllabus.
Chapter 9 of Class 11 Physics explains the mechanical properties of fluids, including the behaviour of liquids and gases, which can flow and are collectively called fluids. This chapter helps students understand key concepts like pressure, viscosity, and surface tension. To make learning easier, NCERT Solutions for class 11 physics are prepared in easy language by our expert faculty. Students can also download the solution PDF from the given below link.
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NCERT Class 11 Physics Chapter 9 Exercise Solutions PDF is available here for free download. These solutions cover all textbook exercise questions with clear, step by step explanations
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
(a) 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
(c) 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
(d) 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 cosine of the angle of contact is the large i.e. angle of contact must be small.
Q 9.2 (e) Explain why
(e) 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 temperatures.
(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 of 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
(a) 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 the 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 thumbs 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
(d) A fluid flowing out of a small hole in a vessel results in a backward thrust on the vessel
Answer:
Through a small area, 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
(e) 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 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
Depth of the ocean is 3 km
The pressure at the bottom of the ocean would be
The above value is much lesser 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:
Maximum Pressure which 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 h Hg
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 flow would be given by
The viscosity of glycerine is
Assuming Laminar flow for a tube of radius r, length l, having pressure difference P across its ends a fluid with viscosity
Reynolds number is given by
Since Reynolds Number is coming out to be 0.3 our Assumption of laminar flow was correct.
Answer:
The speed 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 large at the kink than at the rest of the tube and therefore pressure would be lesser 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 is 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 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
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 attended 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 continuity equation
Q4: A tank filled with water (density
Answer :
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-
Topics of class 11 chapter 9 physics are listed below. These topics are very important for your class 11 exam as well as for competitive exams.
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
Important formulae are listed below. These formulas from Class 11 Physics Chapter 9 are crucial for board exams as well as competitive exams like JEE and NEET.
Stoke's Law:
(where
Terminal Velocity
(
To solve questions from Mechanical Properties of Fluids, first note what is given, like pressure, height, density, or flow speed. Then choose the correct formula, such as Bernoulli’s equation, Pascal’s law, or the formula for viscosity. Use diagrams(FBD) if needed. Keep units consistent, apply the formulas step by step, and solving becomes much easier.
To score well in JEE/NEET, students must go beyond the NCERT in some parts of the Mechanical Properties of Fluids. Below is a comparison showing what NCERT covers and what needs extra focus.
Solids have a fixed shape and volume, while fluids (liquids and gases) can flow and take the shape of their container. Gases, unlike liquids, can expand to fill any available space.
Bernoulli’s principle states that an increase in fluid speed leads to a decrease in pressure or potential energy. It is used in aerodynamics, carburettors, and medical applications like blood flow analysis.
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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