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Mechanical Properties of Fluids NCERT Solutions, Complete PDF Content and Important Questions
NCERT Solutions for Class 11 Physics Chapter 9 PDF Download (2026) + Important Questions: Mechanical Properties of Fluids Welcome to the fascinating world of fluid mechanics! Have you ever wondered how massive airplanes weighing hundreds of tonnes manage to fly? Or why a tiny drop of water is always spherical? What makes honey flow so much slower than water? You will find the answers to all these real-world mysteries in CBSE Class 11 Physics Chapter 9: Mechanical Properties of Fluids . If you are gearing up for the 2026 board exams or competitive entrance tests like JEE Main, JEE Advanced, and NEET, mastering this chapter is crucial. It bridges the gap between basic mechanics and real-world engineering, exploring how liquids and gases behave at rest (fluid statics) and in motion (fluid dynamics).
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ChapterMechanical Properties of Fluids
SubjectPhysics
Class11
BoardCBSE
DifficultyModerate to Hard
Exam Weightage~6-8 Marks
Learning Objectives
- Understand the concept of pressure in fluids and.
- Apply Pascal's Law to hydraulic systems.
- Calculate the variation of pressure with depth and distinguish between absolute and gauge pressure.
- Apply the Equation of Continuity and Bernoulli's Principle to.
- Explain the concept of Viscosity and.
- Use Stokes' Law to determine terminal velocity.
- Understand Surface Tension , surface energy, and.
- Calculate the capillary rise of fluids in narrow tubes.
Key Concepts, Definitions and Formulas
- Pascal's Law: Pressure applied to an enclosed fluid is transmitted undiminished to every portion of the fluid and the walls of the containing vessel
- Variation of Pressure with Depth: $$P = P_a + \rho gh$$ (Where $P_a$ is atmospheric pressure, $\rho$ is density, and $h$ is depth)
- Equation of Continuity: For incompressible fluids, the volume flow rate is constant. $$A_1 v_1 = A_2 v_2$$
- Bernoulli's Principle: For a streamlined, non-viscous, incompressible fluid flow, the sum of pressure, kinetic energy per unit volume, and potential energy per unit volume remains constant. $$P + \frac{1}{2}\rho v^2 + \rho gh = \text{constant}$$
- Viscosity: The internal friction of a fluid that opposes relative motion between its layers
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Full NCERT Solutions
NCERT Q9.1: Explain why: (a) The blood pressure in humans is greater at the feet than at the brain. (b) 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. (c) Hydrostatic pressure is a scalar quantity even though pressure is force divided by area.
Answer: (a) The pressure in a fluid column increases with depth according to the formula $P = \rho gh$. The feet are at a greater depth (lower height) compared to the brain relative to the heart. Therefore, the blood column exerts greater hydrostatic pressure at the feet than at the brain. (b) The density of the atmosphere is not uniform; it is maximum near the surface of the Earth and decreases exponentially with height. Because most of the air mass is concentrated within the first few kilometers, the pressure halves at just 6 km, even though the atmosphere extends for hundreds of kilometers. (c) Pressure is defined as the component of force acting normal (perpendicular) to the surface per unit area. Since hydrostatic pressure acts equally in all directions at a given point inside a fluid, it has no unique direction. Therefore, it is a scalar quantity.
NCERT Q9.2: Explain why: (a) The angle of contact of mercury with glass is obtuse, while that of water with glass is acute. (b) Water on a clean glass surface tends to spread out while mercury on the same surface tends to form drops. (c) Surface tension of a liquid is independent of the area of the surface. (d) Water with detergent dissolved in it should have small angles of contact.
Answer: (a) For mercury and glass, the cohesive forces (between mercury molecules) are much stronger than the adhesive forces (between mercury and glass molecules). Hence, it does not wet glass, forming an obtuse angle. For water and glass, adhesive forces are stronger than cohesive forces, causing water to wet the glass and form an acute angle. (b) Since water has a strong adhesive force with clean glass, it spreads out to maximize the contact area. Mercury has strong cohesive forces, so it tries to minimize its surface area by forming spherical drops. (c) Surface tension is an intrinsic molecular property of the liquid (force per unit length). It depends on the nature of the liquid and its temperature, not on the total surface area. (d) Adding detergent reduces the surface tension of water. This decreases the cohesive force, which in turn significantly reduces the angle of contact. This allows the detergent water to penetrate tiny pores in clothes for better cleaning.
NCERT Q9.3: 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)
Answer: (a) decreases . (As temperature rises, molecular kinetic energy increases, breaking cohesive bonds easily). (b) increases ; decreases . (In gases, higher temp means more molecular collisions and drag. In liquids, higher temp weakens intermolecular bonds, reducing viscosity). (c) shear strain ; rate of shear strain .
NCERT Q9.4: Explain why: (a) To keep a piece of paper horizontal, you should blow over, not under, it. (b) When we try to close a water tap with our fingers, fast jets of water gush through the openings between our fingers. (c) A spinning cricket ball in air does not follow a parabolic trajectory.
Answer: (a) According to Bernoulli's Principle, blowing over the paper increases the air velocity above it, which decreases the pressure above the paper. The higher pressure underneath the paper pushes it upward, keeping it horizontal. (b) By the Equation of Continuity ($Area \times Velocity = \text{constant}$), partially closing the tap with fingers severely reduces the cross-sectional area for the water to flow. To maintain a constant flow rate, the velocity of the water must increase dramatically, creating fast jets. (c) A spinning ball drags air along with it. The velocity of air on one side of the ball increases and on the other side decreases. According to Bernoulli's theorem, this creates a pressure difference across the ball, resulting in a net lateral force (Magnus effect) that makes the ball curve away from a standard parabolic path.
NCERT Q9.5: A 50 kg girl wearing high heel shoes balances on a single heel. The heel is circular with a diameter of 1.0 cm. What is the pressure exerted by the heel on the horizontal floor?
Answer: Given: Mass of the girl, $m = 50 \text{ kg}$. Diameter of the heel, $d = 1.0 \text{ cm} = 0.01 \text{ m}$. Radius of the heel, $r = 0.005 \text{ m}$. Calculation: Force exerted (Weight), $F = mg = 50 \times 9.8 = 490 \text{ N}$. Area of the circular heel, $A = \pi r^2 = 3.14 \times (0.005)^2 = 3.14 \times 0.000025 = 7.85 \times 10^{-5} \text{ m}^2$. Pressure, $P = \frac{F}{A} = \frac{490}{7.85 \times 10^{-5}}$ $P = 6.24 \times 10^6 \text{ Pa}$. Result: The pressure exerted by the heel is $6.24 \times 10^6 \text{ Pa}$ (which is massive, hence why high heels sink into soft ground!).
NCERT Q9.6: Torricelli's barometer used mercury. Pascal duplicated it using French wine of density $984 \text{ kg m}^{-3}$. Determine the height of the wine column for normal atmospheric pressure.
Answer: Given: Normal atmospheric pressure, $P = 1.013 \times 10^5 \text{ Pa}$. Density of wine, $\rho = 984 \text{ kg m}^{-3}$. Acceleration due to gravity, $g = 9.8 \text{ m s}^{-2}$. Calculation: The formula for hydrostatic pressure is $P = \rho gh$. Rearranging for height: $h = \frac{P}{\rho g}$ $h = \frac{1.013 \times 10^5}{984 \times 9.8} = \frac{101300}{9643.2} \approx 10.5 \text{ m}$. Result: The height of the wine column would be approximately 10.5 meters .
Extra Important Questions for 2026 Exams
Important Q1: The working of an airplane wing (aerofoil) is based on: A) Archimedes' Principle B) Pascal's Law C) Bernoulli's Principle D) Stokes' Law
Answer: C) Bernoulli's Principle Explanation: The shape of the aerofoil makes air move faster over the top than the bottom, creating lower pressure above and generating aerodynamic lift.
Important Q2: A liquid will not wet the surface of a solid if its angle of contact is: A) Zero B) Acute (less than 90) C) Right angle (90) D) Obtuse (greater than 90)
Answer: D) Obtuse Explanation: An obtuse angle implies that cohesive forces between liquid molecules are stronger than adhesive forces with the solid.
Important Q3: Two soap bubbles have radii in the ratio 2:1. What is the ratio of excess pressure inside them? A) 1:2 B) 2:1 C) 1:4 D) 4:1
Answer: A) 1:2 Explanation: Excess pressure $P = \frac{4S}{R}$. Therefore, $P \propto \frac{1}{R}$. If $R_1 : R_2 = 2 : 1$, then $P_1 : P_2 = 1 : 2$.
Important Q4: Differentiate between streamline flow and turbulent flow.
Answer: In streamline (steady) flow, every fluid particle passing through a given point follows the exact same path and velocity as the particle before it. In turbulent flow, the motion is irregular, chaotic, and fluid particles form swirling eddies.
Important Q5: Why are drops of liquids usually spherical in shape?
Answer: Due to surface tension, liquids tend to minimize their free surface area. For a given volume, a sphere has the minimum surface area, causing drops to assume a spherical shape.
Important Q6: State Pascal's Law and give one practical application.
Answer: Pascal's Law states that if pressure is applied to an enclosed fluid, it is transmitted undiminished to every portion of the fluid and the walls of the container. Application: Hydraulic lift used to lift cars in service stations.
Important Q7: Why does the viscosity of a gas increase with temperature, unlike liquids?
Answer: In gases, viscosity is due to the collision of molecules. As temperature rises, molecular agitation and collision frequency increase, increasing momentum transfer across layers (higher viscosity). In liquids, viscosity is due to cohesive bonds, which weaken with higher temperature (lower viscosity).
Important Q8: State and prove Bernoulli's Principle for a non-viscous, incompressible fluid in streamline flow.
Answer: Statement: The sum of pressure energy, kinetic energy, and potential energy per unit volume remains constant along a streamline for an ideal fluid. Proof: Consider a tube of varying cross-section. Let fluid enter at area $A_1$ with velocity $v_1$, pressure $P_1$, height $h_1$ and leave at $A_2$ with $v_2$, $P_2$, $h_2$. Mass $\Delta m = \rho A_1 v_1 \Delta t = \rho A_2 v_2 \Delta t$. Work done on fluid by pressure: $W = W_1 - W_2 = P_1 A_1 v_1 \Delta t - P_2 A_2 v_2 \Delta t = (P_1 - P_2) \frac{\Delta m}{\rho}$. Change in Kinetic Energy: $\Delta K = \frac{1}{2} \Delta m (v_2^2 - v_1^2)$. Change in Potential Energy: $\Delta U = \Delta m g (h_2 - h_1)$. By Work-Energy Theorem: $W = \Delta K + \Delta U$ $(P_1 - P_2) \frac{\Delta m}{\rho} = \frac{1}{2} \Delta m (v_2^2 - v_1^2) + \Delta m g (h_2 - h_1)$ Divide by $\Delta m$ and multiply by $\rho$: $P_1 - P_2 = \frac{1}{2}\rho v_2^2 - \frac{1}{2}\rho v_1^2 + \rho gh_2 - \rho gh_1$ $P_1 + \frac{1}{2}\rho v_1^2 + \rho gh_1 = P_2 + \frac{1}{2}\rho v_2^2 + \rho gh_2 = \text{Constant}$.
Important Q9: Derive an expression for the terminal velocity of a small spherical body falling through a viscous medium.
Answer: Let a sphere of radius $r$ and density $\rho$ fall through a fluid of density $\sigma$ and viscosity $\eta$. Three forces act on it:
Important Q10: Describe the phenomenon of capillary rise. Derive the ascent formula for the height of a liquid column in a capillary tube.
Answer: Capillarity is the phenomenon of rise or fall of a liquid inside a fine narrow tube. Derivation: Consider a capillary tube of radius $r$ dipped in a liquid of surface tension $S$ and density $\rho$. Let the angle of contact be $\theta$. The meniscus curves upward, exerting a surface tension force $S$ along the circumference ($2\pi r$) at angle $\theta$. Total upward force = $(S \cos\theta) \times 2\pi r$. This force balances the weight of the liquid column of height $h$. Weight of liquid = Volume $\times$ Density $\times$ $g$ = $(\pi r^2 h) \rho g$. Equating upward and downward forces: $2\pi r S \cos\theta = \pi r^2 h \rho g$ $h = \frac{2S \cos\theta}{r \rho g}$ A mechanic is working in a garage equipped with a hydraulic lift. The lift consists of two pistons. The smaller piston has a cross-sectional area of $10 \text{ cm}^2$, and the larger piston has a cross-sectional area of $3 \text{ m}^2$. The mechanic applies a force of 50 N on the smaller piston.
Important Q11: Which principle of physics governs the working of this hydraulic lift?
Answer: Pascal's Law.
Important Q12: Calculate the pressure transmitted through the hydraulic fluid.
Answer: Area of small piston $A_1 = 10 \text{ cm}^2 = 10 \times 10^{-4} \text{ m}^2 = 10^{-3} \text{ m}^2$. Pressure $P = \frac{F_1}{A_1} = \frac{50}{10^{-3}} = 50,000 \text{ Pa}$.
Important Q13: What is the maximum weight of the car that the larger piston can lift?
Answer: By Pascal's Law, the pressure is transmitted equally. $F_2 = P \times A_2 = 50,000 \times 3 = 150,000 \text{ N}$. The lift can support a car weighing up to 150,000 N. (Directions: Select A if both are true and Reason explains Assertion. Select B if both are true but Reason does not explain Assertion. Select C if Assertion is true, Reason is false. Select D if Assertion is false, Reason is true.)
Important Q14: Assertion (A): The velocity of flow of a liquid is minimum when the cross-sectional area of the pipe is maximum. Reason (R): According to the equation of continuity, the product of the cross-sectional area and velocity is constant.
Answer: A . Both A and R are true, and R is the correct explanation. ($A_1 v_1 = A_2 v_2$, so $v \propto 1/A$).
Important Q15: Assertion (A): A small drop of liquid is always spherical. Reason (R): The excess pressure inside a drop is inversely proportional to its radius.
Answer: B . Both A and R are true, but R does not explain A. The correct reason drops are spherical is because surface tension minimizes the surface area.
FAQ Section
Is Chapter 9 Mechanical Properties of Fluids important for boards?
Yes, it is highly important! It carries about 6 to 8 marks and features guaranteed conceptual questions on Bernoulli's principle and surface tension.
Where can I download the Class 11 Physics NCERT PDF?
You can download the official, rationalized NCERT Class 11 Physics textbook directly from the official NCERT website (ncert.nic.in) under the 'Publications' and 'PDFs' section.
What is the difference between streamline and turbulent flow?
In streamline flow, fluid particles move in smooth, parallel, non-crossing paths. In turbulent flow, the fluid velocity exceeds the critical velocity, causing erratic, swirling, and chaotic motion.
What are the most important derivations in this chapter?
The most critical derivations for the 2026 exams are Bernoulli's Principle, the Ascent Formula (Capillary rise), and Terminal Velocity using Stokes' Law.
What is terminal velocity?
Terminal velocity is the constant, maximum velocity a body achieves when falling through a viscous fluid. It occurs when the downward gravitational force is exactly balanced by the upward buoyant force and viscous drag
More Class 11 Physics Chapters
Ch 1: Units and MeasurementsCh 2: Motion in a Straight LineCh 3: Motion in a PlaneCh 4: Laws of MotionCh 5: Work, Energy and PowerCh 6: System of Particles and Rotational MotionCh 7: GravitationCh 8: Mechanical Properties of SolidsCh 9: Mechanical Properties of FluidsCh 10: Thermal Properties of MatterCh 11: ThermodynamicsCh 12: Kinetic TheoryCh 13: OscillationsCh 14: Waves