NCERT Solutions for Class 11 Physics Chapter 9: Mechanical Properties of Fluids

These Class 11 Physics Chapter 9 solutions cover Mechanical Properties of Fluids with every NCERT exercise (9.1–9.20) reproduced verbatim and solved step by step, with units shown and each numerical answer cross-checked against the NCERT answer key (session 2026–27). The chapter studies pressure, Pascal’s law, Bernoulli’s principle, viscosity and surface tension — the physics of liquids and gases at rest and in motion.

Class: 11 Subject: Physics Chapter: 9 Topic: Mechanical Properties of Fluids Exercises: 9.1–9.20 Session: 2026–27
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Chapter Overview

Liquids and gases can flow, so they are together called fluids. Chapter 9, Mechanical Properties of Fluids, explores how fluids behave at rest and in motion. At rest we study pressure (force per unit area, a scalar), how pressure increases with depth as P = Pa + ρgh, Pascal’s law and its use in hydraulic lifts and brakes, and the working of the barometer and manometer. For fluids in motion we use the equation of continuity (conservation of mass) and Bernoulli’s principle (conservation of energy) to explain Torricelli’s law of efflux, the lift on an aircraft wing and the swing of a cricket ball. Real fluids resist flow through viscosity (Stokes’ law, terminal velocity), and liquid surfaces behave like stretched membranes because of surface tension, which explains drops, bubbles, angle of contact and capillary rise.

Key Concepts & Definitions

Pressure (P): the normal force exerted by a fluid per unit area; a scalar quantity with SI unit pascal (Pa = N m−2).

Pascal’s law: a change in pressure applied to an enclosed fluid is transmitted undiminished to every point of the fluid and to the walls of the container; also, pressure at all points at the same height in a fluid at rest is equal.

Gauge pressure: the excess of absolute pressure over atmospheric pressure, P − Pa = ρgh.

Equation of continuity: for an incompressible fluid in steady flow, Av = constant; flow speed is greater where the cross-section is narrower.

Bernoulli’s principle: for steady, non-viscous, incompressible flow, the sum of pressure energy, kinetic energy and potential energy per unit volume stays constant along a streamline.

Viscosity (η): the internal friction between layers of a fluid in relative motion; ratio of shear stress to strain rate, SI unit Pa·s.

Surface tension (S): the force per unit length (or surface energy per unit area) acting in the plane of a liquid surface, arising from the extra energy of surface molecules.

Angle of contact (θ): the angle between the tangent to the liquid surface and the solid surface, measured inside the liquid; acute when the liquid wets the solid (water on glass), obtuse when it does not (mercury on glass).

Important Formulas

Pressure: P = F/A   (SI unit: pascal, Pa)

Pressure with depth: P = Pa + ρgh  |  Gauge pressure Pg = ρgh

Pascal’s law (hydraulic lift): F2 = (A2/A1) F1

Equation of continuity: A1v1 = A2v2

Bernoulli’s equation: P + ½ρv2 + ρgh = constant

Torricelli’s law (speed of efflux): v = √(2gh)

Stokes’ law: F = 6πηav  |  Terminal velocity vt = 2a2(ρ − σ)g / 9η

Poiseuille’s flow: Q = πP r4 / 8ηL

Surface tension (slider): S = F / 2l

Excess pressure: drop/cavity (Pi − Po) = 2S/r  |  bubble (two surfaces) = 4S/r

Capillary rise: h = 2S cosθ / (ρga)

NCERT Exercise Solutions (9.1–9.20)

All questions below are reproduced verbatim from the NCERT textbook; the solutions are original and the numerical results are verified against the official NCERT answer key. (Take g = 9.8 m s−2 and 1 atm = 1.01 × 105 Pa unless stated otherwise.)

9.1 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) Pressure in a liquid increases with depth as P = Pa + ρgh. The feet are at a much greater height of blood column below the heart than the brain, so the term ρgh is larger at the feet. Hence the blood pressure (which depends on the height of the blood column) is greater at the feet than at the brain. (b) Unlike a liquid, air is highly compressible and its density is not uniform — it is largest near the ground and falls rapidly with height. So pressure does not fall linearly with height. Most of the air’s mass lies in the lower layers; about half of it lies below ~6 km. Therefore the pressure halves within the first ~6 km even though thin air extends well beyond 100 km. (c) Although pressure is defined as force divided by area, the “force” used is only the component of force normal to the area, and in a fluid at rest this normal force is the same in all directions at a point. Pressure has no associated direction, so it is a scalar.

9.2 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. (Put differently, water wets glass while mercury does not.) (c) Surface tension of a liquid is independent of the area of the surface (d) Water with detergent disolved in it should have small angles of contact. (e) A drop of liquid under no external forces is always spherical in shape

ANSWER (a) The angle of contact is decided by the relative strength of cohesive forces (liquid–liquid) and adhesive forces (liquid–solid). For mercury–glass the cohesive force dominates (Ssl > Sla), giving an obtuse angle; for water–glass the adhesive force dominates (Ssl < Sla), giving an acute angle. (b) Water adheres to glass more strongly than its own molecules attract one another, so it spreads to maximise contact (acute angle, wetting). Mercury’s molecules attract each other more strongly than they attract glass, so mercury pulls itself into drops to minimise contact (obtuse angle, non-wetting). (c) Surface tension is the force per unit length (or energy per unit area) and depends only on the nature of the liquid and the temperature, not on how large the surface is. Increasing the area brings more molecules to the surface but does not change the force per unit length, so S stays the same. (d) Detergent is a wetting (surface-active) agent; it lowers the surface tension and reduces the angle of contact. A small angle of contact lets the water penetrate cloth fibres and dirt more effectively, which is why detergents clean well. (e) In the absence of external forces, a liquid drop minimises its surface energy by taking the shape with the least surface area for a given volume. That shape is a sphere, so the drop is spherical.

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) decreases — surface tension usually falls as temperature rises. (b) Viscosity of gases increases with temperature, whereas viscosity of liquids decreases with temperature. (c) For solids the shearing force is proportional to shear strain; for fluids it is proportional to rate of shear strain. (d) The increase in flow speed at a constriction follows conservation of mass (the equation of continuity, Av = constant). (e) Turbulence in the small wind-tunnel model occurs at a greater speed than for an actual plane (a smaller body needs a higher speed to reach the same critical Reynolds number).

9.4 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) The size of the needle of a syringe controls flow rate better than the thumb pressure exerted by a doctor while administering an injection (d) A fluid flowing out of a small hole in a vessel results in a backward thrust on the vessel (e) A spinning cricket ball in air does not follow a parabolic trajectory

ANSWER (a) Blowing over the paper increases the air speed above it, so by Bernoulli’s principle the pressure above drops below the (higher) pressure beneath. This net upward pressure difference holds the paper up; blowing underneath would only push it up further out of the horizontal. (b) By the equation of continuity (Av = constant), narrowing the gaps with the fingers greatly reduces the cross-sectional area, so the water’s speed increases sharply and gushes out as fast jets. (c) From Bernoulli/continuity, flow rate depends very strongly (with the fourth power, by Poiseuille’s law) on the radius of the opening but only linearly on the applied pressure. A small change in needle size therefore controls the flow rate far more effectively than the thumb pressure. (d) The fluid leaving the hole carries momentum forward. By conservation of momentum (Newton’s third law), the vessel experiences an equal and opposite backward thrust — the same principle as rocket propulsion. (e) A spinning ball drags air with it, making the air speed (and hence the pressure) unequal on its two sides. This pressure difference produces a sideways force (the Magnus effect), so the ball deviates from the simple parabolic path of a projectile.

9.5 A 50 kg girl wearing high heel 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 Given: m = 50 kg, diameter = 1.0 cm ⇒ radius r = 0.5 cm = 0.5 × 10−2 m, g = 9.8 m s−2. Area of heel: A = πr2 = π × (0.5 × 10−2)2 = π × 0.25 × 10−4 = 7.85 × 10−5 m2. Force: F = mg = 50 × 9.8 = 490 N. Pressure: P = F/A = 490 / (7.85 × 10−5) = 6.2 × 106 Pa. (Matches the NCERT answer.)

9.6 Toricelli’s barometer used mercury. Pascal duplicated it using French wine of density 984 kg m–3. Determine the height of the wine column for normal atmospheric pressure.

ANSWER Given: ρ = 984 kg m−3, Pa = 1.013 × 105 Pa, g = 9.8 m s−2. The barometer balances the atmosphere with a fluid column: Pa = ρgh. h = Pa / (ρg) = (1.013 × 105) / (984 × 9.8) = (1.013 × 105) / 9643.2 = 10.5 m. The wine column is far taller than the 0.76 m mercury column because wine is about 13.6 times less dense than mercury.

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 Given: maximum stress = 109 Pa, depth h = 3 km = 3 × 103 m, ρsea water ≈ 1.03 × 103 kg m−3, g = 9.8 m s−2. Gauge pressure at that depth: P = ρgh = (1.03 × 103) × 9.8 × (3 × 103) = 3.03 × 107 Pa ≈ 3 × 107 Pa. Conclusion: The pressure at the ocean bed (~3 × 107 Pa) is far smaller than the maximum stress the structure can bear (109 Pa). Hence the structure is suitable.

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 Given: m = 3000 kg, A = 425 cm2 = 425 × 10−4 m2 = 0.0425 m2, g = 9.8 m s−2. By Pascal’s law the pressure on the small piston equals the pressure on the load piston. P = F/A = mg/A = (3000 × 9.8) / 0.0425 = 29400 / 0.0425 = 6.92 × 105 Pa.

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 Given: height of water hw = 10.0 cm, height of spirit hs = 12.5 cm; mercury levels are equal in both arms. Since the mercury surfaces are at the same level, the pressure due to the water column equals the pressure due to the spirit column above mercury: ρw g hw = ρs g hs. Specific gravity of spirit = ρs/ρw = hw/hs = 10.0 / 12.5 = 0.800.

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 New column heights: water = 10.0 + 15.0 = 25.0 cm; spirit = 12.5 + 15.0 = 27.5 cm. Let the difference in mercury levels be x (mercury rises in the arm with the lighter spirit). Balancing pressures at the lower common mercury level: ρw(25.0) = ρs(27.5) + ρHg·x  (in cm of water, with ρs = 0.8 ρw, ρHg = 13.6 ρw). 25.0 = 0.8 × 27.5 + 13.6 x ⇒ 25.0 = 22.0 + 13.6 x ⇒ 13.6 x = 3.0. x = 3.0 / 13.6 = 0.221 cm. The mercury rises in the arm containing spirit; the difference in mercury levels is 0.221 cm.

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 applies only to steady (streamline), non-viscous, incompressible flow. Water in a river rapid is highly turbulent, with rapidly fluctuating velocity and pressure and significant energy loss. The conditions for Bernoulli’s principle are not met, so it cannot be used there.

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

ANSWER No, it does not matter — provided the atmospheric pressure at the two points is the same. Bernoulli’s equation involves the difference of pressures between two points. Replacing each absolute pressure P by the gauge pressure (P − Pa) subtracts the same constant Pa from both sides, leaving the difference unchanged. It would matter only if the atmospheric pressures at the two points were significantly different.

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 × 10–3 kg s–1, what is the pressure difference between the two ends of the tube? (Density of glycerine = 1.3 × 103 kg m–3 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 Given: L = 1.5 m, r = 1.0 cm = 10−2 m, mass flow rate = 4.0 × 10−3 kg s−1, ρ = 1.3 × 103 kg m−3, η = 0.83 Pa·s. Volume flow rate: Q = (mass rate)/ρ = (4.0 × 10−3) / (1.3 × 103) = 3.08 × 10−6 m3 s−1. Poiseuille’s equation: Q = πP r4 / (8ηL) ⇒ P = 8ηLQ / (πr4). P = (8 × 0.83 × 1.5 × 3.08 × 10−6) / (π × (10−2)4) = (3.07 × 10−5) / (3.14 × 10−8) = 9.8 × 102 Pa. Check (Reynolds number): mean speed v = Q/(πr2) = 3.08 × 10−6/(3.14 × 10−4) = 9.8 × 10−3 m s−1. Re = ρv(2r)/η = (1.3 × 103 × 9.8 × 10−3 × 2 × 10−2)/0.83 ≈ 0.3. Since Re « 2000, the flow is indeed laminar.

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 s–1 and 63 m s-1 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 m–3.

ANSWER Given: vupper = 70 m s−1, vlower = 63 m s−1, A = 2.5 m2, ρ = 1.3 kg m−3. Pressure difference (Bernoulli): ΔP = ½ρ(vupper2 − vlower2) = ½ × 1.3 × (702 − 632) = ½ × 1.3 × (4900 − 3969) = ½ × 1.3 × 931 = 605.2 Pa. Lift: F = ΔP × A = 605.2 × 2.5 = 1513 N ≈ 1.5 × 103 N.

9.15 Figures 9.20(a) and (b) refer to the steady flow of a (non-viscous) liquid. Which of the two figures is incorrect? Why?

ANSWER Figure (a) is incorrect. In the figure the liquid column is shown taller (higher pressure) at the constriction. But by the equation of continuity, the flow speed is larger at a constriction (smaller cross-section). By Bernoulli’s equation, where the speed is larger the pressure must be smaller, so the liquid column should be shorter at the narrow part, not taller. Hence figure (a) violates Bernoulli’s principle and is incorrect.

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 min–1, what is the speed of ejection of the liquid through the holes?

ANSWER Given: tube area A1 = 8.0 cm2 = 8.0 × 10−4 m2; 40 holes of diameter 1.0 mm ⇒ radius = 0.5 mm = 0.5 × 10−3 m; tube flow speed v1 = 1.5 m min−1 = 1.5/60 = 0.025 m s−1. Total hole area: A2 = 40 × πr2 = 40 × π × (0.5 × 10−3)2 = 40 × π × 0.25 × 10−6 = 3.14 × 10−5 m2. Continuity: A1v1 = A2v2 ⇒ v2 = A1v1/A2 = (8.0 × 10−4 × 0.025) / (3.14 × 10−5) = (2.0 × 10−5) / (3.14 × 10−5) = 0.64 m s−1.

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 × 10–2 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 Given: weight supported F = 1.5 × 10−2 N, length of slider l = 30 cm = 0.30 m. A soap film has two surfaces, so the surface tension balances the weight over a total length 2l: F = S × 2l. S = F / (2l) = (1.5 × 10−2) / (2 × 0.30) = (1.5 × 10−2) / 0.60 = 2.5 × 10−2 N m−1.

9.18 Figure 9.21 (a) shows a thin liquid film supporting a small weight = 4.5 × 10–2 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 The weight that a film supports depends only on the surface tension and the length of the slider, W = S × 2l, and not on the area of the film. In figures (b) and (c) the same liquid is used at the same temperature and the slider length is the same as in (a). Therefore the weight supported in (b) and in (c) is the same as in (a): 4.5 × 10−2 N in each case.

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 × 10–1 N m–1. The atmospheric pressure is 1.01 × 105 Pa. Also give the excess pressure inside the drop.

ANSWER Given: r = 3.00 mm = 3.00 × 10−3 m, S = 4.65 × 10−1 N m−1, Pa = 1.01 × 105 Pa. A drop has a single liquid surface, so the excess pressure is 2S/r. Excess pressure: Pi − Po = 2S/r = (2 × 4.65 × 10−1) / (3.00 × 10−3) = 0.93 / 0.003 = 310 Pa. Total (absolute) pressure inside: Pi = Pa + 2S/r = 1.01 × 105 + 310 = 1.0131 × 105 Pa. Since the data are correct to three significant figures, Pi ≈ 1.01 × 105 Pa.

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 (20 °C) is 2.50 × 10–2 N m–1? 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 Given: r = 5.00 mm = 5.00 × 10−3 m, S = 2.50 × 10−2 N m−1, depth h = 40.0 cm = 0.40 m, relative density 1.20 ⇒ ρ = 1.20 × 103 kg m−3, Pa = 1.01 × 105 Pa, g = 9.8 m s−2. Soap bubble (two surfaces): excess pressure = 4S/r = (4 × 2.50 × 10−2) / (5.00 × 10−3) = 0.1 / 0.005 = 20.0 Pa. Air bubble in the liquid (one surface): excess pressure = 2S/r = (2 × 2.50 × 10−2) / (5.00 × 10−3) = 0.05 / 0.005 = 10.0 Pa. Pressure outside the air bubble: Po = Pa + ρgh = 1.01 × 105 + (1.20 × 103 × 9.8 × 0.40) = 1.01 × 105 + 4704 = 1.057 × 105 Pa. Pressure inside the air bubble: Pi = Po + 2S/r = 1.057 × 105 + 10 = 1.057 × 105 Pa ≈ 1.06 × 105 Pa (the excess 10 Pa is negligible to three significant figures).

Extra Practice Questions

Short Answer Type Questions

Q1. Why is pressure a scalar quantity?

ANSWERIn a fluid at rest the force on any area is always normal to it and is the same in all directions at a point. Pressure has only magnitude and no fixed direction, so it is a scalar.

Q2. State Pascal’s law and give one application.

ANSWERA pressure applied to an enclosed fluid is transmitted undiminished to every part of the fluid and the container walls. It is used in the hydraulic lift, where a small force on a small piston balances a large load on a large piston.

Q3. Why does the viscosity of a liquid fall but that of a gas rise with temperature?

ANSWERIn a liquid, viscosity is due to intermolecular attraction, which weakens as molecules move faster at higher temperature, so η falls. In a gas, viscosity is due to the transfer of momentum between layers by moving molecules; higher temperature increases molecular motion and this transfer, so η rises.

Q4. On what factors does the terminal velocity of a sphere falling through a viscous fluid depend?

ANSWERFrom vt = 2a2(ρ − σ)g / 9η, it depends directly on the square of the radius and on the density difference between sphere and fluid, and inversely on the viscosity of the fluid.

Q5. Why does the excess pressure inside a soap bubble equal 4S/r while inside a liquid drop it is 2S/r?

ANSWERA liquid drop has only one liquid–air surface, giving excess pressure 2S/r. A soap bubble is a thin film with two liquid–air surfaces (inner and outer), so the surface effect is doubled, giving 4S/r.

Long Answer Type Questions

Q1. State and explain Bernoulli’s principle, and use it to explain the lift on an aircraft wing.

ANSWERBernoulli’s principle states that for the steady, streamline flow of a non-viscous, incompressible fluid, the sum of pressure energy, kinetic energy and potential energy per unit volume is constant along a streamline: P + ½ρv2 + ρgh = constant. It is a statement of conservation of energy for fluid flow. An aircraft wing (aerofoil) is shaped so that air travels faster over the curved upper surface than along the flatter lower surface. By Bernoulli’s equation, the faster flow above the wing has lower pressure and the slower flow below has higher pressure. This pressure difference acts upward over the wing area, producing the dynamic lift that supports the weight of the aircraft.

Q2. Explain the phenomenon of capillary rise and derive the expression for the height to which a liquid rises in a capillary tube.

ANSWERWhen a narrow tube is dipped in a wetting liquid such as water, the liquid rises in the tube because of surface tension and an acute angle of contact. The liquid surface (meniscus) inside the tube is concave, so the pressure just below the meniscus is less than atmospheric by 2S cosθ/a, where a is the tube radius and θ the contact angle. To restore balance, liquid rises until the pressure of the raised column ρgh equals this deficit. Equating, ρgh = 2S cosθ/a, which gives the capillary rise h = 2S cosθ / (ρga). The rise is greater for a thinner tube (smaller a) and for liquids of higher surface tension. For a non-wetting liquid such as mercury, cosθ is negative and the liquid is depressed instead of rising.

Q3. Describe how pressure varies with depth in a fluid and explain the hydrostatic paradox.

ANSWERConsider a fluid of density ρ at rest. For a vertical column of base area A and height h, the difference between the upward force at the bottom and the downward force at the top balances the weight of the fluid, giving P2 − P1 = ρgh. Taking the top surface open to the atmosphere, the pressure at depth h is P = Pa + ρgh, so pressure increases linearly with depth and is the same at all points at the same horizontal level. Because this expression contains only the depth and not the area or shape of the container, vessels of different shapes filled to the same height show the same liquid level when joined at the bottom, even though they hold different amounts of liquid. This counter-intuitive result is called the hydrostatic paradox.

MCQs & Assertion–Reason

1. The SI unit of pressure, the pascal, is equal to:

(a) N m    (b) N m−2    (c) N m2    (d) kg m s−2

2. The pressure at a depth h below the free surface of a liquid of density ρ (open to the atmosphere) is:

(a) ρgh    (b) Pa − ρgh    (c) Pa + ρgh    (d) Pa/ρgh

3. A hydraulic lift works on the principle of:

(a) Archimedes’ principle    (b) Bernoulli’s principle    (c) Pascal’s law    (d) equation of continuity

4. The equation of continuity Av = constant is a statement of conservation of:

(a) energy    (b) momentum    (c) mass    (d) charge

5. According to Bernoulli’s principle, where a fluid flows faster in a horizontal pipe, the pressure is:

(a) higher    (b) lower    (c) unchanged    (d) zero

6. The speed of efflux of a liquid from a small hole at depth h below the open surface is:

(a) gh    (b) 2gh    (c) √(gh)    (d) √(2gh)

7. The viscous drag on a small sphere of radius a moving with speed v through a fluid of viscosity η is (Stokes’ law):

(a) 6πηav    (b) 4πηav    (c) 6πηa2v    (d) πηav

8. The SI unit of coefficient of viscosity is:

(a) Pa    (b) Pa s    (c) N m−1    (d) m2 s−1

9. The excess pressure inside a spherical liquid drop of radius r and surface tension S is:

(a) S/r    (b) 2S/r    (c) 4S/r    (d) S/2r

10. The angle of contact of water with clean glass is:

(a) obtuse    (b) exactly 90°    (c) acute    (d) exactly 180°

Answer key: 1-(b), 2-(c), 3-(c), 4-(c), 5-(b), 6-(d), 7-(a), 8-(b), 9-(b), 10-(c).

For each Assertion–Reason question, choose: (A) Both true and the Reason correctly explains the Assertion; (B) Both true but the Reason is not the correct explanation; (C) Assertion true, Reason false; (D) Assertion false, Reason true.

A-R 1. Assertion: Pressure is a scalar quantity.

Reason: The force used in defining pressure is the component normal to the area, and at a point in a fluid at rest it is the same in all directions.

A-R 2. Assertion: A small force applied on the smaller piston of a hydraulic lift can balance a heavy car on the larger piston.

Reason: Pressure applied to an enclosed fluid is transmitted undiminished to all parts of the fluid.

A-R 3. Assertion: The viscosity of a gas increases as its temperature is raised.

Reason: The viscosity of a liquid also increases with rise in temperature.

A-R 4. Assertion: Water rises higher in a capillary tube of smaller bore.

Reason: The capillary rise is inversely proportional to the radius of the tube.

A-R 5. Assertion: The excess pressure inside a soap bubble is greater than that inside a liquid drop of the same radius and surface tension.

Reason: A soap bubble has two liquid surfaces while a drop has only one.

Answer key: 1-(A), 2-(A), 3-(C), 4-(A), 5-(A).

Common Mistakes to Avoid

Watch out for these

  • Treating pressure as a vector — it is a scalar; only the normal component of force is used.
  • Confusing gauge pressure (ρgh) with absolute pressure (Pa + ρgh) in numericals.
  • Using 2S/r for a soap bubble — a bubble has two surfaces, so use 4S/r; a drop or an air bubble in liquid has one surface (2S/r).
  • Forgetting that a slider film has two surfaces, so S = F/2l, not F/l.
  • Applying Bernoulli’s equation to turbulent or viscous flow — it holds only for steady, non-viscous, incompressible flow.
  • Not converting units (cm2 to m2, mm to m, m min−1 to m s−1) before substituting.
  • Thinking liquid pressure depends on the shape or amount of liquid — it depends only on depth (hydrostatic paradox).

Exam tips for this chapter

Memorise the key formulas (P = Pa + ρgh, Av = constant, Bernoulli’s equation, v = √(2gh), Stokes’ law, terminal velocity, excess-pressure 2S/r and 4S/r, capillary rise) and always write them before plugging in numbers. In numericals, list the given data with units, convert everything to SI first, then substitute — this is how the NCERT solutions are scored. For “explain why” questions, name the principle (Bernoulli, Pascal, continuity, surface tension) explicitly and give one clear reason. Remember the count of surfaces (1 for a drop or air bubble, 2 for a soap bubble or film) — it is the most common slip in surface-tension problems.

Frequently Asked Questions

What is Class 11 Physics Chapter 9 about?

Chapter 9, Mechanical Properties of Fluids, studies the behaviour of liquids and gases at rest and in motion — pressure and its variation with depth, Pascal’s law and hydraulic machines, the equation of continuity, Bernoulli’s principle, viscosity and Stokes’ law, and surface tension with capillary rise.

How many exercises are there in Class 11 Physics Chapter 9?

There are 20 numbered exercises (9.1 to 9.20), mixing conceptual “explain why” questions with numerical problems on pressure, hydraulics, Bernoulli’s principle, viscosity and surface tension. All 20 are solved step by step on this page.

Why is the excess pressure inside a soap bubble 4S/r?

A soap bubble is a thin film with two liquid–air surfaces (inner and outer). Each contributes 2S/r, so the total excess pressure is 4S/r, twice that of a single-surface liquid drop or an air bubble inside a liquid, which is 2S/r.

Are these Class 11 Physics Chapter 9 solutions free?

Yes. All solutions are free and follow the official NCERT textbook for session 2026–27, with every numerical answer verified against the NCERT answer key.

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