Exercises — Mechanical Properties of Fluids

1. 9.13 marksNCERT Cl-11 Physics Part II, Ch9 Exercises, 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.

   Hint. For (a) and (b), think about the weight of the column of fluid above the point. For (b), ask whether air has the same density all the way up. For (c), ask whether pressure at a point picks out any particular direction.

   Step 1 — (a) Blood pressure at the feet. Pressure in a fluid at rest increases with depth as p = p₀ + ρgh. The feet lie roughly 1.5 m below the brain, so the blood there carries the extra weight of that whole column above it. Since the additional pressure is ρgh with ρ ≈ 1060 kg m⁻³, this comes to around 1.5 × 10⁴ Pa more at the feet than at the brain.

   Step 2 — (b) Atmospheric pressure halving in 6 km. A liquid is very nearly incompressible, so its density stays constant with depth. Air is not — it is compressible, and the layers near the ground are squeezed by the weight of everything above them.

   The density of air therefore falls off as you rise, so the upper 94 km of atmosphere contains far less mass than the lowest 6 km. Because pressure counts the weight above you rather than the height above you, most of the atmosphere's weight sits in that first few kilometres, which is why half the pressure is gone by 6 km.

   Step 3 — (c) Why hydrostatic pressure is a scalar. Force is a vector and area has an orientation, so the ratio looks as though it should be a vector. But in a fluid at rest, the pressure at a point is the same in every direction — turn a tiny test surface any way you like and the force per unit area on it is unchanged.

   A quantity that has no preferred direction cannot be a vector, since a vector must point somewhere. Pressure has magnitude only, so it is a scalar.

   ✦ (a) the extra weight of the blood column, ρgh; (b) air is compressible so most of its mass lies low down; (c) pressure at a point is the same in all directions, so it has no direction to be a vector.

   Where students slip. Answering (b) with "gravity gets weaker with height". Gravity changes negligibly over 6 km. The real cause is the compressibility of air, which packs most of the atmosphere's mass into the lowest few kilometres.

2. 9.25 marksNCERT Cl-11 Physics Part II, Ch9 Exercises, 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. (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 dissolved in it should have small angles of contact. (e) A drop of liquid under no external forces is always spherical in shape

   Hint. Four of these five turn on one comparison: is the force between liquid molecules and solid molecules (adhesive) stronger or weaker than the force between liquid molecules themselves (cohesive)? For (c) and (e), think about what surface tension is per unit of.

   Step 1 — The one idea behind (a), (b) and (d). Every liquid-solid pair has two competing forces: cohesion between liquid molecules, and adhesion between liquid and solid molecules. Whichever wins decides the shape the liquid takes.

   Step 2 — (a) Angles of contact. For mercury on glass the cohesive force between mercury atoms is much stronger than the adhesion to glass, so the mercury pulls itself together and the surface curves away from the glass, giving an obtuse angle of contact.

   For water on glass adhesion to the glass wins over cohesion, so the water is drawn along the surface and the angle is acute.

   Step 3 — (b) Spreading against beading. This is the same comparison seen from the outside. Water spreads because adhesion to glass is stronger, so increasing the contact area is energetically favourable. Mercury beads up because cohesion dominates, and it minimises contact with the glass by pulling into drops.

   Step 4 — (c) Independence from area. Surface tension is defined as force per unit length of a line drawn in the surface, or equivalently energy per unit area. Both definitions are ratios per unit, so the total surface can be made larger or smaller without changing the value. It depends on the nature of the liquid and on temperature, not on how much surface there is.

   Step 5 — (d) Detergent and the angle of contact. A detergent must make water penetrate the fine spaces between fibres, and liquid rises into a narrow gap only when it wets the surface. Wetting means a small angle of contact, so detergents are designed to reduce the angle of contact of water — which is why they work.

   Step 6 — (e) Why a free drop is spherical. With no external force acting, the only thing shaping the drop is surface tension, which acts to minimise surface area. For a given volume, the sphere is the shape with the least possible surface area, so that is what the drop settles into.

   ✦ (a) cohesion beats adhesion for mercury, adhesion wins for water; (b) the same comparison; (c) surface tension is per unit length, so area does not enter; (d) small angle means wetting, which is what a detergent needs; (e) a sphere minimises area for a given volume.

   Where students slip. Saying in (c) that a bigger surface has more molecules so more tension. Surface tension is a force per unit length — doubling the surface doubles both the force and the length, leaving the ratio unchanged.

3. 9.35 marksNCERT Cl-11 Physics Part II, Ch9 Exercises, 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) (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)

   Hint. For (b), remember that liquids and gases carry momentum between layers by completely different mechanisms — one by intermolecular forces, the other by molecules crossing between layers. For (d), ask which principle gives you the speed and which then gives you the pressure.

   Step 1 — (a) decreases. Raising the temperature increases the kinetic energy of the molecules and weakens the net inward pull on a surface molecule, so surface tension falls as temperature rises.

   Step 2 — (b) gases: increases; liquids: decreases. These go opposite ways because the mechanism is different in each. In a liquid, viscosity comes from intermolecular attraction between layers, and heating weakens that attraction, so viscosity decreases.

   In a gas the molecules are far apart and viscosity comes from molecules crossing between layers and carrying momentum with them. Heating makes them move faster, so more momentum is transferred and viscosity increases.

   Step 3 — (c) shear strain; rate of shear strain. A solid resists being deformed by a fixed amount, so for a solid the shearing stress is proportional to the shear strain itself. A fluid offers no permanent resistance to a shape change but does resist the speed of that change, so for a fluid the shearing stress is proportional to the rate of shear strain. This is the essential difference between a solid and a fluid.

   Step 4 — (d) conservation of mass. The speeding-up at a constriction follows from the equation of continuity, Av = constant, which is a statement of conservation of mass. Bernoulli's principle then tells you what the pressure does as a consequence — so continuity gives the speed and Bernoulli gives the pressure.

   Step 5 — (e) greater. Turbulence sets in at a fixed value of the Reynolds number, which depends on the product of speed and size. The wind-tunnel model is smaller than the real aircraft, so to reach the same Reynolds number it must be tested at a greater speed.

   ✦ (a) decreases (b) increases; decreases (c) shear strain; rate of shear strain (d) conservation of mass (e) greater.

   Where students slip. Answering (d) with Bernoulli's principle. Bernoulli explains the pressure drop that accompanies the speeding-up, but the speeding-up itself is forced by continuity — mass has to keep flowing at the same rate through a smaller area.

4. 9.45 marksNCERT Cl-11 Physics Part II, Ch9 Exercises, 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) 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

   Hint. Parts (a), (b) and (e) are Bernoulli; (c) is Poiseuille's fourth-power dependence on radius; (d) is Newton's third law with momentum. For (c), ask what happens to the flow rate if the radius halves.

   Step 1 — (a) Blowing over the paper. Blowing over the top raises the air speed there, and by Bernoulli's principle a higher speed means a lower pressure. The still air underneath is then at the higher pressure, so there is a net upward force that holds the paper horizontal. Blowing underneath would do the opposite and push it down.

   Step 2 — (b) Jets between the fingers. By the equation of continuity, Av = constant. Partly closing the tap with your fingers greatly reduces the area available, so the speed must rise by the same factor to pass the same volume of water each second. The result is fast jets through the gaps.

   Step 3 — (c) Needle size against thumb pressure. For laminar flow through a narrow tube, Poiseuille's equation gives the volume flow rate as

   Q = πΔp r⁴ / (8ηL)

   The flow rate is proportional to the pressure difference but to the fourth power of the radius. Halving the needle's radius therefore cuts the flow to one sixteenth, which is far more leverage than a doctor's thumb can supply. That is why the needle size controls the rate.

   Step 4 — (d) Backward thrust. The fluid leaving the hole carries momentum forwards. Since no external force acts on the vessel-plus-fluid system, its total momentum is conserved, so the vessel must gain equal and opposite momentum backwards. Equivalently, by Newton's third law the escaping fluid pushes back on the vessel.

   Step 5 — (e) The spinning cricket ball. A spinning ball drags a layer of air around with it. On one side that dragged air moves with the airflow and on the other against it, so the air speed is higher on one side than the other. By Bernoulli's principle the faster side is at lower pressure, giving a sideways force — the Magnus effect — which curves the ball away from a parabola.

   ✦ (a) faster air above means lower pressure above; (b) continuity forces the speed up; (c) flow rate goes as r⁴, so radius dominates; (d) momentum conservation, or the third law; (e) spin makes the air speed unequal on the two sides, so pressure is unequal.

   Where students slip. Explaining (c) by saying a narrower needle simply resists more. The point is quantitative: the fourth-power dependence means a small change in radius overwhelms any change in applied pressure.

5. 9.52 marksNCERT Cl-11 Physics Part II, Ch9 Exercises, Q9.5

   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?

   Hint. Balancing on a single heel means the whole weight passes through that one small circle. Convert the diameter to a radius in metres before squaring.

   Step 1 — Find the force on the floor. The girl balances on one heel, so the entire weight acts through it:

   F = mg = 50 × 9.8 = 490 N

   Step 2 — Find the area of the heel. The diameter is 1.0 cm, so the radius is 0.5 cm = 0.005 m:

   A = πr² = π(0.005)² = 7.854 × 10⁻⁵ m²

   Step 3 — Compute the pressure.

   p = F/A = 490/(7.854 × 10⁻⁵) = 6.24 × 10⁶ Pa

   Step 4 — Appreciate the size of that number. This is over sixty times atmospheric pressure, from a person of ordinary weight. The reason is entirely the tiny area — the same 490 N spread over a flat shoe sole of about 150 cm² would give only about 3.3 × 10⁴ Pa, roughly two hundred times less. It is why stiletto heels damage soft floors while a much heavier person in flat shoes does not.

   ✦ p ≈ 6.24 × 10⁶ Pa (about 6.2 × 10⁶ N m⁻²).

   Where students slip. Using the diameter in place of the radius, which makes the area four times too large and the pressure four times too small. Halve the diameter first, then square.

6. 9.62 marksNCERT Cl-11 Physics Part II, Ch9 Exercises, Q9.6

   Torricelli'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.

   Hint. A barometer balances atmospheric pressure against the weight of a fluid column, so set p = ρgh and solve for h. Normal atmospheric pressure is 1.013 × 10⁵ Pa.

   Step 1 — Write the balance condition. In a barometer the atmospheric pressure is supported entirely by the column of liquid:

   p = ρgh

   Step 2 — Rearrange for the height.

   h = p/(ρg)

   Step 3 — Substitute, using normal atmospheric pressure.

   h = (1.013 × 10⁵)/(984 × 9.8) = (1.013 × 10⁵)/(9643) = 10.5 m

   Step 4 — Compare with mercury, and see why mercury is used. Mercury has a density of 13.6 × 10³ kg m⁻³, about fourteen times that of wine, so its column is about fourteen times shorter — the familiar 0.76 m.

   Since h is inversely proportional to density, a low-density liquid needs an absurdly tall tube. Pascal's wine barometer had to be over ten metres high, which is exactly why every practical barometer uses the densest convenient liquid.

   ✦ h ≈ 10.5 m of wine.

   Where students slip. Reaching for the mercury height of 0.76 m and scaling it by the wrong ratio. Work from p = ρgh directly; height and density are inversely proportional, so the lighter liquid needs the taller column.

7. 9.73 marksNCERT Cl-11 Physics Part II, Ch9 Exercises, Q9.7

   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.

   Hint. Work out the pressure the sea actually exerts at 3 km depth and compare it with the quoted limit. Sea water has a density of about 1.03 × 10³ kg m⁻³.

   Step 1 — Compute the gauge pressure at the sea bed. The pressure due to the water column alone is

   p = ρgh = (1.03 × 10³)(9.8)(3000) = 3.03 × 10⁷ Pa

   Step 2 — Add atmospheric pressure for the absolute value. The atmosphere presses on the sea surface too, so the total is

   p_total = 3.03 × 10⁷ + 1.01 × 10⁵ ≈ 3.04 × 10⁷ Pa

   The atmospheric contribution is only about 0.3% here, which is why it is often neglected at such depths.

   Step 3 — Compare with the design limit. The structure is built for 10⁹ Pa, and the sea exerts about 3.0 × 10⁷ Pa. Since 3.0 × 10⁷ is roughly 33 times smaller than 10⁹, the structure is comfortably within its limit.

   ✦ Yes — the pressure is about 3.0 × 10⁷ Pa, well below the 10⁹ Pa the structure can withstand, so it is suitable.

   Where students slip. Using the density of fresh water, 1.0 × 10³ kg m⁻³. Sea water is denser at about 1.03 × 10³ kg m⁻³; here it changes little, but in problems near a design limit it matters.

8. 9.82 marksNCERT Cl-11 Physics Part II, Ch9 Exercises, Q9.8

   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?

   Hint. By Pascal's law the pressure is transmitted undiminished through the fluid, so the pressure at the small piston equals the pressure at the large one. That means you never need the small piston's area.

   Step 1 — Apply Pascal's law. Pressure applied to an enclosed fluid is transmitted undiminished to every part of the fluid. So the pressure the small piston must bear is exactly the pressure needed at the large piston:

   p_small = p_large

   This is why the area of the smaller piston is not given — it is not needed.

   Step 2 — Compute the pressure at the load-carrying piston. F = mg = 3000 × 9.8 = 2.94 × 10⁴ N A = 425 cm² = 425 × 10⁻⁴ m² = 4.25 × 10⁻² m²

   p = F/A = (2.94 × 10⁴)/(4.25 × 10⁻²) = 6.92 × 10⁵ Pa

   Step 3 — Note what a hydraulic lift actually multiplies. The pressure is the same on both pistons; what differs is the force, because the areas differ. A small force on a small piston produces the same pressure, which then acts over the large piston's much bigger area to give a large force. Pressure is transmitted, force is multiplied.

   ✦ Maximum pressure ≈ 6.92 × 10⁵ Pa.

   Where students slip. Trying to find the small piston's area first. Pascal's law makes the pressures equal, so the answer follows from the large piston alone — the missing area is missing on purpose.

9. 9.93 marksNCERT Cl-11 Physics Part II, Ch9 Exercises, Q9.9

   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?

   Hint. The mercury levels being equal is the whole point — it means the pressure at the top of the mercury is the same on both sides. Write that equality and the mercury never enters the calculation.

   Step 1 — Use the fact that the mercury levels are equal. If the mercury stands at the same height in both arms, the pressure at the mercury surface must be identical on the two sides. Otherwise the mercury would move until it was.

   Step 2 — Write the pressure on each side. Each mercury surface carries atmospheric pressure plus the weight of the liquid column above it:

   Water side: p₀ + ρ_w g h_w Spirit side: p₀ + ρ_s g h_s

   Step 3 — Equate them. The atmospheric term p₀ and the g cancel from both sides, which is why neither value is needed:

   ρ_w h_w = ρ_s h_s ρ_w (10.0) = ρ_s (12.5)

   Step 4 — Solve for the specific gravity. Specific gravity is the density relative to water, so

   ρ_s/ρ_w = 10.0/12.5 = 0.8

   Step 5 — Check it is sensible. The spirit column is taller than the water column, and a taller column of a lighter liquid is needed to produce the same pressure. A specific gravity below 1 is therefore exactly what we should expect.

   ✦ Specific gravity of spirit = 0.8.

   Where students slip. Writing the ratio the wrong way up as 12.5/10.0 = 1.25, which would make the spirit denser than water. The taller column must belong to the *lighter* liquid, so the answer has to come out below 1.

10. 9.105 marksNCERT Cl-11 Physics Part II, Ch9 Exercises, Q9.10

    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)

    Hint. Add the extra liquid to each arm first, then compare the two new column pressures. The mercury will shift until the difference in its own levels makes up the imbalance.

    Step 1 — Find the new column heights. Water: 10.0 + 15.0 = 25.0 cm Spirit: 12.5 + 15.0 = 27.5 cm

    Step 2 — Compare the pressures the two columns now exert. Taking water's density as 1 and spirit's as 0.8 (from Q9.9), and measuring pressures in centimetres of water:

    Water side: 25.0 × 1 = 25.0 cm of water Spirit side: 27.5 × 0.8 = 22.0 cm of water

    The water side is now the heavier by 25.0 − 22.0 = 3.0 cm of water.

    Step 3 — Let the mercury restore balance. Because the two sides no longer balance, mercury is pushed down on the water side and up on the spirit side until the difference in mercury levels supplies the missing 3.0 cm of water pressure. If that difference is h, measured in centimetres:

    13.6 × h = 3.0 h = 3.0/13.6 = 0.22 cm

    Step 4 — Note why the answer is so small. Mercury is 13.6 times denser than water, so it takes only a very short mercury column to balance a much taller water column. That large density is exactly why mercury is chosen for this kind of tube.

    ✦ The mercury levels differ by about 0.22 cm (roughly 2 mm), standing lower in the water arm.

    Where students slip. Forgetting to scale the spirit column by its specific gravity of 0.8 before comparing. Comparing 25.0 cm against 27.5 cm directly reverses which side is heavier and gives the wrong direction as well as the wrong size.

11. 9.112 marksNCERT Cl-11 Physics Part II, Ch9 Exercises, Q9.11

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

    Hint. List the assumptions Bernoulli's equation is derived under, then ask which of them a river rapid satisfies.

    Step 1 — Recall what Bernoulli's equation assumes. The derivation requires the flow to be:

    • steady — the velocity at any fixed point does not change with time • streamline (laminar) — the layers do not mix • non-viscous — no energy is lost to internal friction • incompressible — the density is constant

    Step 2 — Check a river rapid against that list. Water in a rapid is visibly churning, with eddies, whirlpools and white water. That is turbulent flow, not streamline flow, and the velocity at any given point is changing all the time, so the flow is not steady either.

    Step 3 — Note what turbulence does to the energy. Bernoulli's equation is a statement of energy conservation along a streamline. In turbulent flow a significant amount of mechanical energy is dissipated as heat and sound through internal friction, so the total is not conserved along the path.

    ✦ No. A rapid is turbulent and unsteady rather than streamline and steady, and it dissipates energy through internal friction, so the assumptions behind Bernoulli's equation do not hold.

    Where students slip. Answering yes on the grounds that water is incompressible. Incompressibility is only one of four assumptions, and it is the failure of the steady and streamline conditions that rules Bernoulli out here.

12. 9.122 marksNCERT Cl-11 Physics Part II, Ch9 Exercises, Q9.12

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

    Hint. Gauge pressure and absolute pressure differ by a constant. Ask what happens to a constant that appears on both sides of an equation.

    Step 1 — Write down how the two pressures are related. Absolute pressure and gauge pressure differ by atmospheric pressure, which is a constant:

    p_absolute = p_gauge + p_atmospheric

    Step 2 — See what Bernoulli's equation actually uses. Bernoulli's equation compares two points on a streamline:

    p₁ + ½ρv₁² + ρgh₁ = p₂ + ½ρv₂² + ρgh₂

    What matters is the difference between p₁ and p₂, since the equation is an equality between two sides that each contain a pressure term.

    Step 3 — Substitute and cancel. Replacing each absolute pressure by its gauge value adds p_atmospheric to both sides of the equation. The same constant on both sides cancels, leaving the equation unchanged.

    Step 4 — State the one condition. It does not matter, provided the same choice is used at every point. Mixing an absolute pressure at one point with a gauge pressure at another leaves an uncancelled 10⁵ Pa in the working, which is a large error.

    ✦ No, it makes no difference, because the constant atmospheric term appears on both sides and cancels — as long as you use the same convention throughout.

    Where students slip. Concluding that absolute pressure must always be used because it is the "real" pressure. Bernoulli's equation only ever compares pressures, so any constant offset applied consistently cancels out.

13. 9.135 marksNCERT Cl-11 Physics Part II, Ch9 Exercises, Q9.13 — the laminar-flow check needs the Reynolds number, which is NOT in the 2026-27 chapter

    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].

    Hint. You are given a mass per second but Poiseuille's equation needs a volume per second — divide by the density first. For the bracketed check you need the Reynolds number, which this chapter no longer contains; the formula is given in the solution.

    Step 1 — Convert the mass flow rate to a volume flow rate. Poiseuille's equation is written in terms of volume per second, so divide the given mass rate by the density:

    Q = (4.0 × 10⁻³)/(1.3 × 10³) = 3.08 × 10⁻⁶ m³ s⁻¹

    Step 2 — Apply Poiseuille's equation. For steady laminar flow through a horizontal tube of radius r and length L:

    Δp = 8ηLQ/(πr⁴)

    Substituting η = 0.83 Pa s, L = 1.5 m, r = 1.0 cm = 0.01 m:

    Δp = (8 × 0.83 × 1.5 × 3.08 × 10⁻⁶)/(π × (0.01)⁴) = (3.07 × 10⁻⁵)/(3.14 × 10⁻⁸) = 976 Pa

    Δp ≈ 9.8 × 10² Pa

    Step 3 — The bracketed check, and a note about it. The Reynolds number does not appear anywhere in the 2026-27 chapter, so the tool this part of the question asks for is no longer taught. Here it is:

    Re = ρvD/η

    First find the average flow speed:

    v = Q/A = (3.08 × 10⁻⁶)/(π × (0.01)²) = 9.79 × 10⁻³ m s⁻¹

    Then, with D = 2r = 0.02 m:

    Re = (1.3 × 10³ × 9.79 × 10⁻³ × 0.02)/0.83 = 0.31

    Step 4 — Interpret it. Flow is laminar for Re below about 1000 and turbulent above roughly 2000. A value of 0.31 is smaller than that threshold by a factor of thousands, so the assumption of laminar flow is very safely correct — which is what justified using Poiseuille's equation in Step 2.

    ✦ Δp ≈ 9.8 × 10² Pa, and with Re ≈ 0.31 the flow is firmly laminar, so the assumption holds.

    Where students slip. Putting the mass flow rate straight into Poiseuille's equation without dividing by the density. The equation needs cubic metres per second, and skipping the conversion inflates the answer by a factor of 1300.

14. 9.143 marksNCERT Cl-11 Physics Part II, Ch9 Exercises, Q9.14

    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.

    Hint. The wing is essentially horizontal, so the height term in Bernoulli's equation drops out and you are left with a pressure difference set by the two speeds. Multiply that pressure difference by the area to get a force.

    Step 1 — Apply Bernoulli's equation across the wing. The upper and lower surfaces are at essentially the same height, so the ρgh terms cancel and Bernoulli's equation reduces to

    p₁ + ½ρv₁² = p₂ + ½ρv₂²

    Rearranging for the pressure difference between the lower and upper surfaces:

    Δp = ½ρ(v_upper² − v_lower²)

    Step 2 — Substitute the speeds.

    Δp = ½ × 1.3 × (70² − 63²) = 0.65 × (4900 − 3969) = 0.65 × 931 = 605 Pa

    Step 3 — Convert the pressure difference into a force. Pressure is force per unit area, so multiplying by the wing area gives the lift:

    F = Δp × A = 605 × 2.5 = 1513 N

    Step 4 — Check the direction is upward. The air moves faster over the upper surface, so by Bernoulli the pressure there is lower. The higher pressure underneath therefore pushes the wing up, which is why the lift is upward and why aerofoils are shaped to speed the flow over the top.

    ✦ Lift ≈ 1.5 × 10³ N, directed upward.

    Where students slip. Squaring the difference of the speeds rather than taking the difference of the squares. (70 − 63)² = 49 is not the same as 70² − 63² = 931, and the error makes the lift nearly twenty times too small.

15. 9.153 marksNCERT Cl-11 Physics Part II, Ch9 Exercises, Q9.15 — depends on Fig. 9.20, read from the PDF at 560dpi

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

    Hint. Work out where the liquid moves fastest, then use Bernoulli to say where the pressure must be lowest. The height of liquid in each vertical tube is a direct reading of the pressure at that point.

    Step 1 — Find where the flow is fastest. Both figures show a pipe that narrows to a throat and then widens again. By the equation of continuity, Av = constant, so the speed is greatest where the cross-section is smallest — that is, at the throat.

    Step 2 — Deduce where the pressure is lowest. The pipe is horizontal, so Bernoulli's equation reduces to

    p + ½ρv² = constant

    A larger v therefore requires a smaller p. Since the speed is greatest at the throat, the pressure must be lowest at the throat.

    Step 3 — Read the vertical tubes. Each vertical tube is a manometer: the height of liquid standing in it is a direct measure of the pressure at that point in the pipe. Lower pressure means a shorter column.

    So the tube standing on the throat must show the shortest column of the two.

    Step 4 — Compare that with the figures. In Fig. 9.20(b) the column at the throat is shorter than the one on the wide section, which is exactly right.

    In Fig. 9.20(a) the column at the throat is the taller one, which would mean the pressure is highest where the liquid moves fastest. That contradicts Bernoulli's principle.

    ✦ Figure 9.20(a) is incorrect, because it shows the highest pressure at the narrowest part where the speed is greatest, whereas the pressure there must be the lowest.

    Where students slip. Assuming the pressure must be greatest where the pipe is narrowest because the liquid is "squeezed". Narrowing raises the speed, and by Bernoulli higher speed goes with lower pressure, not higher.

16. 9.163 marksNCERT Cl-11 Physics Part II, Ch9 Exercises, Q9.16

    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?

    Hint. Apply the equation of continuity between the tube and the holes, remembering that the total area of the holes is 40 times the area of one. Convert the speed from metres per minute to metres per second.

    Step 1 — Convert the given flow speed to SI units.

    v_tube = 1.5 m min⁻¹ = 1.5/60 = 0.025 m s⁻¹

    Step 2 — Find the volume flow rate in the tube. A_tube = 8.0 cm² = 8.0 × 10⁻⁴ m²

    Q = A_tube × v_tube = (8.0 × 10⁻⁴)(0.025) = 2.0 × 10⁻⁵ m³ s⁻¹

    Step 3 — Find the combined area of the holes. Each hole has diameter 1.0 mm, so radius 0.5 mm = 0.5 × 10⁻³ m. There are 40 of them:

    A_holes = 40 × π(0.5 × 10⁻³)² = 40 × 7.854 × 10⁻⁷ = 3.14 × 10⁻⁵ m²

    Step 4 — Apply continuity to get the ejection speed. All the liquid entering the tube must leave through the holes, so

    v_holes = Q/A_holes = (2.0 × 10⁻⁵)/(3.14 × 10⁻⁵) = 0.64 m s⁻¹

    Step 5 — Sense-check the result. The holes' combined area is smaller than the tube's, so the liquid must speed up — and indeed 0.64 m s⁻¹ is about 25 times the tube speed of 0.025 m s⁻¹, matching the area ratio.

    ✦ Speed of ejection ≈ 0.64 m s⁻¹.

    Where students slip. Using the area of a single hole instead of all forty. The continuity equation needs the total area through which the liquid leaves, and using one hole makes the answer forty times too large.

17. 9.172 marksNCERT Cl-11 Physics Part II, Ch9 Exercises, Q9.17

    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?

    Hint. A soap film in air has two surfaces, front and back, and each one pulls on the slider. That factor of two is the whole question.

    Step 1 — Count the surfaces. A soap film is a thin sheet of liquid with air on both sides, so it has two surfaces. Each surface exerts a force along the slider, so the total upward force is twice what a single surface would give.

    Step 2 — Write the force balance. The film supports the weight, so

    W = S × (2L)

    where L is the length of the slider and S the surface tension.

    Step 3 — Solve for the surface tension.

    S = W/(2L) = (1.5 × 10⁻²)/(2 × 0.30) = (1.5 × 10⁻²)/0.60 = 2.5 × 10⁻² N m⁻¹

    Step 4 — Confirm the units make sense. Surface tension is a force per unit length, so N m⁻¹ is right. The value is also typical for a soap solution, which supports the working.

    ✦ Surface tension S = 2.5 × 10⁻² N m⁻¹.

    Where students slip. Dividing by L rather than 2L, which doubles the answer. A film has two surfaces; a single liquid surface such as the free top of water in a beaker would not carry the factor of two.

18. 9.183 marksNCERT Cl-11 Physics Part II, Ch9 Exercises, Q9.18 — depends on Fig. 9.21, read from the PDF

    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.

    Hint. Write down what the supported weight actually depends on. Then check which of those quantities differ between the three figures — and which do not.

    Step 1 — Write what the film can support. The weight held up by a film with two surfaces is

    W = S × 2L

    where L is the length of the horizontal edge in contact with the slider, and S is the surface tension.

    Step 2 — Identify what this depends on. Only two things appear: the surface tension S and the length L. Neither the area of the film nor its shape appears anywhere.

    Step 3 — Compare the three figures. The question states the liquid and the temperature are the same throughout, so S is unchanged. Reading the figures, all three films have the same edge length of 40 cm — they differ only in how tall they are and, in (c), in being pointed at the top rather than square.

    Since L is the same and S is the same, W must be the same.

    Step 4 — Explain it physically. Surface tension is a property of the surface itself, not of how much surface there is. Stretching a film to a larger area does not make each unit of its edge pull harder — the liquid simply draws more molecules into the surface, keeping the force per unit length constant. So a taller or differently shaped film of the same width supports exactly the same weight.

    ✦ Both (b) and (c) support the same weight, 4.5 × 10⁻² N, because the supported weight depends only on the edge length and the surface tension, neither of which changes.

    Where students slip. Assuming a bigger film supports more weight. Surface tension is force per unit length, so only the length of the supporting edge matters — the area of the film is irrelevant.

19. 9.193 marksNCERT Cl-11 Physics Part II, Ch9 Exercises, Q9.19

    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.

    Hint. A liquid drop has only one surface, unlike a soap bubble which has two. Get the excess pressure first, then add atmospheric pressure to find the absolute pressure inside.

    Step 1 — Choose the right formula. A drop of liquid has a single surface separating liquid from air, so the excess pressure inside it is

    p_excess = 2S/r

    A soap bubble would have two surfaces and carry 4S/r instead — mixing the two is the classic error here.

    Step 2 — Compute the excess pressure. With S = 4.65 × 10⁻¹ N m⁻¹ and r = 3.00 mm = 3.00 × 10⁻³ m:

    p_excess = (2 × 0.465)/(3.00 × 10⁻³) = 0.930/(3.00 × 10⁻³) = 310 Pa

    Step 3 — Find the absolute pressure inside. The inside pressure exceeds the outside by that amount, so

    p_inside = p_atmospheric + p_excess = 1.01 × 10⁵ + 310 = 1.0131 × 10⁵ Pa

    Step 4 — Note the relative sizes. The excess of 310 Pa is only about 0.3% of atmospheric pressure, so the pressure inside the drop is barely different from outside. This is why a millimetre-sized drop feels no different from its surroundings, while a very small drop, with r in the denominator, would have a far larger excess.

    ✦ Excess pressure ≈ 310 Pa; pressure inside the drop ≈ 1.013 × 10⁵ Pa.

    Where students slip. Using 4S/r because mercury drops are often pictured alongside soap bubbles. A drop has one surface and takes 2S/r; only a bubble in air, with an inner and an outer surface, takes 4S/r.

20. 9.205 marksNCERT Cl-11 Physics Part II, Ch9 Exercises, Q9.20

    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).

    Hint. The two halves of this question need different formulas. A soap bubble in air has two surfaces; an air bubble inside a liquid has only one. For the second part remember to add the pressure of the liquid above it.

    Step 1 — Excess pressure in the soap bubble in air. A soap bubble has an inner and an outer surface, so

    p_excess = 4S/r = (4 × 2.50 × 10⁻²)/(5.00 × 10⁻³) = 0.100/(5.00 × 10⁻³) = 20 Pa

    Step 2 — Recognise that the second case is different. An air bubble inside a liquid has only one surface, the boundary between the air and the surrounding soap solution. So its excess pressure is

    p_excess = 2S/r = (2 × 2.50 × 10⁻²)/(5.00 × 10⁻³) = 10 Pa

    This halving is the point of the question.

    Step 3 — Find the pressure of the liquid at that depth. The relative density is 1.20, so ρ = 1.20 × 10³ kg m⁻³, and the depth is 40.0 cm = 0.400 m:

    p_depth = p_atm + ρgh = 1.01 × 10⁵ + (1.20 × 10³)(9.8)(0.400) = 1.01 × 10⁵ + 4704 = 1.057 × 10⁵ Pa

    Step 4 — Add the excess pressure of the bubble. The pressure inside the bubble exceeds the pressure of the liquid immediately around it by 10 Pa:

    p_inside = 1.057 × 10⁵ + 10 = 1.0572 × 10⁵ Pa

    ✦ Excess pressure in the soap bubble in air = 20 Pa; pressure inside the air bubble at 40.0 cm depth ≈ 1.06 × 10⁵ Pa.

    Where students slip. Using 4S/r for the air bubble submerged in the solution. Two surfaces exist only when there is liquid film with air on both sides; an air cavity inside a liquid has a single interface and takes 2S/r.

Solutions written by the tuition.in editorial team and checked against the NCERT Class 11 Physics Part II textbook, Reprint 2026-27 (keph202.pdf, 22 pages), cross-checked against the official CBSE curriculum 2026-27, Subject Code 042, Unit VII — one end-of-chapter Exercises set (20 questions, 9.1-9.20). Fig. 9.20 (Q9.15) and Fig. 9.21 (Q9.18) were rendered from the PDF and read visually, Fig. 9.20 at 560dpi because the answer turns on which manometer column is taller. Note: Reynolds number does not appear in the 2026-27 chapter, yet Q9.13 asks the reader to check whether the flow is laminar — see that solution.. Questions are referenced from the NCERT textbook for identification.