Chapter 9: Mechanical Properties of Fluids Long Questions | physics Class 11 CBSE

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Chapter 9: Mechanical Properties of Fluids Long Questions | physics Class 11 CBSE | ByteNotes.in

Long Answer

Hard difficulty • Structured explanation

Hard

Question 1

Long Form

Derive Bernoulli's equation for steady, incompressible, non-viscous fluid flow and state its physical significance. Also state the conditions under which it is not valid.

Consider an incompressible fluid flowing steadily through a pipe of varying cross-section at varying heights; let regions 1 (area A₁, height h₁, speed v₁, pressure P₁) and 2 (area A₂, height h₂, speed v₂, pressure P₂) be two sections.
Work done on fluid at left end in time Δt: W₁ = P₁A₁v₁Δt = P₁ΔV; work done by fluid at right end: W₂ = P₂ΔV; net work = (P₁ − P₂)ΔV.
Change in kinetic energy: ΔK = ½ρΔV(v₂² − v₁²); change in potential energy: ΔU = ρgΔV(h₂ − h₁); by work-energy theorem: (P₁ − P₂)ΔV = ΔK + ΔU.
Rearranging: P₁ + ½ρv₁² + ρgh₁ = P₂ + ½ρv₂² + ρgh₂, or in general P + ½ρv² + ρgh = constant.
Physically, this is energy conservation per unit volume: pressure energy plus kinetic energy plus potential energy remains constant along a streamline for ideal flow.
The equation fails for viscous fluids (energy is lost as heat), compressible fluids (elastic energy is ignored), and for turbulent or non-steady flows where velocity and pressure fluctuate with time.

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Long Answer

Medium difficulty • Structured explanation

Medium

Question 2

Long Form

Explain Pascal's Law with proof. Describe the working of a hydraulic lift based on this law and calculate the mechanical advantage.

Pascal's Law states that pressure in a fluid at rest is the same at all points at the same height, and external pressure applied to an enclosed fluid is transmitted undiminished equally in all directions.
Proof: Consider a small right-angled prismatic fluid element ABC-DEF with forces normal to each face; resolving equilibrium conditions gives Pb/Ab = Pc/Ac = Pa/Aa, proving pressure is direction-independent.
For a horizontal bar of fluid in equilibrium, forces at its two ends must balance, proving pressure is equal at all points on the same horizontal plane in a static fluid.
In a hydraulic lift, two pistons of areas A₁ (small) and A₂ (large) are connected by fluid; a small force F₁ on A₁ creates pressure P = F₁/A₁, transmitted throughout the fluid.
The large piston of area A₂ experiences an upward force F₂ = PA₂ = F₁A₂/A₁; by changing F₁ the platform moves up or down.
Mechanical advantage = F₂/F₁ = A₂/A₁; since A₂ >> A₁, a small force produces a very large lifting force — this principle is also used in hydraulic brakes where equal pressure is applied to all four wheels.

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Long Answer

Hard difficulty • Structured explanation

Hard

Question 3

Long Form

Derive the expression for terminal velocity of a sphere falling through a viscous fluid using Stokes' Law. Explain the forces acting on the sphere during its descent.

When a sphere of radius a and density ρ falls through a viscous fluid of density σ and viscosity η, three forces act: weight W = (4/3)πa³ρg (downward), buoyant force Fb = (4/3)πa³σg (upward), and Stokes' viscous drag F = 6πηavt (upward).
Initially the sphere accelerates downward because weight exceeds the opposing forces; as speed increases, the viscous drag increases proportionally.
At terminal velocity vt the net force is zero: weight = buoyancy + viscous drag, giving (4/3)πa³ρg = (4/3)πa³σg + 6πηavt.
Solving: 6πηavt = (4/3)πa³(ρ − σ)g, therefore vt = 2a²(ρ − σ)g/(9η).
Terminal velocity is proportional to the square of the sphere's radius — a larger raindrop falls faster than a smaller one.
Terminal velocity is inversely proportional to the viscosity η of the medium; in a more viscous medium the sphere reaches a lower terminal speed. This is why fine water droplets in clouds fall very slowly.

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Long Answer

Hard difficulty • Structured explanation

Hard

Question 4

Long Form

Discuss the origin of surface tension at the molecular level. Define surface energy and surface tension, and establish the relation between them using a film on a movable bar.

A molecule in the interior of a liquid is surrounded by neighbours on all sides and experiences attractive forces from all directions, giving it a large negative potential energy; molecules at the surface are surrounded only from below and have higher (less negative) potential energy — approximately half that of an interior molecule.
This excess surface energy means a liquid tends to contract its surface to minimum area; increasing surface area requires work to be done against intermolecular forces.
Consider a thin liquid film stretched on a rectangular frame with one movable bar of length l; the film has two surfaces, so the effective length is 2l.
Moving the bar by a small distance d increases the surface area by 2dl; work done = F·d; if S is the surface energy per unit area, the extra energy stored = S(2dl).
From energy conservation: Fd = S(2dl), giving S = F/(2l) — surface tension equals force per unit length and also equals surface energy per unit area.
Surface tension S has SI unit N m⁻¹ and decreases with temperature (like viscosity), falling to zero at the critical temperature of the liquid.

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Long Answer

Medium difficulty • Structured explanation

Medium

Question 5

Long Form

Analyse the variation of pressure with depth in a fluid. Using this, explain the working of a mercury barometer and define the units of pressure: atm, bar, and torr.

For a fluid at rest in a container, consider a cylindrical element of height h and base area A; vertical equilibrium gives (P₂ − P₁)A = ρhAg, hence P₂ − P₁ = ρgh.
Setting the top surface at atmospheric pressure Pa, pressure at depth h below it is P = Pa + ρgh — pressure increases linearly with depth.
Torricelli's mercury barometer: a mercury-filled tube is inverted into a trough; mercury vapour pressure above the column is negligible (≈ 0), so the atmosphere supports the column: Pa = ρgh.
At sea level h ≈ 76 cm of mercury, giving Pa = 1.013 × 10⁵ Pa; 1 atm is defined as this standard atmospheric pressure.
1 bar = 10⁵ Pa (slightly less than 1 atm); 1 torr = 133 Pa is equivalent to 1 mm of Hg (named after Torricelli) and is used in medicine and physiology.
The gauge pressure Pg = P − Pa = ρgh is what instruments like tyre gauges and blood pressure gauges (sphygmomanometers) actually measure, not the absolute pressure.

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Long Answer

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Hard

Question 6

Long Form

Compare the behaviour of liquids and gases with respect to compressibility, density variation with pressure, and viscosity variation with temperature.

Compressibility: Liquids are largely incompressible — their volume changes very little with large changes in pressure; gases are highly compressible and their volume changes significantly with pressure changes.
Density: The density of a liquid is nearly constant at all pressures (treated as constant for most calculations); the density of a gas varies greatly with pressure and temperature (e.g., density of air 1.29 kg m⁻³ vs interstellar space ~10⁻²⁰ kg m⁻³).
Free surface: A liquid has a definite volume and forms a free surface in a container; a gas has no definite volume and expands to occupy the entire container with no free surface.
Viscosity with temperature: Viscosity of liquids decreases with temperature because thermal energy reduces intermolecular cohesive forces; viscosity of gases increases with temperature because random molecular motion increases momentum exchange between layers.
Shear response: Both liquids and gases (fluids) offer very little resistance to shear stress — their shearing stress is about a million times smaller than that of solids; unlike solids, fluids cannot sustain shear stress in equilibrium.
Pressure transmission: Both liquids and gases transmit pressure in all directions (Pascal's Law); but gases exhibit much larger density variations with depth, making the expression P = Pa + ρgh less accurate for gases over large height differences.

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Long Answer

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Hard

Question 7

Long Form

Explain capillary rise and depression. Derive the formula for height of liquid column in a capillary tube and discuss factors affecting it.

When a narrow capillary tube is inserted in a liquid with acute angle of contact (e.g., water in glass), the meniscus is concave; the pressure just below the curved surface is less than atmospheric, causing the liquid to rise.
The pressure difference across the curved meniscus of radius r = a/cos θ is ΔP = (2S cos θ)/a, where a is the tube radius and θ is the angle of contact.
At equilibrium, the weight of the risen liquid column balances this pressure difference: hρg = (2S cos θ)/a, giving h = 2S cos θ/(ρga).
For a liquid with obtuse angle of contact (e.g., mercury in glass), cos θ is negative, so h is negative — the liquid is depressed below the outside level.
Factors affecting capillary rise: h increases as tube radius a decreases (narrower tube = greater rise), h increases with surface tension S, and h decreases as fluid density ρ increases.
An example from the chapter: for water in a capillary of radius a = 0.05 cm, using S = 0.073 N m⁻¹, h ≈ 2.98 cm, demonstrating how surface tension drives significant capillary action in biological systems like water transport in plants.

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Long Answer

Medium difficulty • Structured explanation

Medium

Question 8

Long Form

Explain the concept of viscosity using the model of a liquid between parallel plates. Define coefficient of viscosity and describe laminar flow in a pipe.

Consider a fluid (e.g., oil) enclosed between two horizontal glass plates; the bottom plate is fixed and the top plate moves with constant velocity v; the fluid layer touching each plate has the same velocity as that plate.
Velocity increases uniformly from zero (at the fixed plate) to v (at the moving plate); each intermediate layer is pulled forward by the one above and backward by the one below, resulting in an internal tangential (shear) force between layers.
Unlike solids where shear stress depends on shear strain, in fluids the stress depends on the rate of shear strain (v/l); the coefficient of viscosity η = (F/A)/(v/l) = Fl/(vA).
In laminar flow in a pipe, fluid velocity is maximum along the central axis and decreases to zero at the walls; the velocity profile is parabolic on a cylindrical surface.
The layers of liquid slide over one another like pages of a book; this smooth, ordered type of flow is called laminar (or viscous) flow and occurs at speeds below the critical speed.
Above the critical speed, flow becomes turbulent — disordered and chaotic — and Bernoulli's equation (which assumes non-viscous flow) no longer applies.

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Long Answer

Hard difficulty • Structured explanation

Hard

Question 9

Long Form

Analyse the pressure inside drops, cavities, and bubbles using surface tension. Why must extra air pressure be blown into a soap bubble to form it?

For a spherical liquid drop of radius r in equilibrium: if radius increases by Δr, extra surface energy = 8πrΔrSla; this energy comes from work done by the pressure difference (Pi − Po): W = (Pi − Po)4πr²Δr.
Equating: (Pi − Po)4πr²Δr = 8πrΔrSla, giving excess pressure inside a liquid drop: Pi − Po = 2Sla/r.
A liquid cavity (air bubble in liquid) also has one liquid-air surface and the same excess pressure = 2Sla/r inside the cavity relative to outside.
A soap bubble has two liquid-air interfaces (inner and outer surfaces); applying the surface energy argument to both surfaces gives excess pressure inside a soap bubble: Pi − Po = 4Sla/r.
Since Pi − Po = 4Sla/r for a bubble, a positive gauge pressure must exist inside relative to the atmosphere outside; as the radius increases (bubble grows), the required excess pressure decreases as 1/r.
To form a soap bubble, extra air must be blown in to create this internal excess pressure; if blown too hard the bubble bursts, so a small but precise extra pressure above atmospheric is needed — this is why you blow steadily rather than sharply.

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Long Answer

Hard difficulty • Structured explanation

Hard

Question 10

Long Form

Using the equation of continuity and Bernoulli's principle, explain Torricelli's law and the speed of efflux when (a) the tank is open to atmosphere, and (b) the tank is under pressure P >> Pa.

Consider a tank of liquid with a small hole of area A₁ at height y₁ and free surface of area A₂ >> A₁ at height y₂; by equation of continuity v₂ = (A₁/A₂)v₁ ≈ 0, so the free surface is approximately at rest.
Applying Bernoulli's equation between the free surface (P = P_above, v₂ ≈ 0, height y₂) and the hole (P = Pa, speed v₁, height y₁): P + ρgy₂ = Pa + ½ρv₁² + ρgy₁.
Case (a) — Open tank (P = Pa): simplifies to ½ρv₁² = ρg(y₂ − y₁) = ρgh, giving v₁ = √(2gh); this is Torricelli's Law — efflux speed equals free-fall speed from height h.
Case (b) — Tank under high pressure P >> Pa: v₁ = √(2gh + 2(P − Pa)/ρ); when P >> Pa the term 2gh is negligible, so v₁ ≈ √(2(P − Pa)/ρ) — the efflux speed is controlled by the container pressure, not the height.
Rocket propulsion operates on this principle: high internal pressure drives the gas out at high velocity, producing thrust by Newton's Third Law.
In both cases the efflux speed increases with the pressure difference (P − Pa) and decreases as fluid density ρ increases; the shape of the container has no effect on efflux speed, only the depth h and pressure P matter.

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Chapter 9: Mechanical Properties of Fluids | Long Questions

Derive Bernoulli's equation for steady, incompressible, non-viscous fluid flow and state its physical significance. Also state the conditions under which it is not valid.

Explain Pascal's Law with proof. Describe the working of a hydraulic lift based on this law and calculate the mechanical advantage.

Derive the expression for terminal velocity of a sphere falling through a viscous fluid using Stokes' Law. Explain the forces acting on the sphere during its descent.

Discuss the origin of surface tension at the molecular level. Define surface energy and surface tension, and establish the relation between them using a film on a movable bar.

Analyse the variation of pressure with depth in a fluid. Using this, explain the working of a mercury barometer and define the units of pressure: atm, bar, and torr.

Compare the behaviour of liquids and gases with respect to compressibility, density variation with pressure, and viscosity variation with temperature.

Explain capillary rise and depression. Derive the formula for height of liquid column in a capillary tube and discuss factors affecting it.

Explain the concept of viscosity using the model of a liquid between parallel plates. Define coefficient of viscosity and describe laminar flow in a pipe.

Analyse the pressure inside drops, cavities, and bubbles using surface tension. Why must extra air pressure be blown into a soap bubble to form it?

Using the equation of continuity and Bernoulli's principle, explain Torricelli's law and the speed of efflux when (a) the tank is open to atmosphere, and (b) the tank is under pressure P >> Pa.

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