A gas at pressure P is compressed suddenly in an insulated cylinder. Its bulk modulus during this compression is:

Zero, since no heat is exchanged

Water flows steadily through a horizontal pipe whose cross-section narrows. In the narrow section:

The speed is greater and the pressure is lower

Both the speed and the pressure are greater

The speed is lower and the pressure is greater

The speed is greater and the pressure is unchanged

Mechanical Properties of Solids and Fluids — Elasticity, Pressure, Flow and Surface Tension

This chapter looks long because it names many phenomena, but it is generated by a single quantity: force divided by area. Call it stress and you get elasticity; call it pressure and you get fluid statics; let it vary along a streamline and you get Bernoulli; let it act along a boundary line instead of over an area and you get surface tension. Read the whole chapter as one idea appearing in four costumes and the formula count collapses. The other habit worth building here is dimensional suspicion: nearly every result below is a ratio of two areas, two densities or two radii, so if an answer does not scale the way the geometry says it should, it is wrong before you check the arithmetic.

Elasticity — Stress, Strain and the Three Moduli

Stress is the restoring force per unit area, F/A, measured in N/m² or pascal; strain is the fractional deformation and is therefore dimensionless. Hooke's law says that up to a limit the two are proportional, and the constant of proportionality is a modulus of elasticity. Everything in this section is one of three moduli, distinguished only by what shape change the force produces: a change of length, a change of volume, or a change of shape at constant volume.

The three moduli and what each one measures
Modulus	Definition	Applies to
Young’s modulus Y	Longitudinal stress / longitudinal strain = FL/(A·ΔL)	Solids only — a wire or rod being stretched
Bulk modulus B	−ΔP / (ΔV/V); its reciprocal 1/B is compressibility	Solids, liquids and gases — the only modulus a fluid has in full
Shear (rigidity) modulus η or G	Tangential stress / angular shear θ	Solids only — a fluid cannot resist shear, which is why it flows
Poisson’s ratio σ	Lateral strain / longitudinal strain; theoretically −1 to 0.5, practically 0.2–0.4	A ratio, not a modulus — it has no unit

🎯 Why steel is "more elastic" than rubber, which sounds backwards

In everyday speech elastic means stretchy, so students confidently answer "rubber". In physics elasticity means the strength of the restoring tendency, measured by Y: a large Y means the material fights back hard and returns to shape for a given stress. Steel has Y ≈ 2 × 10¹¹ Pa, rubber about 10⁶ Pa, so steel is roughly a hundred thousand times more elastic. The everyday word is describing the opposite property — how much the material yields — which is low elasticity. Two useful consequences follow. For the same stress, steel strains far less, which is why load-bearing structures are steel and not rubber. And a large Y with a large ultimate strength is what makes a material suitable for a cable, while a small Y with a large elastic range is what makes rubber suitable for a shock absorber — the rubber stores energy over a long stretch and gives it back.

🧠 Scaling ΔL without computing it: length up, radius squared down

Rearranging Y = FL/(AΔL) gives ΔL = FL/(AY) ∝ L/r², and most objective questions are answered from that proportionality alone. Double the length and the extension doubles; double the radius and it falls to a quarter, not a half — the squared dependence on radius is where marks are lost. Two standard set-ups follow directly. Wires joined in series carry the same tension F and their extensions add, so the thinner and longer wire stretches more and the total is ΣFLᵢ/(AᵢY). Wires in parallel share the load in proportion to A·Y/L and have the same extension, so the stiffer wire takes the larger share of the force. Recognising which case you are in takes one glance at the diagram and decides the whole problem.

The stress–strain curve for a metal names the limits of all this. Up to the proportional limit the graph is straight and Hooke's law holds. A little beyond lies the elastic limit or yield point, past which the body no longer returns fully and a permanent set remains. The curve then rises to the ultimate tensile strength, after which it falls — the wire thins locally, or necks, so the true stress at that neck keeps climbing even as the applied load drops, and it breaks at the fracture point. Work done in stretching is stored as elastic potential energy: U = ½ F ΔL, or per unit volume u = ½ × stress × strain = ½ Y (strain)², which is the area under the stress–strain graph.

⚠️ A gas has two different bulk moduli, and the question tells you which

Compress a gas slowly, letting heat escape, and PV = constant gives B_isothermal = P. Compress it fast, with no heat exchange, and PVᵞ = constant gives B_adiabatic = γP, which is larger by the factor γ. So the same gas is stiffer when squeezed quickly — and the words "suddenly", "rapidly" or "insulated" in the stem are the signal to use γP. This is not a detail: it is exactly the correction Laplace made to Newton's calculation of the speed of sound. Newton assumed the compressions in a sound wave were isothermal and got about 280 m/s; treating them as adiabatic gives v = √(γP/ρ) ≈ 332 m/s, which matches measurement. The same two-moduli distinction therefore connects this chapter to waves and to thermodynamics, and a question can be set from any of the three directions.

Fluid Statics — Pressure, Pascal and Archimedes

Pressure at depth h in a liquid of density ρ open to the atmosphere is P = P₀ + ρgh. Read that formula for what it does not contain: no area, no volume, no shape. Pressure therefore depends only on the vertical depth below the free surface, and all points at the same depth in a connected fluid are at the same pressure. The absolute value P is what appears in gas laws; the gauge pressure ρgh is what a tyre gauge or a manometer reads, and mixing the two is the commonest slip in numerical work.

🎯 The hydrostatic paradox: why a wide vessel does not press harder

Three vessels of wildly different shape, filled to the same height, exert exactly the same pressure on the base — and if the base areas are equal, the same force too, even though the vessels hold very different weights of water. Students find this impossible, because the wide vessel is obviously heavier. The resolution is that the extra weight in a flaring vessel is carried by the slanted walls, not the base: the walls push inward and downward on the liquid, and the liquid pushes outward and upward on the walls, so part of the load never reaches the bottom. In a vessel that narrows outward the walls do the opposite and the base can feel a force greater than the weight of the liquid it holds. Pascal demonstrated the effect by bursting a barrel with a thin tall pipe of water — the height, not the amount, sets the pressure.

Pascal's law states that a pressure applied to an enclosed fluid is transmitted undiminished to every part of it. A hydraulic lift uses this: F₁/A₁ = F₂/A₂, so a small force on a narrow piston lifts a large load on a wide one, with the force multiplied by A₂/A₁. Nothing is gained for free — the narrow piston must move A₂/A₁ times as far, so the work is the same, exactly as with a lever. Archimedes' principle then says a body immersed in a fluid experiences an upward buoyant force equal to the weight of the fluid it displaces, F_B = ρ_fluid·V_displaced·g, acting at the centre of buoyancy.

Floating, sinking and apparent weight
Situation	Condition	Result to quote
Floats partly submerged	ρ_body < ρ_liquid	Fraction submerged = ρ_body / ρ_liquid
Floats fully submerged	ρ_body = ρ_liquid	Neutral buoyancy — stays wherever placed
Sinks	ρ_body > ρ_liquid	Apparent weight = W(1 − ρ_liquid/ρ_body)
Ice cube floating in water	ρ_ice = 0.9 ρ_water	9/10 submerged; on melting the level does not change

🧠 The melting-ice family of questions, settled once

A floating ice cube displaces a volume of water whose weight equals the weight of the ice. When it melts it becomes exactly that weight of water, which occupies exactly the volume it was displacing, so the level is unchanged. Now vary it. If the ice contains a trapped stone, the stone while afloat displaces water equal to its own weight, but after melting it sinks and displaces only its own volume, which is less — so the level falls. If the ice contains a trapped air bubble or a cork, nothing changes: the level stays the same, because the bubble contributes no weight either way. And if the cube floats in a liquid denser than water, such as mercury, the meltwater is less dense than the mercury it displaced, so the level rises. One principle, four answers, and the stem always tells you which case it is.

Flow, Viscosity and Surface Tension

For steady flow of an incompressible liquid the equation of continuity is A₁v₁ = A₂v₂ — mass in equals mass out, so a narrow pipe means fast flow. Add energy conservation along a streamline and you get Bernoulli's equation, P + ½ρv² + ρgh = constant, which is nothing but the work–energy theorem written per unit volume. Its consequence is the one students must internalise: where the speed is high, the pressure is low.

Bernoulli in the exam — the standard applications
Application	Mechanism	Result
Venturimeter	Constriction speeds the flow and drops the pressure	Flow rate from the manometer height difference
Torricelli’s law (hole in a tank)	Efflux behaves like free fall through the head h	v = √(2gh) — independent of the liquid’s density
Aerofoil lift	Faster flow over the curved upper surface	Lower pressure above ⇒ net upward force
Spinning ball (Magnus effect)	Spin drags air, speeding one side and slowing the other	Sideways force — the swing and the curve ball
Atomiser and Bunsen burner	A fast jet lowers the pressure at a side opening	Liquid or air is drawn up into the stream

⚠️ Bernoulli assumes an ideal fluid — three conditions that are often violated

The equation is derived for flow that is steady, incompressible, non-viscous and streamline. Drop any one and it fails. Real water in a long pipe loses pressure steadily along the pipe even at constant cross-section, because viscosity dissipates energy — a fact Bernoulli alone would deny. Fast air is compressible, so aerodynamics near the speed of sound needs more than Bernoulli. And once flow becomes turbulent there are no well-defined streamlines to apply it along. In the exam this appears as a conceptual question — "which assumption is violated" — and as a warning: never use Bernoulli across a pump, a fan or a region of turbulence, because energy is being added or destroyed there. The dividing line between smooth and turbulent flow is the Reynolds number R = ρvD/η, which is dimensionless; below about 1000 flow is laminar, above about 2000 turbulent.

Viscosity is a fluid's internal friction: F = −ηA(dv/dx), with η in Pa·s. A sphere of radius r moving slowly through it feels Stokes' drag F = 6πηrv, which grows with speed until it balances weight minus buoyancy. The body then falls at a constant terminal velocity v_t = (2/9)·r²(ρ − σ)g/η, where ρ is the sphere's density and σ the fluid's. Note the r² — the single most examined feature of the result.

📖 Why raindrops do not arrive at lethal speed

A drop falling freely from a cloud a kilometre up would arrive at roughly 140 m/s and be dangerous. It does not, because drag rises with speed until it cancels the weight and the drop settles to its terminal velocity — a few metres per second for ordinary rain. Since v_t ∝ r², the size dependence is dramatic and visible: fine drizzle drops are about a fifth the radius of large raindrops and so fall about twenty-five times slower, which is exactly why mist appears to hang in the air while a downpour hammers down. The same r² explains why dust stays airborne for hours, why fog does not fall, and why a parachute works — it does not slow you by weighing less but by multiplying the effective area, and hence the drag, until the terminal speed is survivable.

Surface tension S is the force per unit length along a line in the surface, or equivalently the surface energy per unit area — the two definitions have the same units, J/m² = N/m, which is worth checking once. A curved surface therefore squeezes what is inside it, and the excess pressure follows: 2S/r inside a liquid drop, 4S/r inside a soap bubble (two surfaces), and 2S/r inside an air bubble in a liquid (one surface). In a capillary tube the liquid rises to h = 2S cosθ/(rρg), where θ is the contact angle.

🎯 Why water climbs a capillary and mercury sinks in it

The contact angle θ settles it. Water wets glass because water–glass adhesion beats water–water cohesion, giving θ < 90°, cosθ positive, a concave meniscus and a rise. Mercury does the opposite: cohesion within mercury beats adhesion to glass, θ ≈ 140°, cosθ negative, the meniscus is convex and the level falls. The same sign flip explains why water spreads on clean glass but beads on a waxed car, and why a detergent works — it lowers S, letting water wet and penetrate fabric it would otherwise sit on top of. Two exam-ready riders. Since h ∝ 1/r, a finer tube lifts liquid higher, which is how sap reaches the top of tall plants and how a blotting paper drinks ink. And if the tube is too short to allow the full h, the liquid does not spurt out of the top — it simply adjusts its radius of curvature and stops at the rim, which is a favourite conceptual trap.

🧠 Two soap bubbles joined by a tube: the small one always loses

Excess pressure inside a bubble is 4S/r, so the smaller bubble has the higher internal pressure. Connect two of unequal size and air flows from the small one into the large one: the small bubble shrinks and the large one grows, until the small has collapsed. This is the reverse of the intuitive guess that they will equalise, and it is asked in exactly this form. The same 1/r reasoning answers the family: a drop of radius R split into n identical droplets gains surface area, so energy must be supplied, ΔU = 4πR²S(n^1/3 − 1); merging droplets back into one releases energy, which is why a fine spray warms very slightly as it coalesces, and why mercury spilled on a table gathers itself into beads rather than staying spread.

Key takeaways

ΔL = FL/(AY) ∝ L/r² — doubling the radius quarters the extension. Y for solids, B for all states, shear modulus for solids only.
Elasticity in physics means a STRONG restoring tendency, so steel is far more elastic than rubber.
For a gas B_isothermal = P but B_adiabatic = γP — "suddenly" in the stem means use γP.
P = P₀ + ρgh depends only on vertical depth, not on the vessel’s shape or the amount of liquid.
Fraction submerged = ρ_body/ρ_liquid. A melting ice cube leaves the level unchanged; with a stone inside it falls.
Bernoulli: high speed means low pressure — but only for steady, incompressible, non-viscous streamline flow.
Terminal velocity v_t ∝ r², so a drop of half the radius falls four times slower.
Excess pressure: 2S/r for a drop, 4S/r for a soap bubble. Capillary rise h = 2S cosθ/(rρg) ∝ 1/r.

Practice questions (8)

Attempt each one before opening the answer. Every explanation names the tempting wrong option as well as the right one, because that is where marks are lost.

Two wires of the same material and the same length carry equal loads. The second has twice the radius of the first. The extension of the second wire is:
1. One quarter of the first
2. Half of the first
3. The same as the first
4. Twice the first
Show answer
Answer: A — One quarter of the first
ΔL = FL/(AY) and A = πr², so ΔL ∝ 1/r². Doubling the radius quadruples the area and cuts the extension to a quarter. The tempting "half" comes from treating the dependence as 1/r, forgetting that it is the cross-sectional area, not the radius, that appears in the formula.
A gas at pressure P is compressed suddenly in an insulated cylinder. Its bulk modulus during this compression is:
1. γP
2. P
3. P/γ
4. Zero, since no heat is exchanged
Show answer
Answer: A — γP
"Suddenly" plus "insulated" means adiabatic, where PVᵞ = constant gives B = γP. The distractor P is the isothermal value, correct only for slow compression with heat free to escape. Zero is wrong for a revealing reason: no heat exchange does not mean no resistance to compression — an adiabatic gas is in fact γ times stiffer than an isothermal one.
An ice cube with a small stone frozen inside it floats in a glass of water. When the ice melts completely, the water level:
1. Falls
2. Rises
3. Stays exactly the same
4. Rises first, then falls back
Show answer
Answer: A — Falls
While afloat, the stone is supported by the ice, so it displaces water equal to its own weight. Once the ice melts the stone sinks and displaces only its own volume, which for a body denser than water is less. So the level falls. "Stays the same" is the answer for a plain ice cube, or for one with an air bubble — the stone is what breaks that symmetry.
A block of relative density 0.6 floats in water. The fraction of its volume above the water surface is:
1. 0.4
2. 0.6
3. 0.5
4. 1.67
Show answer
Answer: A — 0.4
The submerged fraction equals ρ_body/ρ_liquid = 0.6, so the fraction above is 1 − 0.6 = 0.4. Option 0.6 is the classic misread: it is the submerged fraction, and the question asked for the part above. The value 1.67 is the inverted ratio, which should be rejected on sight since a fraction cannot exceed 1.
Water flows steadily through a horizontal pipe whose cross-section narrows. In the narrow section:
1. The speed is greater and the pressure is lower
2. Both the speed and the pressure are greater
3. The speed is lower and the pressure is greater
4. The speed is greater and the pressure is unchanged
Show answer
Answer: A — The speed is greater and the pressure is lower
Continuity (Av = constant) raises the speed where the pipe narrows, and Bernoulli then requires the pressure to drop so that P + ½ρv² stays constant. The intuition that a squeezed fluid must be at higher pressure is what makes option two tempting — but that would violate energy conservation, since the extra kinetic energy has to be paid for out of pressure energy.
Two spherical raindrops fall through air at their terminal velocities. The radius of the second is three times that of the first. The ratio v₂/v₁ is:
1. 9
2. 3
3. 27
4. √3
Show answer
Answer: A — 9
Terminal velocity v_t = (2/9)r²(ρ − σ)g/η is proportional to r², so tripling the radius multiplies the speed by 9. The answer 3 assumes a linear dependence, and 27 comes from using volume (r³) instead — a natural slip, since weight does scale as r³, but drag scales as r, and the ratio of the two gives r².
A soap bubble and a spherical water drop have the same radius r, and the same surface tension S. The ratio of the excess pressure inside the bubble to that inside the drop is:
1. 2 : 1
2. 1 : 1
3. 1 : 2
4. 4 : 1
Show answer
Answer: A — 2 : 1
A soap bubble has two liquid–air surfaces so its excess pressure is 4S/r; a drop has one, giving 2S/r. The ratio is therefore 2 : 1. Option 4 : 1 comes from comparing 4S/r against S/r and forgetting the factor 2 in the drop formula; 1 : 1 comes from overlooking the second surface of the bubble altogether.
Water rises to a height h in a capillary tube. If the same tube is taken to a place where g is halved, the rise becomes:
1. 2h
2. h/2
3. h
4. h/4
Show answer
Answer: A — 2h
Capillary rise is h = 2S cosθ/(rρg), so h is inversely proportional to g and halving g doubles the rise. The instinct that weaker gravity should lift less is what makes h/2 attractive, but gravity is the force being fought here, not the force doing the lifting — the lifting is done by surface tension, which has not changed.