AP EAPCET (Engineering) · Physics (JEE & NEET)
Properties of Solids and Liquids
Elasticity, fluid pressure, Pascal and Bernoulli principles, viscosity, surface tension and thermal properties of matter.
Eight concepts. A solid resists being stretched, a liquid refuses to be squeezed and insists on a surface — and almost every mark in the chapter comes from dividing a force by the geometry it acts across and naming what is left.
- AP EAPCET (Engineering)
- Medium level
- 8 concepts
- 5 practice questions
1Stress, strain and Young's modulus
Stress is the load spread over the area carrying it, σ = F/A, and strain is the fractional stretch it produces, ε = ΔL/L. Within the elastic limit Hooke's law says the two are proportional, and the constant of proportionality is Young's modulus: Y = σ/ε = FL/(AΔL). Stress has the units of pressure, strain has none, so Y is quoted in N m⁻² and runs to about 10¹¹ for a metal — near 2 × 10¹¹ for steel, a little over 10¹¹ for copper.
Y belongs to the material, not to the specimen. Cut the wire shorter or draw it thinner and Y does not move — what changes is the stiffness k = YA/L, the newtons per metre you would actually feel. Confusing the two is the standard slip.
Figure. Only the first straight stretch obeys Hooke's law, and only there does the slope mean anything: that slope is Y. Everything to the right of the proportional limit is a different question, and the flat run near the middle is the metal flowing at almost no extra stress.
Reading the stress–strain curve
- Straight firstUp to the proportional limit the curve is a line and its slope is Y.
- Elastic limitA little beyond, the curve bends but the wire still returns to its original length.
- Past yieldIt then flows at almost constant stress, work-hardens to the ultimate point, and breaks.
Young's modulus of a metal wire
A metal wire 2 m long and 1 mm² in cross-section stretches by 1 mm under a 50 N load. Find Y, and the extension the same wire would show under 100 N.
- \sigma = F/A = 50/10^{-6}5 × 10⁷ N m⁻²
- \varepsilon = \Delta L/L = 10^{-3}/25 × 10⁻⁴
- Y = \sigma/\varepsilon1 × 10¹¹ N m⁻²
- Same wire at 100 N: \Delta L \propto F2 mm
Pro tip. The one thing that goes wrong here is the area: 1 mm² is 10⁻⁶ m², not 10⁻³. Using 10⁻³ makes Y come out a thousand times too small, and the answer still looks like a number a wire might have.
A second wire of the same material is twice as long and twice the diameter of the first. Compared with the first it has
- The same Young's modulus and the same stiffness
- The same Young's modulus but twice the stiffness
- Twice the Young's modulus
Y is a property of the metal, not of a particular piece of it, so it is unchanged. Stiffness is k = YA/L: twice the diameter is four times the area, and twice the length halves it again, so k doubles.
2Energy stored in a stretched wire
Stretching a wire is work, and within the elastic limit that work is stored and comes back. The load is not constant while it happens — it grows from zero to F as the wire extends from zero to ΔL — so the work is the average load times the extension, U = ½FΔL, and the factor of one half is the whole of the concept.
Divide by the volume AL and the geometry drops out, leaving the elastic energy density u = ½ × stress × strain, which can equally be written ½Yε² or σ²/2Y. For a given wire U goes as F², so doubling the load stores four times the energy.
Figure. Load against extension is a straight line through the origin, so the work done is the triangle beneath it — half the base times the height, which is exactly ½FΔL. Draw the same picture for a spring and you get ½kx² by the identical argument.
Energy in the same wire
The 2 m, 1 mm² wire of the previous concept is stretched 1 mm by a 50 N load. Find the energy stored and the energy stored per unit volume.
- U = (1/2)F\Delta L = 0.5 \times 50 \times 10^{-3}0.025 J
- Volume = AL = 10^{-6} \times 22 × 10⁻⁶ m³
- u = U/V = 0.025/(2\times10^{-6})1.25 × 10⁴ J m⁻³
- Check: (1/2)\sigma\varepsilon = 0.5 \times 5\times10^{7} \times 5\times10^{-4}1.25 × 10⁴ J m⁻³
Pro tip. Write FΔL instead of ½FΔL and every answer comes out exactly twice too large. The half is there because the load started at zero — the same reason a spring stores ½kx² and not kx².
Two wires of the same material and the same length, one of area A and one of area 2A, are each stretched by the same force F. The energy stored in the thinner wire is
- Half that in the thicker one
- The same as in the thicker one
- Twice that in the thicker one
U = ½FΔL and ΔL = FL/AY, so U = F²L/2AY at fixed force. Halving the area doubles both the extension and the stored energy — the thin wire is the one that stores, and the one that snaps back hardest.
3Pressure with depth, and Pascal's law
A column of liquid of height h presses down with its own weight, so the pressure at depth h below a free surface is P = P₀ + ρgh. Only the depth enters. A wide tank, a narrow tube and a flask that flares outwards all read the same pressure at the same depth, which is the hydrostatic paradox — the vessel's shape and the amount of liquid in it are both irrelevant.
Pascal's law adds that a pressure applied to an enclosed fluid is transmitted undiminished to every part of it. Put two pistons of areas A₁ and A₂ on the same enclosed liquid and the pressure they share forces F₂/F₁ = A₂/A₁, which is the hydraulic lift. Nothing is created: the large piston travels less far in exactly the proportion its force is multiplied.
Figure. The two piston faces are drawn level on purpose: at the same height in the same connected liquid there is no ρgh difference between them, so the pressure really is common and the force ratio really is the area ratio. Stagger them vertically and a ρgh correction appears.
How a hydraulic lift works
- Press the small pistonA force F₁ on area A₁ raises the pressure of the enclosed oil by F₁/A₁.
- Every point feels itPascal's law carries that same extra pressure across to the large piston.
- Collect the forceThe same pressure over the larger area A₂ gives F₂ = F₁A₂/A₁.
Lifting a car by hand
A hydraulic lift has pistons of 10 cm² and 500 cm². An 800 kg car sits on the large piston. Find the force needed on the small one, and how far it must travel to raise the car 5 cm. Take g = 10 m s⁻².
- P = F_2/A_2 = 8000/0.051.6 × 10⁵ Pa
- F_1 = PA_1 = 1.6\times10^{5} \times 10^{-3}160 N
- d_1 = (A_2/A_1)d_2 = 50 \times 0.052.5 m
- Check: F_1d_1 = 160 \times 2.5 against F_2d_2400 J each way
Pro tip. The force is multiplied fifty times and the distance divided by fifty, so the work is identical to the joule. A hydraulic lift is a lever made of oil — and the last row is worth writing out, because a question that quietly asks for the work is asking whether you believed the fifty.
Three vessels — a straight cylinder, a cone widening upwards and a cone narrowing upwards — stand on a table, each filled with water to the same depth. The pressure on the base is
- Greatest for the vessel holding the most water
- The same for all three
- Greatest for the one narrowing upwards
P = P₀ + ρgh depends on depth alone, so all three read the same pressure. The total force on the base does differ, because the bases have different areas — but that is a question about area, not about pressure.
4Buoyancy and floating
The pressure on the underside of a submerged body is larger than on its top, because the underside is deeper. What survives when you add up the pressure over the whole surface is Archimedes' upthrust: F_B = ρ_fluid V_displaced g, the weight of the fluid pushed aside. Note whose density it is — the fluid's, never the body's — and which volume — the submerged part, not the whole body.
A body floats in equilibrium when the upthrust equals its weight, so ρ_body V g = ρ_fluid V_sub g and the fraction submerged is simply the ratio of densities, V_sub/V = ρ_body/ρ_fluid. Fully submerged, the apparent weight is W − F_B, which is why a stone feels lighter under water and why a balance reads low there.
Figure. The block is drawn to scale: eight tenths of its height is below the surface because its density is eight tenths of the water's. The two arrows are drawn the same length because they must be — a floating body's upthrust is its weight, no more and no less.
Float, sink or hover
- Lighter than the fluidρ_body < ρ_fluid: it rises until only ρ_body/ρ_fluid of it is under.
- Exactly equalρ_body = ρ_fluid: it hovers, in equilibrium at any depth.
- Denserρ_body > ρ_fluid: it sinks, and on the bottom weighs W − ρ_fluid Vg.
How much more will it take?
A block of wood of density 800 kg m⁻³ and volume 2 × 10⁻³ m³ floats in water. Find the fraction submerged, and the extra mass that must be placed on it to just submerge it fully. Take g = 10 m s⁻².
- V_{sub}/V = \rho_b/\rho_w = 800/10000.8 submerged
- Weight = \rho_b Vg = 800 \times 2\times10^{-3} \times 1016 N
- Upthrust when fully under = \rho_w Vg20 N
- Extra load = 20 - 16 = 4 N, i.e. (\rho_w-\rho_b)V0.4 kg
Pro tip. The last row is the shortcut: the extra mass is always (ρ_fluid − ρ_body)V, so you never need the two weights separately. The same ratio explains icebergs — ice at 917 against seawater at 1025 floats with 0.895 of itself submerged, so about a tenth shows.
A block of ice floats in a glass of water filled to the brim. When the ice has completely melted, the water level
- Overflows, because ice takes up more room than water
- Is exactly unchanged
- Falls, because the ice shrinks as it melts
While floating, the ice displaces a volume of water whose weight equals the ice's own weight. Melting turns the ice into exactly that weight of water, which fills exactly that volume. Nothing spills and nothing drops.
5Continuity and Bernoulli
An incompressible liquid in steady flow cannot pile up anywhere, so whatever volume enters a pipe per second must leave it per second: A₁v₁ = A₂v₂. Narrow the pipe and the liquid must speed up, in exact inverse proportion to the area.
Bernoulli's equation is energy conservation for that same flow: P + ½ρv² + ρgh stays constant along a streamline, provided the flow is steady, incompressible and non-viscous. Read across the three terms and the consequence is the one worth memorising — where the fluid moves fastest the pressure is lowest, because the kinetic term can only grow at the pressure term's expense.
Figure. Both ratios in this figure are to scale: the narrow section is drawn a quarter as tall as the wide one, and its velocity arrow is drawn four times as long. That is continuity in one glance — the same volume per second squeezed through a quarter of the area has to travel four times as fast.
Using Bernoulli
- Pick two pointsBoth must lie on the same streamline; write P + ½ρv² + ρgh at each.
- Get the speedsContinuity supplies the second speed from the first: v₂ = (A₁/A₂)v₁.
- Equate and solveOn a horizontal pipe the ρgh terms cancel and only the other two survive.
The pressure drop at a constriction
Water flows steadily through a horizontal pipe whose cross-section falls from 8 cm² to 2 cm². The speed in the wide part is 1 m s⁻¹. Find the speed in the narrow part and the pressure difference between them.
- v_2 = (A_1/A_2)v_1 = 4 \times 14 m s⁻¹
- v_2^2 - v_1^2 = 16 - 115 m² s⁻²
- P_1 - P_2 = (1/2)\rho \times 15 = 500 \times 157500 Pa
- As a water column: \Delta P/\rho g = 7500/10^{4}0.75 m
Pro tip. The speed rose four-fold but the pressure drop went as v², so it is fifteen units of ½ρv₁², not four. That last row is how a venturi meter is actually read: a side tube at each section, and the 75 cm difference in their water levels is the measurement.
Water flows steadily up a pipe that both narrows and rises. Compared with the wide low section, the pressure at the narrow high section is
- Higher, since the water has been pushed upwards
- Lower, since both the other two terms have increased
- The same, since Bernoulli's sum is constant
The sum P + ½ρv² + ρgh is what stays constant, not P. Narrowing raises ½ρv² and rising raises ρgh, so P must fall by the total of both. The two effects add here rather than competing.
6Surface tension and excess pressure
A molecule in the bulk of a liquid is pulled equally in every direction; one at the surface has neighbours only below, so the surface behaves like a stretched skin. Surface tension S is the force per unit length along any line drawn in that surface — and equivalently the energy needed per unit area of new surface, since N m⁻¹ and J m⁻² are the same unit. That second reading is the useful one whenever a drop is split or a bubble is blown.
A curved surface cannot balance unless the pressure on its concave side is higher. For a single spherical surface the excess is 2S/r, so a liquid drop in air and an air bubble inside a liquid both give 2S/r. A soap bubble is a thin film with an inner and an outer surface, two of them, so its excess pressure is 4S/r — twice as much, for the same radius and the same liquid.
Figure. Both curves are the real 1/r law for S = 0.03 N m⁻¹, and the solid one sits at exactly twice the dashed one at every radius — that constant factor of two is the soap film's second surface. Read them at 1 mm and you have the ledger's 120 Pa and 60 Pa.
Getting the excess pressure right
- Count the surfacesOne for a drop or a bubble in liquid; two for a soap bubble or any film.
- MultiplyEach spherical surface contributes 2S/r, so a soap bubble comes to 4S/r.
- Higher insideThe excess is always on the concave side — inside a drop, inside a bubble.
Two soap bubbles on one tube
Two soap bubbles of radii 1 mm and 2 mm, blown from a solution with S = 0.03 N m⁻¹, are joined by a tube with a tap. Find the excess pressure in each, decide which way the air goes, and compare with a drop of the same liquid at 1 mm.
- \Delta P_1 = 4S/r_1 = 4 \times 0.03/10^{-3}120 Pa
- \Delta P_2 = 4S/r_2 = 4 \times 0.03/(2\times10^{-3})60 Pa
- Air flows from the higher excess to the lower1 mm bubble shrinks
- Drop of the same liquid at r_1: 2S/r_160 Pa, half
Pro tip. Both traps are in the last two rows. The small bubble empties into the big one, which is the opposite of what intuition says about pressure and size; and using 2S/r for a soap bubble halves every answer, because a film has two faces and you have only counted one.
An air bubble of radius r sits at depth h inside a liquid of surface tension S. The excess of the pressure inside the bubble over the pressure of the liquid immediately outside it is
- 2S/r
- 4S/r
- 2S/r + ρgh
The bubble has one surface, so the excess across it is 2S/r. The ρgh belongs to the liquid outside and is already included in the pressure the bubble is being compared with — adding it again double-counts the depth. The absolute pressure inside is P₀ + ρgh + 2S/r, which is a different quantity.
7Capillary rise
Where a liquid meets a solid it makes a contact angle θ. Water on glass wets it, θ is small, the meniscus curves concave upwards and the liquid climbs the tube; mercury on glass does not wet it, θ is about 140°, cos θ is negative, and the mercury is pushed down instead. The same formula covers both cases and the sign of cos θ is what switches between them.
Balance the vertical pull of surface tension round the circle of contact, 2πrS cos θ, against the weight of the raised column, ρgπr²h, and the height follows: h = 2S cos θ/(ρgr). This is Jurin's law, and the message in it is h ∝ 1/r — the finer the tube, the higher the liquid goes.
Figure. The column inside the tube stands above the free surface outside it, and the whole of that difference is being carried by a ring of surface tension a few molecules wide at the top. Narrow the bore and the ring's length falls as r while the column's weight falls as r², so the height has to rise.
Where the formula comes from
- The pullSurface tension acts along the contact circle of length 2πr, at angle θ to the wall.
- The loadIts vertical component 2πrS cos θ has to hold up a column of weight ρgπr²h.
- CancelSet them equal; πr cancels once and r once more, leaving h = 2S cos θ/ρgr.
Water in a fine capillary
Water of surface tension 0.072 N m⁻¹ and density 1000 kg m⁻³ rises in a capillary of radius 0.2 mm with contact angle 0°. Find the rise, and check it against the force balance. Take g = 10 m s⁻².
- 2S\cos\theta = 2 \times 0.072 \times 10.144 N m⁻¹
- \rho g r = 1000 \times 10 \times 2\times10^{-4}2.0 Pa
- h = 0.144/2.00.072 m = 7.2 cm
- Check: pull 2\pi rS against weight \rho g\pi r^2h9.05 × 10⁻⁵ N both
Pro tip. The last row is the whole derivation run backwards, and it agrees to three figures, so the 7.2 cm is not a formula you half-remembered. Halve the radius and the rise doubles to 14.4 cm — which is why sap, blotting paper and a brick's damp course all work better the finer the pores.
Water would rise 7.2 cm in a capillary, but the tube available is only 4 cm long and stands vertically in the water. The water
- Rises to the top and flows out steadily, a perpetual fountain
- Rises to the top and stops, the meniscus flattening out
- Rises only 4 cm up the tube and then falls back
The liquid cannot overflow — that would be free energy. It fills the tube and then the meniscus adjusts its radius of curvature upwards until 2S/R exactly balances the ρgh the short tube allows. What the formula predicts is the height for a fully curved meniscus, not a pump.
8Viscosity and terminal velocity
A real liquid resists being sheared. Adjacent layers moving at different speeds drag on each other with F = ηA(dv/dx), where dv/dx is the velocity gradient across the flow and η, the coefficient of viscosity, is measured in Pa s. It is the term Bernoulli's equation throws away.
For a small sphere in slow laminar flow that drag takes Stokes' form F = 6πηrv, and it grows with speed. Drop the sphere into a liquid and three forces act: weight down, upthrust up, and viscous drag up and increasing. The sphere accelerates only until they balance, after which it falls at a constant terminal velocity v_t = 2r²(ρ − σ)g/9η. The r² is the striking part — a droplet twice as big falls four times as fast.
Figure. The curve is v = v_t(1 − e^(−t/τ)) — steepest at release, when the drag is still zero and the sphere is in near free fall through the liquid, and flattening onto the dashed asymptote as the drag catches up with the weight. It reaches 63% of v_t in one time constant and never quite arrives.
Getting to the terminal velocity
- Three forcesWeight (4/3)πr³ρg down, upthrust (4/3)πr³σg up, drag 6πηrv up.
- Set the sum to zeroAt terminal velocity the acceleration is zero, so drag = weight − upthrust.
- Solve for vCancel the πr and rearrange: v_t = 2r²(ρ − σ)g/9η.
A steel ball in glycerine
A steel sphere of radius 1 mm and density 7800 kg m⁻³ is dropped into glycerine of density 1260 kg m⁻³ and viscosity 0.83 Pa s. Find its terminal velocity, and that of a sphere of twice the radius. Take g = 10 m s⁻².
- \rho - \sigma = 7800 - 12606540 kg m⁻³
- 2r^2(\rho-\sigma)g = 2\times10^{-6} \times 6540 \times 100.1308
- v_t = 0.1308/(9 \times 0.83)1.75 × 10⁻² m s⁻¹
- Radius doubled, v_t \propto r^27.0 × 10⁻² m s⁻¹
Pro tip. The fourth row never touched η, ρ or g — a ratio question only ever needs the r². And check the sign of ρ − σ before you write anything down: make the sphere less dense than the liquid and v_t comes out negative, which is not an error but a bubble rising at its own terminal speed.
A solid sphere and a hollow sphere of the same outer radius, made of the same metal, are dropped into the same viscous liquid. The hollow one reaches a terminal velocity that is
- The same, since the radius is the same
- Smaller, since its mean density is lower
- Larger, since it is lighter and meets less resistance
The drag 6πηrv depends on the outer radius, which is shared, but the driving force depends on (ρ − σ) with ρ the sphere's mean density. Hollowing it out lowers ρ, lowers ρ − σ, and lowers v_t in proportion. Being lighter is exactly why it falls more slowly here — the opposite of free fall in vacuum.
Notes
- Elasticity and Hooke's law: Stress (force per area) is proportional to strain within the elastic limit. Young's modulus Y=\dfrac{\text{stress}}{\text{strain}}=\dfrac{FL}{A\Delta L}; elastic potential energy per unit volume is \tfrac12\times\text{stress}\times\text{strain}.
- Fluid statics: Pressure at depth h is P=P_0+\rho g h. Pascal's law states pressure applied to an enclosed fluid is transmitted undiminished (basis of hydraulic lifts). A body submerged experiences buoyant force F_B=\rho_{fluid}Vg (Archimedes' principle).
- Bernoulli's principle: For steady, non-viscous, incompressible flow, P+\tfrac12\rho v^2+\rho gh=\text{constant} along a streamline; the continuity equation A_1v_1=A_2v_2 conserves volume flow rate.
- Surface tension and capillarity: Surface tension S is the force per unit length along a liquid surface. Excess pressure inside a spherical drop is \dfrac{2S}{r} and inside a soap bubble \dfrac{4S}{r}; liquid rises in a capillary to height h=\dfrac{2S\cos\theta}{\rho g r}.
- Viscosity: A viscous fluid resists shear with force F=\eta A\dfrac{dv}{dx}. A sphere falling through a viscous medium reaches terminal velocity v_t=\dfrac{2r^2(\rho-\sigma)g}{9\eta} (Stokes' law).
Formulas
- Young's modulus: Y=\dfrac{FL}{A\Delta L}, elastic energy density u=\tfrac12\,\text{stress}\times\text{strain}
- Fluid pressure and buoyancy: P=P_0+\rho gh,\quad F_B=\rho Vg
- Continuity and Bernoulli: A_1v_1=A_2v_2,\quad P+\tfrac12\rho v^2+\rho gh=\text{const}
- Excess pressure: drop \dfrac{2S}{r}, bubble \dfrac{4S}{r}
- Capillary rise: h=\dfrac{2S\cos\theta}{\rho g r}
- Terminal velocity (Stokes): v_t=\dfrac{2r^2(\rho-\sigma)g}{9\eta}
Exam traps & shortcuts
- A soap bubble has two surfaces, so its excess pressure 4S/r is double that of a liquid drop 2S/r of the same radius - a favourite trap in MCQs.
- Terminal velocity scales as r^2, so doubling a droplet's radius quadruples its terminal speed; use ratios to avoid plugging in \eta.
- Where a pipe narrows, speed rises (continuity) and pressure drops (Bernoulli) - remember pressure is lowest where the fluid moves fastest.
Reference tables
Every spherical surface contributes 2S/r, and the excess always sits on the concave side. All that changes between these cases is how many surfaces there are, and whether there is any curvature at all.
| Object | Surfaces | Excess pressure |
|---|---|---|
| Liquid drop in air | 1 | 2S/r |
| Air bubble inside a liquid | 1 | 2S/r |
| Soap bubble in air | 2 | 4S/r |
| Soap film, flat (r → ∞) | 2 | zero |
Every line here should be reconstructible from the concept above it, not merely recalled.
| Quantity | Relation | Watch for |
|---|---|---|
| Young's modulus | Y = FL/AΔL | 1 mm² is 10⁻⁶ m² |
| Stiffness of a wire | k = YA/L | This changes with shape; Y does not |
| Elastic energy | U = ½FΔL, u = ½ × stress × strain | The half — the load started at zero |
| Pressure with depth | P = P₀ + ρgh | Depth only, never the vessel's shape |
| Hydraulic lift | F₂/F₁ = A₂/A₁ | The distance divides by the same ratio |
| Buoyancy | F_B = ρ_fluid V_sub g | The fluid's density, the submerged volume |
| Floating fraction | V_sub/V = ρ_body/ρ_fluid | Only while it floats freely |
| Continuity | A₁v₁ = A₂v₂ | Steady flow of an incompressible liquid |
| Bernoulli | P + ½ρv² + ρgh = constant | Along one streamline, and non-viscous |
| Excess pressure | 2S/r one surface, 4S/r a soap bubble | Count the surfaces before dividing |
| Capillary rise | h = 2S cos θ/ρgr | cos θ < 0 for mercury, so it falls |
| Viscous force | F = ηA(dv/dx) | η in Pa s; it is what Bernoulli drops |
| Stokes drag | F = 6πηrv | Small sphere, slow laminar flow only |
| Terminal velocity | v_t = 2r²(ρ − σ)g/9η | Goes as r²; negative means it rises |
Recap
Read only this the night before.
- Elasticity
- Y = FL/AΔL belongs to the material; stiffness YA/L belongs to the wire. Energy is ½FΔL — never FΔL.
- Statics
- P = P₀ + ρgh, depth only. Pascal multiplies the force and divides the distance by the same factor, so the work is unchanged.
- Floating
- Fraction submerged = ρ_body/ρ_fluid. Fully under, the apparent weight is W − ρ_fluid Vg.
- Flow
- A₁v₁ = A₂v₂, and P + ½ρv² + ρgh is constant along a streamline. Fastest is always lowest in pressure.
- Surfaces
- 2S/r for a drop or a bubble in liquid, 4S/r for a soap bubble's two films. The small bubble empties into the big one.
- Capillary and drag
- h = 2S cos θ/ρgr, so h ∝ 1/r and mercury goes down. v_t = 2r²(ρ − σ)g/9η, so v_t ∝ r².
Practise Properties of Solids and Liquids
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