NCERT Ch 9–10

Mechanical Properties of Solids & Fluids

Elasticity, pressure, buoyancy & flow · Class 11 Physics

1The big idea

Real materials are neither perfectly rigid nor perfectly free. Push on a solid and it deforms a little, then pushes back — that spring-like resistance is elasticity. Fluids (liquids and gases) cannot resist a shape change at all, so instead they transmit and redistribute forces as pressure. This one chapter covers both, and the thread running through it is the same in each half: relate the deforming force to the response, and every result falls out.

For solids the response is a fractional deformation; the ratio of "cause" to "effect" is a modulus. For fluids the response is a pressure field; from it we get buoyancy, flow speeds and the tiny forces at a liquid's surface. Nothing here needs memorising if you keep asking: what is pushing, and what gives way?

2Stress & strain

When you hang a load on a wire it stretches. Two numbers capture what is happening, both defined so that they do not depend on the size of the sample.

Stress is the internal restoring force per unit area of cross-section — how hard the material is working, spread over the area carrying the load:

Stress σ = FA   (unit: N m−2 = pascal, Pa)

Strain is the fractional deformation — the change divided by the original, so it is a pure number with no units:

Longitudinal strain ε = ΔLL

A thick rod and a thin wire of the same steel, carrying loads that give the same stress, stretch by the same fraction. That is the whole point of defining them this way — the numbers describe the material, not the particular specimen.

Hooke's law is the experimental fact that, for small deformations, stress is proportional to strain. The constant of proportionality is the material's modulus of elasticity:

Hooke's law stress = (modulus) × strain

This holds only up to the proportional limit. Push past it and the curve bends; past the elastic limit the body no longer returns to its original shape (permanent, plastic deformation), and eventually it reaches the breaking point.

strain ε → stress σ Hooke's law (linear) limit breaks
A typical stress–strain curve. The straight part is where Hooke's law lives; its slope is the modulus. Beyond the elastic limit the material yields, then breaks.

3The three elastic moduli

There are three ways to deform a solid, and each has its own modulus — but all three are just "stress ÷ strain" for that kind of deformation.

Young's modulus Y — stretching

Pull a wire lengthwise. Young's modulus is tensile stress divided by longitudinal strain:

Young's modulus Y = σε = F / AΔL / L = F LA ΔL

Rearranged, the stretch is ΔL = F LA Y — a long thin wire of a low-Y material stretches most. Steel has a large Y (≈ 2 × 1011 Pa), which is exactly why it barely stretches under load and is used for cables and beams.

Bulk modulus B — squeezing from all sides

Immerse a body in a fluid and raise the pressure by Δp. Its volume shrinks. Bulk modulus is volume stress (the pressure change) divided by volume strain:

Bulk modulus B = −ΔpΔV / V

The minus sign is there because a positive pressure gives a negative ΔV (the volume falls), so B comes out positive. Its reciprocal k = 1/B is the compressibility. Liquids have large B (hard to compress); gases have small B.

Shear modulus η — sliding one face past another

Apply a force parallel to a face while the opposite face is held. The block tilts by a small angle θ. Shear modulus (or modulus of rigidity) is tangential stress divided by shear strain:

Shear modulus η = F / Aθ   (θ in radians)

Only solids resist shear — that is precisely what distinguishes a solid from a fluid. A fluid has η = 0: apply the gentlest sideways force and it just keeps flowing, which is why fluids take the shape of their container.

Remember this

All three moduli are the same recipe — stress ÷ strain — for three kinds of deformation: Y for length, B for volume, η for shape. A big modulus means a stiff material that resists that deformation strongly.

4Energy stored in a stretched body

Stretching a wire stores elastic potential energy, just like compressing a spring. To find it, note that as the wire extends the restoring force grows linearly from 0 to its final value F (Hooke's law), so the average force doing work is F/2, acting over the extension ΔL:

W = (average force) × extension = ½ F ΔL

Now write it per unit volume by dividing by the volume V = A L, and split the fraction into a stress part and a strain part:

WV = ½ F ΔLA L = ½ · FA · ΔLL = ½ × stress × strain

So the total elastic potential energy stored is

Elastic potential energy U = ½ × stress × strain × volume

The energy density (energy per unit volume) is just ½ × stress × strain — the area under the stress–strain line, exactly as ½ k x2 is the area under a spring's force–extension line.

5Fluid pressure & Pascal's law

Move to fluids. A fluid at rest cannot support a shear, so the only force it exerts on any surface is perpendicular to it: pressure. Pressure is force per unit area, P = F/A, and it acts equally in all directions at a point.

Pressure due to a column of fluid. Consider a horizontal patch of area A at depth h below the surface. The column of fluid above it has volume A h, mass ρ A h, and weight ρ A h g. That weight presses down on the patch, so the extra pressure is weight ÷ area:

Gauge pressure at depth h P = ρ A h gA = ρ g h

The area cancels — pressure depends only on depth, not on the shape or width of the vessel (the "hydrostatic paradox"). Including the atmosphere pushing on the surface, the total (absolute) pressure is P = P0 + ρ g h.

h surface (P₀) P = ρgh
Pressure grows with depth as ρ g h and, at any point, pushes equally in every direction.

Pascal's law. A pressure applied anywhere to an enclosed fluid is transmitted undiminished to every part of it. Put a small force on a small piston and it appears as the same pressure under a large piston, where it becomes a large force. That is the hydraulic lift and hydraulic brake:

Hydraulic press (equal pressure) F1A1 = F2A2  ⟹  F2 = F1 A2A1

A force is multiplied by the area ratio — no energy is created, because the large piston moves proportionally less.

6Buoyancy & floating

Because pressure grows with depth, the fluid pushes up harder on the bottom of a submerged body than it pushes down on the top. The net upward force is the buoyant force, and its size is fixed by Archimedes' principle:

Remember this

The buoyant force equals the weight of the fluid the body displaces: FB = ρfluid Vdisplaced g. It does not care what the body is made of — only how much fluid it shoves aside.

A body floats when it can displace enough fluid to match its own weight before it is fully submerged. Set buoyant force = weight for a floating body of density ρb, total volume V, submerged volume Vsub:

ρfluid Vsub g = ρb V g  ⟹  VsubV = ρbρfluid

So the fraction submerged equals the ratio of densities. Ice has density ≈ 917 kg m−3 and water 1000, so ice floats with about 92% of its volume underwater — the famous tip of the iceberg. If ρb > ρfluid the body cannot displace enough even when fully under, so it sinks. The activity below lets you feel this out.

◆ Activity — float or sink?

Water has density 1000 kg m−3 (the dashed line). Drag the block's density and watch it settle. Below 1000 it floats, riding with a submerged fraction of exactly ρblock / ρwater; at or above 1000 it can't displace enough and sinks to the bottom.

7Flowing fluids — continuity & Bernoulli

Now let the fluid move. For an ideal fluid (incompressible, no viscosity) in steady flow, two conservation laws do almost all the work.

Equation of continuity — conservation of mass

The same amount of fluid that enters a pipe must leave it. In a time Δt the fluid at a section of area A moving at speed v sweeps out a volume A v Δt. Since none is created or lost, this volume rate is the same everywhere:

Continuity A v = constant   ⟹   A1 v1 = A2 v2

Narrow the pipe and the fluid must speed up — that is why a thumb over a hose nozzle makes the jet shoot faster, and why a river runs quick through a gorge and slow across a plain.

Bernoulli's equation — conservation of energy

Apply the work–energy theorem to a moving parcel of ideal fluid: the work done by the pressure difference across it changes its kinetic and gravitational potential energy. Per unit volume this gives Bernoulli's beautiful result — pressure energy, kinetic energy and potential energy per unit volume add to a constant along a streamline:

Bernoulli's equation P + ½ ρ v2 + ρ g h = constant

Read it as a trade-off: where a fluid moves faster, its pressure drops (at the same height). That single sentence explains an aeroplane wing (faster air over the curved top → lower pressure → lift), the lift of a spinning ball, and why two ships steaming side by side are drawn together. Each term has units of pressure (Pa) — check: ½ ρ v2 is (kg m−3)(m s−1)2 = kg m−1 s−2 = Pa. ✓

8Viscosity, Stokes & terminal velocity

Real fluids have internal friction: adjacent layers moving at different speeds drag on each other. This is viscosity. The tangential force needed to keep a layer sliding is

Newton's law of viscous flow F = η A ΔvΔx

where ΔvΔx is the velocity gradient across the flow and the coefficient η is the coefficient of viscosity (unit Pa s, also called the poiseuille; honey has a large η, water a small one).

Stokes' law. A small sphere of radius r moving slowly at speed v through a fluid of viscosity η feels a retarding drag

Stokes' law Fdrag = 6π η r v

Terminal velocity. Drop a small sphere (density ρ) into a fluid (density σ). Three forces act: weight down, buoyancy up, and viscous drag up (which grows with speed). The sphere accelerates until the three balance and it settles to a constant terminal velocity vt. Setting up the balance — weight = buoyancy + drag:

43π r3 ρ g = 43π r3 σ g + 6π η r vt

Solve for vt. The 43π r3 g factors out of the two weight-like terms, leaving:

Terminal velocity vt = 2 r2 (ρ − σ) g9 η

Notice vt ∝ r2: a big raindrop falls far faster than a tiny mist droplet, and a fine dust particle drifts down almost imperceptibly. If ρ < σ the sphere rises (a bubble in water).

9Surface tension & capillarity

A molecule deep inside a liquid is pulled equally in all directions by its neighbours. A molecule at the surface has neighbours only below and to the sides, so it feels a net inward pull. The surface therefore behaves like a stretched elastic skin that tries to minimise its area — which is why small drops are spherical (a sphere has the least area for a given volume).

Surface tension T is the force per unit length along the surface (unit N m−1), equal also to the surface energy per unit area (J m−2).

Excess pressure inside a drop and a bubble

The curved skin squeezes inward, so the pressure just inside a drop is higher than outside. Balancing the surface-tension force pulling the two halves of a drop together against the excess-pressure force pushing them apart gives:

Excess pressure — liquid drop (one surface) Δp = 2Tr

A soap bubble has two surfaces (inner and outer film), so it contributes twice the surface-tension force and the excess pressure doubles:

Excess pressure — soap bubble (two surfaces) Δp = 4Tr

Both go as 1/r, so smaller drops and bubbles have higher internal pressure — connect a small bubble to a big one and the small one empties into the big one.

Capillary rise

Dip a thin tube in water and the water climbs up it. The liquid wets the glass and its surface curves into a concave meniscus; surface tension around the circle of contact (length 2π r) pulls the column upward. In equilibrium the upward pull supports the weight of the raised column of height h:

T (2π r) cos θ = (ρ g h)(π r2)

The left side is the vertical component of the surface-tension force (θ = contact angle); the right is the weight of the lifted column. Cancel π r and solve for h:

Capillary rise h = 2T cos θρ g r

The rise is larger for a narrower tube (h ∝ 1/r) — thin capillaries lift water highest, which is how liquids creep through blotting paper, soil and the fine channels of a plant. If the liquid does not wet the tube (mercury in glass, θ > 90°, so cos θ < 0), h is negative — the level is depressed.

Worked example

A steel wire of length 2.0 m and cross-sectional area 0.50 mm2 carries a 5.0 kg load. Young's modulus of steel is Y = 2.0 × 1011 Pa; take g = 10 m s−2. Find the stress, the extension, and the elastic energy stored.

Force on the wire: F = m g = 5.0 × 10 = 50 N. Area: A = 0.50 mm2 = 0.50 × 10−6 m2.

Cross-check: U = ½ F ΔL = ½ × 50 × 10−3 = 2.5 × 10−2 J ✓ — same answer from the "average force × extension" route.

10Common confusions to clear up

11Check yourself

Class 11 Physics · Mechanical Properties of Solids & Fluids · aligned to NCERT Chapters 9–10 · SmartStudy.School