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JEE Main chapters · Class 11 · Physics

Properties of Solids and Liquids: JEE Main Physics formulas and notes

This chapter combines the mechanical properties of solids and fluids with the thermal properties of matter. It is a large chapter, but most of the questions are single formulas: JEE Mains usually asks 3 to 4 questions from it, mostly from surface tension, Bernoulli, viscosity, calorimetry, conduction and radiation. Plan for about 12 hours, and aim for 90 seconds per question.

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Formulas with solved examples

Elasticity

Stress, strain and Hooke’s law

How to apply. Stress is the restoring force per unit area, so it is force divided by the area of cross-section (not the surface area). Strain is the fractional change, so it is a pure number.

Example. A force of 100 N on a wire of cross-section 1 mm² gives a stress of 100 ÷ 10⁻⁶ = 10⁸ Pa.

Young’s modulus

How to apply. Use it for the extension of a wire or rod. When two wires are compared, write the ratio ΔL ∝ L/r² (same material, same force) and cancel everything common.

Example. Steel wire, L = 2 m, r = 0.5 mm, F = 100 N, Y = 2 × 10¹¹: A = 7.85 × 10⁻⁷ m², so ΔL = 100 × 2 ÷ (7.85 × 10⁻⁷ × 2 × 10¹¹) = 1.27 mm.

Bulk modulus and compressibility

How to apply. The volume decreases when the pressure rises, which the minus sign takes care of. Solids and liquids are hard to compress, so B is large; gases have a small B.

Example. Water, B = 2.2 × 10⁹ Pa: an extra pressure of 2.2 × 10⁷ Pa (about 220 atm) shrinks the volume by 2.2 × 10⁷ ÷ 2.2 × 10⁹ = 1%.

Shear modulus (modulus of rigidity)

How to apply. The force acts along the surface and tilts a cube into a parallelogram. The shear strain is the angle θ through which the vertical face turns.

Example. F = 10³ N on a face of 0.1 m² of a block with G = 4 × 10¹⁰ Pa: stress = 10³ ÷ 0.1 = 10⁴ Pa, so θ = 10⁴ ÷ (4 × 10¹⁰) = 2.5 × 10⁻⁷ rad.

Poisson’s ratio and the relations between moduli

How to apply. A wire that gets longer gets thinner. σ gives the fractional decrease in radius for a given fractional increase in length. σ = 0.5 means the volume does not change.

Example. A wire with σ = 0.25 stretched by 1% shrinks in radius by 0.25%. For σ = 0.25, Y = 1.5 B and Y = 2.5 G.

Stress–strain curve

How to apply. Read the curve from left to right. Below the elastic limit the body returns to its shape when the load is removed. Beyond it a permanent set remains. A larger slope means a larger Y; a larger area means a tougher material.

Example. Steel and copper are ductile (large plastic region); glass is brittle. Rubber has no straight part and its loading and unloading curves differ, so it loses energy in each cycle.

Elastic potential energy

How to apply. The force grows from zero, so the work is the average force times the extension: half of F ΔL. For a wire of fixed length and area, U is proportional to the square of the force or of the extension.

Example. The steel wire above: U = ½ × 100 × 1.27 × 10⁻³ = 0.064 J.

Wire as a spring, combinations and own weight

How to apply. Replace each wire by a spring of constant YA/L and then use the spring rules. For a rod stretched by its own weight, the tension varies from Mg at the top to 0 at the bottom, which gives the factor ½.

Example. A steel rod of length 2 m, area 10⁻⁴ m², mass 15.7 kg: ΔL = 15.7 × 10 × 2 ÷ (2 × 10⁻⁴ × 2 × 10¹¹) = 7.85 × 10⁻⁶ m.

Fluid statics

Pressure in a fluid

How to apply. Add ρgh to the pressure at the top as you go down. Compare the pressures at two points at the same level of the same liquid to solve U-tube and barometer problems.

Example. At 10 m under water (g = 10 m/s²): P = 10⁵ + 1000 × 10 × 10 = 2 × 10⁵ Pa, that is 2 atm.

Pascal’s law and the hydraulic lift

How to apply. The same pressure acts on both pistons, so the force is multiplied by the ratio of the areas. The smaller piston moves a larger distance in the same ratio.

Example. Pistons of 10 cm² and 1000 cm²: a force of 50 N gives 5000 N, but the small piston moves 100 cm for each 1 cm of the large one.

U-tube, accelerating liquids and rotation

How to apply. Treat the acceleration as a change in effective gravity. The free surface is always perpendicular to the effective g. For a U-tube, equate the pressures at the interface level.

Example. A tank accelerating at 5 m/s² (g = 10): the surface makes tan θ = 0.5, or 26.6°, with the horizontal. The surface at the back is higher.

Force on a dam or a wall

How to apply. Because the pressure varies, use the average pressure for the force, but the force does not act at the middle. For the torque or the line of action use the centre of pressure at two thirds of the depth.

Example. Dam: water 10 m deep, 20 m wide: F = ½ × 1000 × 10 × 10² × 20 = 10⁷ N.

Archimedes’ principle and buoyancy

How to apply. The buoyant force is the weight of the fluid displaced. For a floating body, equate its weight to this force and read off the fraction submerged. For a hanging body use the apparent weight.

Example. Ice (density 900 kg/m³) in water: 90% is below the surface. A body weighing 10 N in air and 8 N in water has a volume of 2 ÷ (1000 × 10) = 2 × 10⁻⁴ m³.

Floating bodies: levels, lifts and oscillation

How to apply. Ice displaces its own weight of water, and the water it melts into has the same volume. The restoring force on a floating body with vertical sides (constant cross-section A) is the extra buoyant force ρ A g x, which acts like a spring of constant ρAg (small oscillations, no damping).

Example. A wooden cylinder of mass 0.5 kg and cross-section 50 cm² in water: k = 1000 × 50 × 10⁻⁴ × 10 = 50 N/m, so T = 2π√(0.5/50) = 0.63 s.

Fluid flow

Equation of continuity

How to apply. Mass does not pile up in a steady flow, so what flows in per second flows out. For a pipe of changing diameter the speed goes as 1/d².

Example. A pipe whose diameter is halved: the area becomes ¼ so the speed becomes 4 times.

Bernoulli’s theorem

How to apply. Write the three terms at two points on the same streamline and equate them. Use continuity to connect the two speeds. Each term has the dimensions of pressure.

Example. Horizontal pipe, v₁ = 2 m/s, area ratio A₁ : A₂ = 4 : 1, so v₂ = 8 m/s: P₁ − P₂ = ½ × 1000 × (64 − 4) = 3 × 10⁴ Pa.

Torricelli’s law, the range and the venturi meter

How to apply. Apply Bernoulli between the free surface (speed nearly zero) and the hole. After leaving the hole the water is a projectile with horizontal speed v.

Example. A tank filled to 5 m with a hole at 5 m depth: v = √(2 × 10 × 5) = 10 m/s. A hole halfway down a tank of height H throws the water a distance H.

Lift on a wing and the Magnus effect

How to apply. The difference in speed above and below gives a difference in pressure, and so a net upward force. Use Bernoulli for each side of the wing.

Example. Air (1.3 kg/m³) at 70 m/s above and 60 m/s below a wing of area 10 m²: ΔP = ½ × 1.3 × (4900 − 3600) = 845 Pa, so the lift is about 8450 N.

Time to empty a tank

How to apply. The level falls at speed −dh/dt = (a/A)√(2gh), so integrate dh/√h. Valid for a << A.

Example. Tank of area 1 m², hole 1 cm² (10⁻⁴ m²), water 5 m deep: t = 10⁴ × √(2 × 5 ÷ 10) = 10⁴ s.

Viscosity

Viscous force and coefficient of viscosity

How to apply. Between layers of a fluid moving at different speeds, there is a drag proportional to the area and the velocity gradient. For a plate over a thin film use dv/dx = v/thickness.

Example. A plate of 0.1 m² moves at 0.5 m/s over an oil film 1 mm thick with η = 0.2 Pa s: F = 0.2 × 0.1 × 0.5 ÷ 10⁻³ = 10 N.

Stokes’ law and terminal velocity

How to apply. Draw the forces: weight down, buoyancy and Stokes’ drag up. At terminal velocity the net force is zero. The speed grows as r², so a bigger sphere of the same material falls faster.

Example. Sphere r = 1 mm, ρ = 2500 kg/m³ in a liquid of σ = 1000 kg/m³ and η = 0.1 Pa s: v(T) = 2 × 10⁻⁶ × 1500 × 10 ÷ 0.9 = 0.033 m/s.

Reynolds number and the nature of flow

How to apply. Reynolds number is a ratio of inertial to viscous effects. A larger diameter, speed or density, and a smaller viscosity favour turbulence.

Example. Water (η = 10⁻³ Pa s) in a pipe of D = 2 cm with v = 0.1 m/s: R = 1000 × 0.1 × 0.02 ÷ 10⁻³ = 2000, so the flow is on the edge of turbulence.

Poiseuille’s formula

How to apply. The pressure difference P across a narrow tube drives the flow. The very strong dependence on r⁴ explains why a small narrowing of an artery matters. Use the electrical analogy for combinations.

Example. Radius doubled and length doubled at the same pressure: Q becomes 2⁴ ÷ 2 = 8 times.

Surface tension

Surface tension and surface energy

How to apply. The surface of a liquid is like a stretched membrane. The work done in increasing the area is T times the increase in area. Count both sides for a film.

Example. A soap film on a frame with a slider 5 cm long, T = 0.03 N/m: the force on the slider is 2 × 0.03 × 0.05 = 3 × 10⁻³ N.

Excess pressure in a drop or a bubble

How to apply. The curved surface presses on the inside, so the inside pressure is greater. Count how many surfaces the film has: one for a drop, two for a soap bubble.

Example. Soap bubble of radius 1 cm with T = 0.03 N/m: ΔP = 4 × 0.03 ÷ 0.01 = 12 Pa.

Work done in changing the surface area

How to apply. The surface energy is T times the area. The total surface area increases when a drop splits, so work is needed; it decreases when drops merge, so heat is released. The volume is conserved: R³ = n r³.

Example. A drop of radius 1 mm (T = 0.07 N/m) split into 1000 drops: W = 4π × 10⁻⁶ × 0.07 × 9 = 7.9 × 10⁻⁶ J.

Angle of contact and capillary rise

How to apply. Equate the weight of the liquid column, πr²hρg, to the upward pull of the surface tension, 2πrT cos θ. Remember that h is measured from the level outside, and that a shorter tube changes the radius of the meniscus, not the height.

Example. Water (T = 0.072 N/m) in a tube with r = 0.5 mm: h = 2 × 0.072 ÷ (5 × 10⁻⁴ × 1000 × 10) = 2.9 cm.

Capillary variations

How to apply. The same balance of forces as for a tube, but the surface tension acts along the line of contact. For a plate the total length of contact is the perimeter in contact with the liquid.

Example. Weight held up in a tube of r = 0.5 mm with T = 0.07 N/m (θ = 0): 2π × 5 × 10⁻⁴ × 0.07 = 2.2 × 10⁻⁴ N. A thin ring of radius 4 cm needs 4π × 0.04 × 0.07 = 0.035 N more than its weight.

Joining soap bubbles

How to apply. At constant outside pressure the surface-area condition gives r² = r₁² + r₂². For two bubbles in contact, the excess pressures differ by 4T(1/r₂ − 1/r₁) = 4T/R.

Example. Bubbles of radius 3 cm and 4 cm merge in air at constant pressure: r = √(9 + 16) = 5 cm. Two touching bubbles of 3 cm and 6 cm share a film of radius 6 cm.

Thermal expansion and calorimetry

Linear, area and volume expansion

How to apply. All dimensions grow in the same ratio, including the gap or the hole. For large ΔT, use L′ = L(1 + αΔT). The expansion coefficients are very small, so the approximations are fine.

Example. A steel rod 1 m long (α = 1.2 × 10⁻⁵ K⁻¹) heated by 100 °C grows by 1.2 × 10⁻³ m = 1.2 mm.

Expansion of liquids and anomalous water

How to apply. A liquid in a vessel shows only the difference between its expansion and that of the vessel. The real expansion is the sum of the apparent expansion and the expansion of the vessel.

Example. A liquid with γ(real) = 1.8 × 10⁻⁴ °C⁻¹ in a glass vessel with α = 9 × 10⁻⁶ °C⁻¹: γ(apparent) = 1.8 × 10⁻⁴ − 0.27 × 10⁻⁴ = 1.53 × 10⁻⁴ °C⁻¹.

Temperature scales, thermal stress and the pendulum clock

How to apply. If a rod cannot expand, it behaves as if it had been compressed back to its original length, so the stress is Y times the strain αΔT. A hotter pendulum is longer and slower.

Example. Steel rod (A = 1 cm², Y = 2 × 10¹¹ Pa, α = 1.2 × 10⁻⁵) fixed at both ends and heated by 50 °C: F = 2 × 10¹¹ × 10⁻⁴ × 1.2 × 10⁻⁵ × 50 = 1.2 × 10⁴ N. A clock with α = 1.9 × 10⁻⁵ warmed by 10 °C loses 8.2 s per day.

Specific heat and calorimetry

How to apply. In a mixing problem, no heat is lost to the surroundings, so the sum of the heat gained and lost is zero. Include the calorimeter by its water equivalent.

Example. 200 g of water at 80 °C and 300 g at 20 °C: T = (200 × 80 + 300 × 20) ÷ 500 = 44 °C.

Latent heat and change of state

How to apply. Add up the heat for each stage: warming, melting, warming again, boiling. Check that the heat available is enough to complete a stage before you go to the next one.

Example. 10 g of ice at −10 °C to steam at 100 °C: 10 × 0.5 × 10 + 10 × 80 + 10 × 1 × 100 + 10 × 540 = 50 + 800 + 1000 + 5400 = 7250 cal.

Composite rods and gaps

How to apply. Set the two expansions equal (or add them) and cancel ΔT. The ratio of lengths is the inverse of the ratio of α.

Example. Steel (α = 1.2 × 10⁻⁵) and brass (α = 1.8 × 10⁻⁵) rods differ in length by 10 cm at all temperatures: L(steel)/L(brass) = 3/2, so L(steel) = 30 cm and L(brass) = 20 cm.

Heat transfer

Conduction

How to apply. The rate of flow of heat is the temperature difference divided by the thermal resistance, exactly like Ohm’s law with current ↔ heat current and potential ↔ temperature.

Example. Copper rod, L = 1 m, A = 1 cm², ends 100 °C apart: dQ/dt = 400 × 10⁻⁴ × 100 ÷ 1 = 4 W.

Conduction through rods in series and parallel

How to apply. Draw the resistance network and solve it as an electrical circuit. For the junction temperature of two rods, equate the heat current through each.

Example. Two rods of the same size, k₁ = 2k₂, the ends at 100 °C and 0 °C: the junction is at (2 × 100 + 0) ÷ 3 = 66.7 °C.

Radiation: Stefan–Boltzmann law

How to apply. The power radiated depends on the fourth power of the absolute temperature. For a body in a room, subtract the energy it absorbs from the surroundings.

Example. A black body of area 1 m² at 500 K: P = 5.67 × 10⁻⁸ × 500⁴ = 3.5 × 10³ W.

Wien’s displacement law and the spectrum

How to apply. Wien’s law gives the temperature from the colour. The curves at different temperatures never cross, and the area under each grows as T⁴.

Example. λ(m) = 500 nm: T = 2.9 × 10⁻³ ÷ (5 × 10⁻⁷) = 5800 K, close to the temperature of the surface of the Sun.

Newton’s law of cooling

How to apply. The rate of loss of heat is proportional to the excess temperature over the surroundings. For a short interval use the average temperature as in the second form, which avoids the exponential.

Example. A body cools from 80 °C to 60 °C in 10 min, T₀ = 20 °C: 20 ÷ 10 = K × 50, so K = 0.04 min⁻¹. In the next 10 min it falls from 60 °C to 46.7 °C.

Ice on a pond and cooling of a liquid

How to apply. The heat conducted through the ice already formed is what freezes the next layer. As the layer grows, the resistance grows, so freezing slows.

Example. If ice 1 cm thick takes 1 h to form, the layer from 1 cm to 2 cm takes (4 − 1) = 3 h more.

Most asked question types

How do I solve “Extension of wires and the moduli” questions in JEE Main?

A wire or rod carries a load, or two wires are compared. Find the extension, the stress or the ratio of extensions. 1. Write Y = FL/(AΔL). 2. For a ratio, use ΔL ∝ L/r² with the same material and force. 3. For series wires add the extensions; for parallel wires divide the force. Example: Same force and material, lengths L and 2L, radii r and 2r: ΔL ∝ L/r², so the extensions are in the ratio (1 ÷ 1) : (2 ÷ 4) = 2 : 1.

How do I solve “Stress–strain curve and energy stored” questions in JEE Main?

A graph or a stretched wire is given. Find the modulus, the energy stored or compare two materials. 1. Read Y from the slope of the straight part. 2. Energy per unit volume = area under the curve = ½ × stress × strain. 3. Total energy = that × volume. Example: Energy stored in the steel wire above: ½ × 100 N × 1.27 mm = 0.064 J.

How do I solve “Pressure, Pascal’s law and manometers” questions in JEE Main?

A liquid column, U-tube, barometer or hydraulic press is given. Find the pressure, a height or a force. 1. Take the pressure as the same at the same level in the same liquid. 2. Add ρgh going down. 3. For a press use F₁/A₁ = F₂/A₂. Example: 10 m of water adds 1 atm: the total is 2 atm.

How do I solve “Buoyancy and floating” questions in JEE Main?

A body floats or hangs in a liquid. Find the fraction submerged, the apparent weight or what happens when the conditions change. 1. Floating: weight = ρ(fluid) V(in) g. 2. Hanging: apparent weight = W − B. 3. In a lift, g and g′ cancel in the ratio. Example: Ice of density 900 kg/m³ floats in water with 90% submerged.

How do I solve “Continuity and Bernoulli” questions in JEE Main?

A fluid flows through a pipe of changing area or height. Find a speed or a pressure difference. 1. Use A₁v₁ = A₂v₂. 2. Write P + ½ρv² + ρgh at both points. 3. Find ΔP. Example: Speeds 2 m/s and 8 m/s: the pressure drops by 3 × 10⁴ Pa.

How do I solve “Efflux through a hole and the range” questions in JEE Main?

A tank has a hole. Find the speed of the water, the range or the best height of the hole. 1. v = √(2gh) with h below the free surface. 2. R = 2√(h(H − h)). 3. The largest range R = H when h = H/2. Example: Hole 5 m below the surface: v = 10 m/s.

How do I solve “Viscous drag and terminal velocity” questions in JEE Main?

A sphere falls through a viscous liquid, or a plate slides over a film. Find the force or the terminal speed. 1. Draw weight, buoyancy and 6πηrv. 2. At the terminal speed they balance. 3. Use v(T) ∝ r² for a comparison. Example: r = 1 mm sphere in liquid of η = 0.1 Pa s: v(T) = 0.033 m/s.

How do I solve “Surface tension: force and energy” questions in JEE Main?

A film, a ring or a splitting drop is given. Find the force or the work done against surface tension. 1. Count the surfaces. 2. W = T × change in area. 3. Conserve volume when drops split or join. Example: A drop of radius 1 mm (T = 0.07 N/m) split into 1000 drops needs 7.9 × 10⁻⁶ J.

How do I solve “Excess pressure in a bubble or a drop” questions in JEE Main?

A drop or a soap bubble is given. Find the excess pressure or compare two bubbles. 1. Drop: 2T/R. 2. Soap bubble: 4T/R. 3. The smaller bubble has the greater pressure. Example: A soap bubble of 1 cm radius has an excess pressure of 12 Pa.

How do I solve “Capillary rise” questions in JEE Main?

A tube or two plates dip in a liquid. Find the height of the liquid, the angle of contact or the change in the rise. 1. h = 2T cos θ/(rρg). 2. Use h ∝ 1/r for a comparison. 3. θ > 90°: it is depressed. Example: Water in a tube of radius 0.5 mm rises 2.9 cm.

How do I solve “Thermal expansion” questions in JEE Main?

A rod, plate or liquid is heated. Find the change in length, area or volume, or the stress when it is held fixed. 1. ΔL = LαΔT, β = 2α, γ = 3α. 2. A hole expands like the material. 3. Held fixed: stress = YαΔT. Example: 1 m steel rod heated by 100 °C grows by 1.2 mm.

How do I solve “Calorimetry and latent heat” questions in JEE Main?

Hot and cold bodies are mixed, or ice is melted by hot water. Find the final temperature or the mass melted. 1. Heat lost = heat gained. 2. Add the latent heat when a state changes. 3. Check that each stage is complete. Example: 200 g of water at 80 °C mixed with 300 g at 20 °C gives 44 °C.

How do I solve “Conduction through rods and slabs” questions in JEE Main?

Rods are joined in series or parallel, or a slab conducts heat. Find the rate of flow, the junction temperature or the effective k. 1. R = L/kA for each rod. 2. Series: add R. Parallel: add the heat currents. 3. Use the same current in series to find the junction temperature. Example: Two rods with k₁ = 2k₂, ends at 100 °C and 0 °C: the junction is at 66.7 °C.

How do I solve “Radiation, Wien’s law and Newton’s cooling” questions in JEE Main?

A hot body radiates or cools. Find the power radiated, the temperature or the time of cooling. 1. P = eσAT⁴ with T in kelvin. 2. λ(m) T = 2.9 × 10⁻³ m K. 3. For cooling use the average-temperature form. Example: A body cooling from 80 °C to 60 °C in 10 min goes from 60 °C to 46.7 °C in the next 10 min.

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Simulations for Properties of Solids and Liquids

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