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Stretch a rubber band or compress a spring and the object changes shape. Remove the force, and if it returns to its original form, the material is elastic. The chapter that begins with deforming solids moves into fluids, where materials cannot resist shear and flow continuously. These Properties of Solids and Liquids JEE notes cover elasticity, stress and strain, elastic moduli, fluid pressure, buoyancy, Bernoulli's theorem, viscosity, surface tension, and capillarity with important concepts and examples for quick revision.
Properties of Solids and Liquids JEE Notes: Elasticity, Stress and Strain
Solids resist changes in their shape and size because of internal restoring forces. When an external force tries to deform a body, the internal forces oppose this change and try to bring the body back to its original state.
Deforming Force and Restoring Force
Deforming force: An external force that changes the shape or size of a body.
Restoring force: The internal force developed inside the body that opposes deformation and tries to restore the original shape.
Elasticity: The property of a material by which it regains its original shape and size after the deforming force is removed.
Plasticity: The property of a material by which it does not return to its original shape after removal of force. Clay is a common example of a plastic material.
Stress and Strain
To compare deformation in different objects, we use stress and strain instead of only force and change in length.
Stress: The restoring force developed per unit cross-sectional area of a deformed body.
$$\text{Stress}=\frac{F}{A}$$
The SI unit of stress is pascal (Pa) or $$\mathrm{N/m^2}$$.
Strain: The fractional change produced in the dimensions of a body due to applied stress.
Strain has no units because it is a ratio.
| Type of Stress | Formula | Type of Strain | Formula |
|---|---|---|---|
| Longitudinal stress | $$\frac{F}{A}$$ | Longitudinal strain | $$\frac{\Delta L}{L}$$ |
| Bulk stress | $$-\Delta P$$ | Volume strain | $$\frac{\Delta V}{V}$$ |
| Shearing stress | $$\frac{F}{A}$$ | Shearing strain | $$\tan\phi\approx\phi$$ |
Here, $$F$$ is the applied force, $$A$$ is the cross-sectional area, $$L$$ is the original length, $$V$$ is the original volume and $$\phi$$ represents the angle of shear.
Hooke's Law
Within the elastic limit, stress is directly proportional to strain.
$$\text{Stress}=E\times\text{Strain}$$
Here, $$E$$ represents the modulus of elasticity.
The stress-strain graph remains linear within the proportional limit. The value of modulus depends on the material and represents its stiffness.
Elastic Moduli
Different types of deformation require different elastic constants.
| Modulus | Formula | Application |
|---|---|---|
| Young's Modulus (Y) | $$Y=\frac{F/A}{\Delta L/L}=\frac{FL}{A\Delta L}$$ | Stretching or compression along one direction |
| Bulk Modulus (B) | $$B=-\frac{\Delta P}{\Delta V/V}=-\frac{V\Delta P}{\Delta V}$$ | Uniform compression from all directions |
| Shear Modulus (G) | $$G=\frac{F/A}{\tan\phi}$$ | Tangential deformation |
Important JEE points:
- A higher Young's modulus means the material is stiffer.
- Steel has a much higher Young's modulus than rubber, which means steel undergoes less deformation for the same force.
- The negative sign in bulk modulus keeps its value positive because pressure increase causes volume decrease.
- Fluids cannot resist shear stress, therefore their shear modulus is zero.
Worked Example
A steel wire of length 2 m and cross-sectional area $$2\times10^{-6}\,\mathrm{m^2}$$ is stretched by a force of 400 N. If Young's modulus of steel is $$2\times10^{11}\,\mathrm{Pa}$$, find the elongation.
Using:
$$Y=\frac{FL}{A\Delta L}$$
Therefore:
$$\Delta L=\frac{FL}{AY}$$
$$\Delta L=\frac{400\times2}{(2\times10^{-6})(2\times10^{11})}$$
$$\Delta L=2\times10^{-3}\,\mathrm{m}$$
$$\Delta L=2\,\mathrm{mm}$$
Poisson's Ratio
When a wire is stretched, its length increases but its diameter decreases. The ratio describing this lateral deformation compared to longitudinal deformation is called Poisson's ratio.
$$\sigma_P=-\frac{\text{lateral strain}}{\text{longitudinal strain}}$$
It is a dimensionless quantity. The negative sign indicates that lateral and longitudinal strains have opposite directions.
- For stable isotropic materials, Poisson's ratio lies between 0 and 0.5.
- Steel has a value close to 0.29.
- Rubber has a value close to 0.49.
- Cork has a value close to zero, which is why it works well as a bottle stopper.
Elastic Potential Energy
The work done in deforming a body within its elastic limit is stored as elastic potential energy.
| Quantity | Formula |
|---|---|
| Energy per unit volume | $$u=\frac{1}{2}\times\text{stress}\times\text{strain}$$ |
| Energy in a wire | $$U=\frac{1}{2}F\Delta L$$ |
| Alternative form | $$U=\frac{F^2L}{2AY}$$ |
Worked Example
A steel wire is stretched by a force of 400 N and elongates by 2 mm. Find the elastic energy stored.
$$U=\frac{1}{2}F\Delta L$$
$$U=\frac{1}{2}\times400\times2\times10^{-3}$$
$$U=0.4\,J$$
Pressure in Fluids, Pascal's Law and Buoyancy
A fluid is a substance that can flow, which includes both liquids and gases. Unlike solids, fluids cannot resist shear stress and continuously deform when a tangential force is applied.
Pressure in Fluids
Pressure: The force acting normally per unit area of a surface.
$$P=\frac{F}{A}$$
The SI unit of pressure is pascal (Pa).
Hydrostatic Pressure
The pressure inside a stationary liquid increases with depth because the weight of the liquid above contributes to the pressure.
$$P=P_0+\rho gh$$
Here:
- $$P_0$$ is the pressure at the surface.
- $$\rho$$ is the density of the fluid.
- $$g$$ is acceleration due to gravity.
- $$h$$ is the depth below the surface.
Important properties of hydrostatic pressure:
- Pressure increases linearly with depth.
- Pressure is the same at all points on the same horizontal level in a connected fluid.
- Atmospheric pressure at sea level is approximately $$1.013\times10^5\,\mathrm{Pa}$$.
Worked Example
A diver is 20 m below the surface of seawater having density $$1025\,\mathrm{kg/m^3}$$. Find the pressure experienced by the diver.
$$P=P_0+\rho gh$$
$$P=1.013\times10^5+1025\times9.8\times20$$
$$P=3.022\times10^5\,\mathrm{Pa}$$
The pressure is approximately 3 atmospheres.
Pascal's Law
Pascal's law states that pressure applied to an enclosed fluid is transmitted equally and undiminished in all directions.
This principle is used in hydraulic machines such as hydraulic lifts and hydraulic brakes.
$$\frac{F_1}{A_1}=\frac{F_2}{A_2}$$
Therefore:
$$F_2=F_1\frac{A_2}{A_1}$$
Worked Example
A hydraulic lift has a small piston area of $$5\,\mathrm{cm^2}$$ and a large piston area of $$500\,\mathrm{cm^2}$$. Find the force required on the small piston to lift a 1000 kg car.
Weight of car:
$$F_2=mg=1000\times9.8$$
$$F_1=F_2\frac{A_1}{A_2}$$
$$F_1=9800\times\frac{5}{500}$$
$$F_1=98\,N$$
Archimedes' Principle and Buoyant Force
When an object is partially or completely immersed in a fluid, the fluid applies an upward force called buoyant force or upthrust.
According to Archimedes' principle, the buoyant force is equal to the weight of the fluid displaced by the object.
$$F_b=\rho_fV_{sub}g$$
Here:
- $$\rho_f$$ is the density of fluid.
- $$V_{sub}$$ is the submerged volume.
| Condition | Result |
|---|---|
| $$\rho_{body}<\rho_{fluid}$$ | Body floats |
| $$\rho_{body}=\rho_{fluid}$$ | Neutral buoyancy |
| $$\rho_{body}>\rho_{fluid}$$ | Body sinks |
For a floating body:
$$\frac{V_{sub}}{V_{total}}=\frac{\rho_{body}}{\rho_{fluid}}$$
The apparent weight of an immersed object is:
$$W_{apparent}=mg-\rho_fVg$$
Worked Example
An ice cube of density $$900\,\mathrm{kg/m^3}$$ floats in water of density $$1000\,\mathrm{kg/m^3}$$. Find the fraction of ice above water.
Fraction submerged:
$$\frac{V_{sub}}{V_{total}}=\frac{900}{1000}=0.9$$
Fraction above water:
$$1-0.9=0.1$$
Therefore, 10% of the ice cube remains above water.
Fluid Dynamics: Continuity Equation and Bernoulli's Theorem
Fluid dynamics studies the motion of fluids. The ideal fluid model assumes that the fluid is incompressible, non-viscous, and flows steadily.
Streamline Flow
A streamline represents the path followed by a fluid particle during steady flow. At any point on a streamline, the velocity of fluid remains constant with time.
Laminar flow: Smooth and orderly flow where fluid layers move without mixing.
Turbulent flow: Irregular flow containing mixing and eddies.
Equation of Continuity
The equation of continuity is based on conservation of mass. The amount of fluid entering a pipe per second must equal the amount leaving it.
$$A_1v_1=A_2v_2$$
The product $$Av$$ represents volume flow rate.
$$Q=Av$$
Important conclusion:
- When pipe area decreases, fluid velocity increases.
- When pipe area increases, fluid velocity decreases.
Worked Example
Water flows through a pipe that narrows from diameter 4 cm to 2 cm. If velocity in the wider section is 1 m/s, find the velocity in the narrower section.
$$A_1v_1=A_2v_2$$
Since area is proportional to diameter squared:
$$\pi(2)^2\times1=\pi(1)^2v_2$$
$$v_2=4\,m/s$$
Bernoulli's Theorem
Bernoulli's theorem represents conservation of energy in a flowing fluid. The total energy per unit volume remains constant along a streamline.
$$P+\frac{1}{2}\rho v^2+\rho gh=\text{constant}$$
Between two points:
$$P_1+\frac{1}{2}\rho v_1^2+\rho gh_1=P_2+\frac{1}{2}\rho v_2^2+\rho gh_2$$
The main conclusion from Bernoulli's theorem is:
- Higher fluid velocity corresponds to lower pressure.
- Lower fluid velocity corresponds to higher pressure.
This explains aeroplane lift, perfume sprayers, and the motion of spinning balls.
| Special Case | Formula |
|---|---|
| Torricelli's theorem | $$v=\sqrt{2gh}$$ |
| Venturi effect | $$P_1-P_2=\frac{1}{2}\rho(v_2^2-v_1^2)$$ |
Worked Example
A tank contains water up to a height of 5 m. Find the velocity of water escaping through a small hole near the bottom.
$$v=\sqrt{2gh}$$
$$v=\sqrt{2\times9.8\times5}$$
$$v=\sqrt{98}$$
$$v=9.9\,m/s$$
JEE Tip: Bernoulli's equation is easier to remember when viewed as conservation of energy: pressure energy + kinetic energy + potential energy remains constant.
Viscosity, Stokes' Law and Poiseuille's Formula
Real fluids have internal friction between their layers. This property of fluids that opposes relative motion between adjacent layers is called viscosity.
Honey has higher viscosity compared to water because its layers offer greater resistance to flow.
Newton's Law of Viscosity
According to Newton's law of viscosity, the viscous force between two fluid layers is proportional to the velocity gradient.
$$F=-\eta A\frac{dv}{dy}$$
Here:
- $$\eta$$ is the coefficient of viscosity.
- $$A$$ is the area of contact between layers.
- $$\frac{dv}{dy}$$ is the velocity gradient.
The SI unit of viscosity is:
$$\mathrm{Pa\cdot s}$$
The CGS unit is poise.
Stokes' Law
When a small spherical object moves through a viscous fluid, it experiences a resistive force called viscous drag.
$$F=6\pi\eta rv$$
This relation is valid for small spheres moving slowly through fluids with laminar flow.
Terminal Velocity
When a sphere falls through a viscous fluid, it initially accelerates. As velocity increases, viscous drag also increases. Eventually, the drag balances the effective weight, and the sphere falls with constant velocity called terminal velocity.
$$v_t=\frac{2r^2(\rho_s-\rho_f)g}{9\eta}$$
From this relation:
- Terminal velocity increases with the square of radius.
- Terminal velocity increases with density difference.
- Terminal velocity decreases when viscosity increases.
Worked Example
A steel ball of radius $$1\,mm$$ and density $$7800\,kg/m^3$$ falls through glycerine of density $$1260\,kg/m^3$$. If viscosity is $$0.83\,Pa\cdot s$$, find terminal velocity.
$$v_t=\frac{2r^2(\rho_s-\rho_f)g}{9\eta}$$
$$v_t=\frac{2(10^{-3})^2(7800-1260)(9.8)}{9(0.83)}$$
$$v_t=0.0172\,m/s$$
Therefore, terminal velocity is approximately:
$$1.72\,cm/s$$
Poiseuille's Formula
For steady laminar flow of a viscous fluid through a cylindrical pipe:
$$Q=\frac{\pi\Delta P r^4}{8\eta L}$$
Important observation:
- Flow rate depends on the fourth power of radius.
- Doubling the radius increases flow rate by 16 times.
Reynolds Number
Reynolds number predicts whether fluid flow will be smooth or turbulent.
$$Re=\frac{\rho vD}{\eta}$$
| Reynolds Number | Flow Type |
|---|---|
| $$Re<2000$$ | Laminar flow |
| $$2000 | Transition region |
| $$Re>4000$$ | Turbulent flow |
Surface Tension and Capillarity
Molecules inside a liquid experience attractive forces from all directions. However, molecules at the surface experience a net inward force, causing the surface to behave like a stretched membrane.
Surface Tension
Surface tension is the force acting per unit length on the surface of a liquid.
$$S=\frac{F}{l}$$
The SI unit of surface tension is N/m.
Surface Energy
The energy required to increase the surface area of a liquid is called surface energy.
$$E=S\Delta A$$
Excess Pressure Due to Surface Tension
| Surface | Excess Pressure |
|---|---|
| Liquid drop | $$\Delta P=\frac{2S}{R}$$ |
| Soap bubble | $$\Delta P=\frac{4S}{R}$$ |
| Liquid meniscus | $$\Delta P=\frac{2S}{R}$$ |
A soap bubble has two surfaces, which is why its excess pressure is twice that of a liquid drop.
Worked Example
Find the excess pressure inside a soap bubble of radius $$0.01\,m$$ if surface tension is $$0.03\,N/m$$.
$$\Delta P=\frac{4S}{R}$$
$$\Delta P=\frac{4(0.03)}{0.01}$$
$$\Delta P=12\,Pa$$
Angle of Contact
The angle between the tangent to the liquid surface and the solid surface at the point of contact is called angle of contact.
| Angle | Behaviour |
|---|---|
| $$\theta<90^\circ$$ | Liquid wets the surface and rises |
| $$\theta>90^\circ$$ | Liquid does not wet the surface and falls |
| $$\theta=0^\circ$$ | Complete wetting |
Capillary Rise
The rise or fall of liquid in a narrow tube due to surface tension is called capillarity.
$$h=\frac{2S\cos\theta}{\rho gr}$$
Important points:
- Water rises in glass because angle of contact is less than 90°.
- Mercury is depressed because its angle of contact is greater than 90°.
- Smaller tube radius gives greater capillary rise.
Worked Example
Water rises 10 cm in a capillary tube of radius $$0.5\,mm$$. Find surface tension if angle of contact is zero.
$$S=\frac{h\rho gr}{2\cos\theta}$$
$$S=\frac{0.1\times1000\times9.8\times0.5\times10^{-3}}{2}$$
$$S=0.245\,N/m$$
Properties of Solids and Liquids Formula Sheet
| Concept | Formula |
|---|---|
| Young's modulus | $$Y=\frac{FL}{A\Delta L}$$ |
| Bulk modulus | $$B=-\frac{V\Delta P}{\Delta V}$$ |
| Shear modulus | $$G=\frac{F/A}{\phi}$$ |
| Elastic energy | $$U=\frac{1}{2}F\Delta L$$ |
| Hydrostatic pressure | $$P=P_0+\rho gh$$ |
| Buoyant force | $$F_b=\rho_fV_{sub}g$$ |
| Continuity equation | $$A_1v_1=A_2v_2$$ |
| Bernoulli equation | $$P+\frac{1}{2}\rho v^2+\rho gh=constant$$ |
| Stokes' law | $$F=6\pi\eta rv$$ |
| Terminal velocity | $$v_t=\frac{2r^2(\rho_s-\rho_f)g}{9\eta}$$ |
| Surface tension | $$S=\frac{F}{l}$$ |
| Capillary rise | $$h=\frac{2S\cos\theta}{\rho gr}$$ |
JEE Important Points, Common Mistakes and Quick Revision
Points JEE Repeatedly Tests
- Stress is force per unit area, while strain is a dimensionless ratio representing deformation.
- Hooke's law is valid only within the elastic limit where stress is directly proportional to strain.
- Young's modulus represents stiffness. A higher value means the material undergoes less deformation for the same applied force.
- Fluids cannot resist shear stress, therefore their shear modulus is zero.
- Pressure inside a fluid increases linearly with depth and is independent of the shape of the container.
- Pascal's law is the working principle behind hydraulic lifts and hydraulic brakes.
- Buoyant force depends on the volume of fluid displaced, not the total volume of the object unless it is completely immersed.
- In continuity equation, a smaller area means higher fluid velocity.
- Bernoulli's theorem shows that faster-moving fluid has lower pressure.
- Terminal velocity depends on radius squared, density difference, and viscosity of the fluid.
- Surface tension causes droplets to become spherical because a sphere has minimum surface area.
- A soap bubble has twice the surface compared to a liquid drop, giving twice the excess pressure.
- Capillary rise depends on surface tension, angle of contact, density, gravity and radius of the tube.
Common Mistakes to Avoid
- Confusing stress and pressure: Stress is related to internal restoring force in solids, while pressure is force applied normally on a fluid surface.
- Forgetting the negative sign in bulk modulus: The negative sign ensures bulk modulus remains positive because pressure increase reduces volume.
- Using total volume instead of submerged volume: Buoyant force depends only on the volume of fluid displaced.
- Using the wrong pressure formula for bubbles: A soap bubble has two surfaces, so its excess pressure is different from a liquid drop.
- Ignoring radius dependence in Poiseuille's law: Flow rate changes with the fourth power of radius.
- Applying Bernoulli's theorem to viscous or turbulent flow: It applies only for steady, incompressible, non-viscous flow.
- Missing the angle of contact in capillary problems: Water and mercury behave differently because their contact angles are different.
- Mixing units: Always convert quantities into SI units before substitution.
Quick Revision Notes for Properties of Solids and Liquids
- Elasticity is the ability of a body to regain its original shape after removal of deforming force.
- Stress is force per unit area and strain is fractional deformation.
- Three elastic moduli are Young's modulus, bulk modulus and shear modulus.
- Fluids have zero shear modulus because they cannot resist shear deformation.
- Elastic energy stored in a wire depends on the applied force and extension.
- Hydrostatic pressure increases with depth according to the relation between pressure, density, gravity and height.
- Pascal's law explains force multiplication in hydraulic machines.
- Archimedes' principle explains buoyancy and floating conditions.
- Continuity equation represents conservation of mass in fluid flow.
- Bernoulli's theorem connects pressure, velocity and height of a flowing fluid.
- Viscosity represents resistance offered by fluid layers against relative motion.
- Stokes' law helps determine drag force and terminal velocity.
- Surface tension arises due to molecular forces acting at the liquid surface.
- Capillary rise depends on surface tension and tube radius.
Exam focus: Numerical questions from this chapter are commonly based on Young's modulus, pressure in fluids, buoyancy, Bernoulli's theorem, terminal velocity and capillary rise. Understanding the formula and practising numerical applications is essential for scoring well in JEE Physics.
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