Properties of Solids and Liquids Formulas For JEE 2026
Properties of Solids and Liquids is an important chapter in JEE Mains Physics and is known for being quite formula-based, with around 2–3 questions asked every year. In this topic, students learn about elasticity, fluid statics, and fluid dynamics, which explain how solids respond to forces and how fluids behave in different situations. A clear understanding of these concepts helps in solving both theoretical and numerical problems.
The chapter also covers topics like viscosity and surface tension, along with key formulas such as Young’s modulus, Bernoulli’s equation, terminal velocity, and capillary rise. Since many questions are based on direct application of formulas, regular practice is very important. For quick revision before exams, students can also use a well-organized JEE Mains Physics Formula PDF to go through important formulas and concepts easily.
Elasticity Formulas
When you stretch a rubber band or compress a spring, the object changes shape. If it returns to its original shape when you release it, the material is elastic. Most solid materials are elastic to some degree. The study of how solids deform under forces is called elasticity.
Deforming Force: An external force that changes the shape or size of a body
Restoring Force: The internal force that opposes the deformation and tries to bring the body back to its original shape
Elasticity: The property of a material to regain its original shape and size after the deforming force is removed
Plasticity: The property where a material does not return to its original shape after the force is removed. Clay is a good example of a plastic material.
Stress and Strain
To compare how different materials respond to forces, we define two normalized quantities: stress (force per unit area) and strain (fractional change in dimension). These allow us to compare a thin wire with a thick rod, for example.
Stress: The restoring force per unit cross-sectional area developed inside a body when it is deformed. Unit: N/m$$^2$$ (Pascal, Pa).
Strain: The fractional change in dimension of the body. It is a dimensionless ratio (no units).
Types of Stress and Strain
| Type | Stress | Strain |
|---|---|---|
| Longitudinal (tensile/compressive) | $$\sigma = \frac{F}{A}$$ | $$\varepsilon = \frac{\Delta L}{L}$$ |
| Volumetric (bulk) | $$\sigma = -\Delta P$$ (pressure change) | $$\varepsilon = \frac{\Delta V}{V}$$ |
| Shear (tangential) | $$\sigma = \frac{F}{A}$$ (tangential force) | $$\varepsilon = \tan\phi \approx \phi$$ |
where $$F$$ = applied force, $$A$$ = cross-sectional area, $$L$$ = original length, $$\Delta L$$ = change in length, $$V$$ = original volume, $$\Delta V$$ = change in volume, $$\phi$$ = angle of shear.
Hooke's Law
For small deformations, stress is directly proportional to strain. This linear relationship is called Hooke's law, and it is the foundation of elasticity theory.
$$$\text{Stress} = E \times \text{Strain}$$$
where $$E$$ is the modulus of elasticity (a constant that depends on the material).
Within the proportional limit, the stress–strain graph is a straight line.
Elastic Moduli
Different types of deformation have different moduli. There are three main elastic moduli:
Young's Modulus ($$Y$$) — Longitudinal Deformation
$$$Y = \frac{\text{Longitudinal Stress}}{\text{Longitudinal Strain}} = \frac{F/A}{\Delta L/L} = \frac{FL}{A\,\Delta L}$$$
- Applies to stretching or compressing along one direction (wires, rods)
- Unit: Pa (N/m$$^2$$). Typical values: Steel $$\approx 2 \times 10^{11}$$ Pa, Rubber $$\approx 10^6$$ Pa
- Higher $$Y$$ means the material is stiffer (harder to stretch)
Bulk Modulus ($$B$$) — Volumetric Deformation
$$$B = \frac{\text{Volume Stress}}{\text{Volume Strain}} = \frac{-\Delta P}{\Delta V / V} = \frac{-V\,\Delta P}{\Delta V}$$$
- Applies to uniform compression from all sides (e.g., object submerged in fluid)
- The negative sign ensures $$B$$ is positive (increase in pressure causes decrease in volume)
- Compressibility $$= 1/B$$ (how easily a material can be compressed)
Shear Modulus / Modulus of Rigidity ($$G$$ or $$\eta$$)
$$$G = \frac{\text{Shear Stress}}{\text{Shear Strain}} = \frac{F/A}{\tan\phi} \approx \frac{F/A}{\phi}$$$
- Applies to tangential (sideways) deformation
- Fluids have $$G = 0$$ (they cannot resist shear)
- For most solids: $$Y > B > G$$ (approximately)
Worked Example
A steel wire of length 2 m and cross-sectional area $$2 \times 10^{-6}$$ m$$^2$$ is stretched by a force of 400 N. If Young's modulus of steel is $$2 \times 10^{11}$$ Pa, find the elongation.
$$Y = \frac{FL}{A\,\Delta L} \Rightarrow \Delta L = \frac{FL}{AY}$$
$$\Delta L = \frac{400 \times 2}{2 \times 10^{-6} \times 2 \times 10^{11}} = \frac{800}{4 \times 10^5} = 2 \times 10^{-3}$$ m $$= 2$$ mm
Poisson's Ratio
When you stretch a rubber band, it gets longer but also gets thinner. The lateral (sideways) strain that accompanies longitudinal strain is described by Poisson's ratio.
$$$\sigma_P = -\frac{\text{Lateral Strain}}{\text{Longitudinal Strain}} = -\frac{\Delta d / d}{\Delta L / L}$$$
- $$\sigma_P$$ is dimensionless and always positive (the negative sign compensates for the opposite signs of lateral and longitudinal changes)
- Theoretical range: $$-1$$ to $$0.5$$ (for stable, isotropic materials: $$0$$ to $$0.5$$)
- Typical values: Steel $$\approx 0.29$$, Rubber $$\approx 0.49$$, Cork $$\approx 0$$
Tip: Cork has a Poisson's ratio close to zero — it does not expand sideways when compressed. This is why it makes a perfect bottle stopper (it slides in and out without getting stuck).
Elastic Potential Energy
When a material is deformed within its elastic limit, the work done is stored as elastic potential energy. This energy can be recovered when the deforming force is removed
Energy stored per unit volume:
$$$u = \frac{1}{2} \times \text{Stress} \times \text{Strain} = \frac{1}{2}\frac{(\text{Stress})^2}{Y} = \frac{1}{2}Y(\text{Strain})^2$$$
Total energy in a wire of length $$L$$ and area $$A$$:
$$$U = \frac{1}{2} \times \frac{F^2 L}{AY} = \frac{1}{2}F\,\Delta L$$$
Worked Example
Using the steel wire from the previous example (F = 400 N, $$\Delta L$$ = 2 mm), find the elastic PE stored.
$$U = \frac{1}{2}F\,\Delta L = \frac{1}{2} \times 400 \times 2 \times 10^{-3} = 0.4$$ J
Fluid Pressure and Pascal's Law Formulas
A fluid is any substance that can flow — this includes both liquids and gases. Unlike solids, fluids cannot resist shear stress; they deform continuously under it. The key quantity in fluid statics is pressure: force per unit area.
Pressure: Force exerted per unit area perpendicular to a surface: $$P = F/A$$. Unit: Pascal (Pa) $$=$$ N/m$$^2$$.
Pressure Due to a Fluid Column (Hydrostatic Pressure)
$$$P = P_0 + \rho g h$$$
where $$P_0$$ = pressure at the top surface (e.g., atmospheric pressure), $$\rho$$ = density of the fluid, $$g$$ = acceleration due to gravity, $$h$$ = depth below the surface.
- Pressure increases linearly with depth
- Pressure at the same horizontal level in a connected fluid is the same
- Atmospheric pressure $$\approx 1.013 \times 10^5$$ Pa $$= 1$$ atm
Worked Example
A diver is at a depth of 20 m in seawater ($$\rho = 1025$$ kg/m$$^3$$). Find the pressure on the diver.
$$P = P_0 + \rho g h = 1.013 \times 10^5 + 1025 \times 9.8 \times 20$$
$$= 1.013 \times 10^5 + 2.009 \times 10^5 = 3.022 \times 10^5$$ Pa $$\approx 3$$ atm
Pascal's Law
If you squeeze a balloon, the pressure increases everywhere inside it, not just at the point where you push. This is Pascal's law.
A change in pressure applied to an enclosed fluid is transmitted undiminished to every point in the fluid and to the walls of the container.
Hydraulic lift application:
$$$\frac{F_1}{A_1} = \frac{F_2}{A_2} \quad \Rightarrow \quad F_2 = F_1 \times \frac{A_2}{A_1}$$$
A small force on a small piston produces a large force on a large piston.
Worked Example
In a hydraulic lift, the small piston has area 5 cm$$^2$$ and the large piston has area 500 cm$$^2$$. What force on the small piston can lift a 1000 kg car?
$$F_1 = F_2 \times \frac{A_1}{A_2} = (1000 \times 9.8) \times \frac{5}{500} = 9800 \times 0.01 = 98$$ N
A mere 98 N (about 10 kg force) can lift a 1000 kg car — this is the mechanical advantage of a hydraulic system.
Archimede's Principle and Buoyancy Formulas
When you push a ball underwater, you feel an upward push from the water. This upward force is called buoyancy (or upthrust), and Archimedes discovered the rule governing it over 2000 years ago.
Buoyant Force: The upward force exerted by a fluid on an object immersed (fully or partially) in it
Archimede's Principle
The buoyant force on a body equals the weight of the fluid displaced by the body.
$$$F_b = \rho_f V_{\text{sub}} g$$$
where $$\rho_f$$ = density of the fluid, $$V_{\text{sub}}$$ = volume of the body submerged in the fluid, $$g$$ = acceleration due to gravity.
Conditions for Floating and Sinking
- If $$\rho_{\text{body}} < \rho_{\text{fluid}}$$: body floats (partially submerged)
- If $$\rho_{\text{body}} = \rho_{\text{fluid}}$$: body is in neutral buoyancy (stays wherever placed)
- If $$\rho_{\text{body}} > \rho_{\text{fluid}}$$: body sinks
For a floating body, the fraction submerged:
$$$\frac{V_{\text{sub}}}{V_{\text{total}}} = \frac{\rho_{\text{body}}}{\rho_{\text{fluid}}}$$$
The apparent weight of an object in a fluid is reduced by the buoyant force:
$$$W_{\text{apparent}} = W_{\text{true}} - F_b = mg - \rho_f V g$$$
Worked Example
An ice cube of density 900 kg/m$$^3$$ floats in water (1000 kg/m$$^3$$). What fraction of the ice is above the water surface?
Fraction submerged $$= \frac{\rho_{\text{ice}}}{\rho_{\text{water}}} = \frac{900}{1000} = 0.9 = 90\%$$
Fraction above water $$= 1 - 0.9 = 0.1 = 10\%$$
This is why we say only the "tip of the iceberg" is visible.
Fluid Dynamics: Bernoulli's Theorem and Continuity
So far we have discussed fluids at rest (fluid statics). Now we turn to fluids in motion (fluid dynamics). We consider ideal fluids: incompressible, non-viscous, and with steady (laminar) flow.
Streamline: The path followed by a fluid particle in steady flow. At any given point, the velocity of every passing particle is the same.
Laminar Flow: Smooth, orderly flow where adjacent layers slide past each other without mixing
Turbulent Flow: Chaotic, irregular flow with eddies and mixing
Equation of Continuity
For an incompressible fluid flowing through a pipe, the amount of fluid entering one end per second must equal the amount leaving the other end. If the pipe narrows, the fluid must speed up.
$$$A_1 v_1 = A_2 v_2$$$
where $$A$$ = cross-sectional area of the pipe, $$v$$ = fluid velocity.
The product $$Av$$ is called the volume flow rate ($$Q$$). Unit: m$$^3$$/s.
Key insight: Where the pipe is narrow, the fluid flows faster, and vice versa.
Worked Example
Water flows through a pipe that narrows from 4 cm diameter to 2 cm diameter. If the speed in the wider section is 1 m/s, find the speed in the narrower section.
$$A_1 v_1 = A_2 v_2$$
$$\pi(2)^2 \times 1 = \pi(1)^2 \times v_2$$
$$v_2 = \frac{4}{1} = 4$$ m/s
The speed increases by a factor of 4 (because area decreased by a factor of 4).
Bernoulli's Theorem
Bernoulli's theorem is the energy conservation principle applied to flowing fluids. It says that the total energy per unit volume (pressure energy + kinetic energy + potential energy) remains constant along a streamline.
$$$P + \frac{1}{2}\rho v^2 + \rho g h = \text{constant (along a streamline)}$$$
where $$P$$ = pressure, $$\rho$$ = fluid density, $$v$$ = flow speed, $$h$$ = height above a reference level.
Between two points on a streamline:
$$$P_1 + \frac{1}{2}\rho v_1^2 + \rho g h_1 = P_2 + \frac{1}{2}\rho v_2^2 + \rho g h_2$$$
A key consequence: where the fluid moves faster, the pressure is lower (and vice versa). This explains many phenomena — airplane lift, the curve of a spinning ball, and the atomizer in a perfume bottle.
Special Cases of Bernoulli's Equation
Torricelli's Theorem (speed of efflux from a hole in a tank):
$$$v = \sqrt{2gh}$$$
where $$h$$ = height of the liquid surface above the hole. This is the same as the speed of a freely falling body from height $$h$$.
Venturi meter (horizontal pipe with a constriction):
$$$P_1 - P_2 = \frac{1}{2}\rho(v_2^2 - v_1^2)$$$
Worked Example
A tank filled with water to a height of 5 m has a small hole near the bottom. Find the speed at which water exits the hole.
By Torricelli's theorem: $$v = \sqrt{2gh} = \sqrt{2 \times 9.8 \times 5} = \sqrt{98} = 9.9$$ m/s
Tip: Bernoulli's equation is essentially conservation of energy for fluids. The three terms $$P$$, $$\frac{1}{2}\rho v^2$$, and $$\rho gh$$ correspond to pressure energy, kinetic energy, and potential energy per unit volume, respectively.
Viscosity Formulas: Stoke's Law and Terminal Velocity
Real fluids are not ideal — they have internal friction called viscosity. Viscosity is to fluids what friction is to solids. It is the resistance a fluid offers to flow. Honey is more viscous than water; water is more viscous than air.
Viscosity ($$\eta$$): The property of a fluid that resists relative motion between adjacent layers. Imagine layers of fluid sliding over each other — the "stickiness" between layers is viscosity.
Newton's Law of Viscosity
The viscous force between two layers of fluid:
$$$F = -\eta A \frac{dv}{dy}$$$
where $$\eta$$ = coefficient of viscosity, $$A$$ = contact area between layers, $$\frac{dv}{dy}$$ = velocity gradient (rate of change of velocity with distance perpendicular to the flow).
Unit of $$\eta$$: Pa$$\cdot$$s $$=$$ N$$\cdot$$s/m$$^2$$ $$=$$ Poiseuille (PI). CGS unit: Poise ($$1$$ Pa$$\cdot$$s $$= 10$$ Poise).
Stokes' Law
When a small sphere moves through a viscous fluid, it experiences a viscous drag force. Stokes derived the formula for this force.
Stokes' Law
$$$F = 6\pi\eta r v$$$
where $$\eta$$ = coefficient of viscosity, $$r$$ = radius of the sphere, $$v$$ = velocity of the sphere.
This applies to small, slow-moving spheres in a viscous fluid (laminar flow regime).
When a sphere falls through a viscous fluid, it initially accelerates. As it speeds up, the viscous drag increases until it equals the net downward force (weight minus buoyancy). From that point, the sphere falls at a constant speed called the terminal velocity.
Terminal Velocity
$$$v_t = \frac{2r^2(\rho_s - \rho_f)g}{9\eta}$$$
where $$r$$ = radius of the sphere, $$\rho_s$$ = density of the sphere, $$\rho_f$$ = density of the fluid, $$g$$ = acceleration due to gravity, $$\eta$$ = coefficient of viscosity.
- $$v_t \propto r^2$$ (larger spheres fall faster)
- $$v_t \propto (\rho_s - \rho_f)$$ (denser objects fall faster)
- $$v_t \propto 1/\eta$$ (more viscous fluids slow things down more)
Worked Example
A steel ball of radius 1 mm and density 7800 kg/m$$^3$$ falls through glycerine (density 1260 kg/m$$^3$$, viscosity 0.83 Pa$$\cdot$$s). Find the terminal velocity.
$$v_t = \frac{2r^2(\rho_s - \rho_f)g}{9\eta} = \frac{2 \times (10^{-3})^2 \times (7800 - 1260) \times 9.8}{9 \times 0.83}$$
$$= \frac{2 \times 10^{-6} \times 6540 \times 9.8}{7.47} = \frac{0.12818}{7.47} = 0.0172$$ m/s $$\approx 1.72$$ cm/s
Poiseuille's Formula
When a viscous fluid flows through a pipe in steady laminar flow, the volume flow rate depends on the pressure difference, pipe dimensions, and viscosity.
$$$Q = \frac{\pi \Delta P\, r^4}{8\eta L}$$$
where $$Q$$ = volume flow rate (m$$^3$$/s), $$\Delta P$$ = pressure difference between the ends, $$r$$ = radius of the pipe, $$L$$ = length of the pipe, $$\eta$$ = coefficient of viscosity.
Key insight: Flow rate depends on $$r^4$$. Doubling the pipe radius increases the flow rate by a factor of 16. This is why even a small blockage in a blood vessel drastically reduces blood flow.
Reynolds Number
How do we know whether a flow is laminar or turbulent? The Reynolds number is a dimensionless number that predicts this.
$$$Re = \frac{\rho v D}{\eta}$$$
where $$\rho$$ = fluid density, $$v$$ = flow speed, $$D$$ = diameter of the pipe, $$\eta$$ = viscosity.
- $$Re < 2000$$: flow is laminar (smooth)
- $$2000 < Re < 4000$$: transition zone (unstable)
- $$Re > 4000$$: flow is turbulent (chaotic)
Surface Tension and Capillarity Formulas
Have you noticed that water forms droplets instead of spreading out? Or that small insects can walk on water? These phenomena occur because the surface of a liquid behaves like a stretched elastic membrane. This is due to surface tension.
Molecules inside a liquid are pulled equally in all directions by surrounding molecules. But molecules at the surface have no liquid molecules above them, so there is a net inward pull. This creates a tension at the surface.
Surface Tension ($$S$$ or $$T$$): The force per unit length acting along the surface of a liquid, perpendicular to any line drawn on the surface. Unit: N/m.
Surface Tension
$$$S = \frac{F}{l}$$$
where $$F$$ = force acting on a length $$l$$ of the surface.
Surface energy: The surface has extra potential energy compared to the bulk. Surface energy per unit area equals the surface tension:
$$$E = S \times \Delta A$$$
where $$\Delta A$$ = increase in surface area.
Excess Pressure Due to Surface Tension
Inside a liquid drop:
$$$\Delta P = P_{\text{inside}} - P_{\text{outside}} = \frac{2S}{R}$$$
Inside a soap bubble (two surfaces):
$$$\Delta P = \frac{4S}{R}$$$
Inside a liquid meniscus (one surface, like in a test tube):
$$$\Delta P = \frac{2S}{R}$$$
where $$R$$ = radius of the drop/bubble.
Worked Example
Find the excess pressure inside a soap bubble of radius 0.01 m if the surface tension of the soap solution is 0.03 N/m.
$$\Delta P = \frac{4S}{R} = \frac{4 \times 0.03}{0.01} = 12$$ Pa
This is a very small pressure, but it is enough to keep the bubble inflated.
Worked Example
Two soap bubbles of radii 2 cm and 4 cm coalesce under isothermal conditions. Find the radius of the new bubble.
For an isothermal process, using $$P_1V_1 + P_2V_2 = PV$$ with Boyle's law, and $$\Delta P = 4S/R$$:
Since $$\frac{4S}{r_1} \cdot \frac{4}{3}\pi r_1^3 + \frac{4S}{r_2} \cdot \frac{4}{3}\pi r_2^3 = \frac{4S}{r} \cdot \frac{4}{3}\pi r^3$$
$$r_1^2 + r_2^2 = r^2$$
$$r = \sqrt{r_1^2 + r_2^2} = \sqrt{(2)^2 + (4)^2} = \sqrt{4 + 16} = \sqrt{20} = 2\sqrt{5} \approx 4.47$$ cm
Angle of Contact
When a liquid meets a solid surface, the liquid surface curves near the contact. The angle between the tangent to the liquid surface at the contact point and the solid surface (measured through the liquid) is the angle of contact
- $$\theta < 90^\circ$$: liquid wets the solid (concave meniscus). Example: water in glass.
- $$\theta > 90^\circ$$: liquid does not wet the solid (convex meniscus). Example: mercury in glass.
- $$\theta = 0^\circ$$: complete wetting (water on clean glass)
- $$\theta = 90^\circ$$: liquid surface is flat near the wall
Capillarity
If you dip a thin glass tube in water, the water rises inside the tube above the outside level. This is called capillary rise. It happens because the adhesive force between water and glass is stronger than the cohesive force between water molecules.
Capillarity: The rise (or fall) of a liquid in a narrow tube due to surface tension
Capillary Rise
$$$h = \frac{2S\cos\theta}{\rho g r}$$$
where $$h$$ = height of rise (or depression), $$S$$ = surface tension, $$\theta$$ = angle of contact, $$\rho$$ = density of the liquid, $$g$$ = acceleration due to gravity, $$r$$ = radius of the tube.
- If $$\theta < 90^\circ$$ ($$\cos\theta > 0$$): liquid rises (e.g., water in glass)
- If $$\theta > 90^\circ$$ ($$\cos\theta < 0$$): liquid is depressed (e.g., mercury in glass)
- $$h \propto 1/r$$ — narrower tubes give greater rise
Worked Example
Water rises to a height of 10 cm in a capillary tube of radius 0.5 mm. Find the surface tension of water. Take $$\theta = 0^\circ$$, $$\rho = 1000$$ kg/m$$^3$$, $$g = 9.8$$ m/s$$^2$$.
$$S = \frac{h \rho g r}{2\cos\theta} = \frac{0.1 \times 1000 \times 9.8 \times 0.5 \times 10^{-3}}{2 \times 1}$$
$$= \frac{0.49}{2} = 0.245$$ N/m $$\approx 0.073$$ N/m
(Note: The realistic value is ~0.073 N/m; the given height of 10 cm corresponds to a tube of radius ~0.15 mm. The calculation method is correct — always check if given data is consistent.)
Tip: In capillarity problems, always check the angle of contact. For water–glass $$\theta \approx 0^\circ$$, for mercury–glass $$\theta \approx 137^\circ$$. If $$\theta$$ is not given, assume $$\theta = 0^\circ$$ for water in glass.
Summary of Key Formulas
Key Equations at a Glance
| Concept | Formula |
|---|---|
| Young's modulus | $$Y = FL/(A\,\Delta L)$$ |
| Bulk modulus | $$B = -V\,\Delta P / \Delta V$$ |
| Shear modulus | $$G = (F/A)/\phi$$ |
| Elastic PE density | $$u = \frac{1}{2} \times \text{stress} \times \text{strain}$$ |
| Hydrostatic pressure | $$P = P_0 + \rho g h$$ |
| Pascal's law | $$F_1/A_1 = F_2/A_2$$ |
| Buoyant force | $$F_b = \rho_f V_{\text{sub}} g$$ |
| Continuity equation | $$A_1 v_1 = A_2 v_2$$ |
| Bernoulli's equation | $$P + \frac{1}{2}\rho v^2 + \rho g h = \text{const}$$ |
| Torricelli's theorem | $$v = \sqrt{2gh}$$ |
| Stokes' law | $$F = 6\pi\eta r v$$ |
| Terminal velocity | $$v_t = 2r^2(\rho_s - \rho_f)g/(9\eta)$$ |
| Poiseuille's equation | $$Q = \pi\Delta P\,r^4/(8\eta L)$$ |
| Excess pressure (drop) | $$\Delta P = 2S/R$$ |
| Excess pressure (bubble) | $$\Delta P = 4S/R$$ |
| Capillary rise | $$h = 2S\cos\theta/(\rho g r)$$ |
Tip: This topic covers a wide range of concepts. JEE frequently asks numerical problems on Young's modulus, Bernoulli's theorem, terminal velocity, and capillary rise. Practice the formulas with numbers — simply knowing the formula is not enough; you must be comfortable substituting and computing.
Properties of Solids and Liquids Formulas For JEE 2026: Conclusion
Properties of Solids and Liquids is a crucial chapter in JEE Physics that combines conceptual understanding with strong formula application. Topics like elasticity, fluid mechanics, viscosity, and surface tension are not only important for exams but also help in building a deeper understanding of real-world physical phenomena. Since many questions are directly formula-based, students who regularly practice problems can easily score well in this section.
To excel in this chapter, it is important to focus on clarity of concepts along with consistent revision. Practicing numerical problems, revising key formulas, and understanding their applications can significantly improve accuracy and speed. With the right preparation strategy, this chapter can become one of the most scoring and time-saving sections in the JEE exam.
