Mechanical Properties of Solids Class 11 Notes - IIT JEE | NEET

Mechanical Properties of Solids is Chapter 8 of the current NCERT Class 11 Physics book (Chapter 9 in older books), between Gravitation and Mechanical Properties of Fluids. In the JEE Main 2026 syllabus, it is the solids part of the unit Properties of Solids and Liquids. These Mechanical Properties of Solids Class 11 Notes follow the NCERT chapter and add the IIT JEE and NEET problem types.

Mechanical Properties of Solids Class 11 Notes: Chapter Overview and Syllabus

Check the latest CBSE, NTA (JEE Main), JEE Advanced and NMC (NEET) syllabi for the exact lists. The number of questions changes every year, so use previous years’ papers to see the trend.

Tip from the eSaral Physics team: Treat a wire as a spring with the force constant $k = \dfrac{YA}{L}$. Then every problem on a stretched wire, on wires in series and parallel and on elastic energy is a spring problem that you already know.

Mechanical Properties of Solids Class 11 Notes for IIT JEE and NEET

India's Best Exam Preparation for Class 11th - Download Now

India's Best Exam Preparation for Class 11th - Download Now

India's Best Exam Preparation for Class 11th - Download Now

India's Best Exam Preparation for Class 11th - Download Now

Elasticity, Plasticity and Deformation in Mechanical Properties of Solids Class 11 Notes

• A body is elastic if it regains its original shape and size when the deforming force is removed. A body is plastic if it keeps the deformation. A perfectly rigid body does not deform at all (an ideal case).
• When a body is deformed, a restoring force develops inside it, which is equal and opposite to the deforming force at equilibrium.
• The elastic limit is the largest stress up to which the body regains its original shape. Steel is more elastic than rubber, because for the same deforming force, it deforms less and returns to its original shape.

Stress and Strain in Mechanical Properties of Solids Class 11 Notes

Stress is the restoring force per unit area (the unit is the pascal, $1\ \text{Pa} = 1\ \text{N m}^{-2}$, and the dimensions are $[\text{ML}^{-1}\text{T}^{-2}]$). Strain is the fractional change in the shape or the size (it has no unit and no dimensions).

Hooke’s Law and the Elastic Moduli in Mechanical Properties of Solids Class 11 Notes

Hooke’s law: within the elastic limit (more exactly, within the proportional limit), the stress is proportional to the strain. The constant of proportionality is the modulus of elasticity.

All three have the unit Pa. The reciprocal of the bulk modulus is the compressibility, $k = \dfrac{1}{B}$. Solids have all three moduli. Liquids and gases have only the bulk modulus, because they cannot resist a shear, so $G = 0$ for them. For a gas, $B = P$ for an isothermal change, and $B = \gamma P$ for an adiabatic change.

Typical values of $Y$: steel about $2.0\times10^{11}$ Pa, iron about $1.9\times10^{11}$ Pa, copper about $1.1\times10^{11}$ Pa and aluminium about $0.7\times10^{11}$ Pa. Rubber has a much smaller value. A large $Y$ means that the material is stiffer.

Stretching of a wire: for a wire of length $L$, radius $r$ and Young’s modulus $Y$ under a load $F$:

$$\Delta L = \frac{FL}{AY} = \frac{FL}{\pi r^{2}Y}, \qquad \text{force constant } k = \frac{F}{\Delta L} = \frac{YA}{L}$$

For wires of the same material and the same load, $\Delta L \propto \dfrac{L}{r^{2}}$. For wires (or springs) in series, $\dfrac{1}{k_{eq}} = \sum\dfrac{1}{k_i}$, and in parallel, $k_{eq} = \sum k_i$.

A wire that hangs under its own weight (length $L$, density $\rho$, mass $M$):

$$\Delta L = \frac{MgL}{2AY} = \frac{\rho g L^{2}}{2Y}$$

Stress–Strain Curve in Mechanical Properties of Solids Class 11 Notes

For a metal wire that is stretched until it breaks, the stress–strain curve has these regions:

• A ductile material (copper, steel) has a long plastic region and can be drawn into wires. A brittle material (glass, cast iron) breaks soon after the elastic limit.
• An elastomer (rubber) can be stretched to a very large strain, and it does not obey Hooke’s law. Its loading and unloading curves are different, which is called elastic hysteresis, and the area of the loop is the energy that is lost as heat.
• The breaking stress is the stress at which a wire breaks. A safety factor is used in design, so that the working stress is much lower than the breaking stress.

Poisson’s Ratio and the Relations Between the Moduli in Mechanical Properties of Solids Class 11 Notes

When a wire is stretched, it becomes thinner. Poisson’s ratio is the ratio of the lateral strain to the longitudinal strain:

$$\sigma = \frac{\text{lateral strain}}{\text{longitudinal strain}}, \qquad -1 < \sigma < 0.5$$

For most metals, $\sigma$ is between about $0.2$ and $0.4$. For an isotropic material:

$$Y = 3B(1 - 2\sigma) = 2G(1 + \sigma), \qquad \sigma = \frac{3B - 2G}{2(3B + G)}$$

Elastic Potential Energy in Mechanical Properties of Solids Class 11 Notes

The work that is done to stretch a wire is stored in it as elastic potential energy:

$$U = \tfrac{1}{2}F\,\Delta L = \tfrac{1}{2}\frac{YA}{L}(\Delta L)^{2}, \qquad u = \frac{U}{AL} = \tfrac{1}{2}\times\text{stress}\times\text{strain} = \tfrac{1}{2}Y\varepsilon^{2} = \frac{\sigma^{2}}{2Y}$$

Here $u$ is the energy density (energy per unit volume). The energy is the area under the force–extension graph.

Thermal Stress and Applications of Elasticity in Mechanical Properties of Solids Class 11 Notes

Thermal stress: if a rod of length $L$, cross-section $A$ and coefficient of linear expansion $\alpha$ is held between rigid walls and heated by $\Delta T$, it cannot expand, so a stress develops in it:

$$\text{stress} = Y\alpha\,\Delta T, \qquad F = YA\alpha\,\Delta T$$

Applications:

• Cranes and ropes: a rope is made of many thin strands, because a thin wire is easier to bend, and the breaking load depends on the total area. The safety factor is chosen from the breaking stress.
• Bridges and buildings: the beams and the columns are designed so that the stress is below the elastic limit. For a beam of length $l$, breadth $b$ and depth $d$ that is loaded at the middle with $W$, the depression is $\delta = \dfrac{Wl^{3}}{4bd^{3}Y}$, so a deeper beam bends much less, which is why girders have an I-shape.
• Springs in a vehicle and rubber in tyres use the elastic properties.

IIT JEE and NEET Problem Tips in Mechanical Properties of Solids Class 11 Notes

• Use $k = YA/L$. A wire is a spring, and a wire cut into two halves has a stiffness that is double for each half.
• For the same material and load, compare using $\Delta L \propto \dfrac{L}{r^{2}}$, and remember that a doubled radius gives four times the area.
• Energy problems: $U = \tfrac{1}{2}F\,\Delta L$ and $u = \tfrac{1}{2}\sigma\varepsilon$. When a wire is stretched by a hanging mass, only the stored energy is $\tfrac{1}{2}F\Delta L$, and the rest of the work of gravity is lost as kinetic energy and heat.
• Hydraulic and bulk problems: use $\dfrac{\Delta V}{V} = -\dfrac{\Delta P}{B}$, and remember the negative sign for a compression.
• Check the units: $Y$, $B$ and $G$ are in Pa, and the radius or the area must be in SI units.

Key Formulas in Mechanical Properties of Solids Class 11 Notes

Solved Examples from Mechanical Properties of Solids Class 11 Notes

Example 1 (stretching of a wire): A steel wire of length 2 m and cross-section 1 mm$^2$ carries a load of 100 N ($Y = 2\times10^{11}$ Pa). Find the extension.

$\Delta L = \dfrac{FL}{AY} = \dfrac{100\times 2}{10^{-6}\times 2\times10^{11}} = 10^{-3}$ m $= 1$ mm.

Example 2 (a wire of a given radius): A wire of radius 0.5 mm and length 1 m carries a mass of 10 kg ($g = 10$ m/s$^2$, $Y = 2\times10^{11}$ Pa). Find the extension.

$A = \pi(0.5\times10^{-3})^{2} \approx 7.85\times10^{-7}$ m$^{2}$, and $F = 100$ N. $\Delta L = \dfrac{100\times 1}{7.85\times10^{-7}\times 2\times10^{11}} \approx 6.4\times10^{-4}$ m, which is about 0.64 mm.

Example 3 (breaking load): A wire of cross-section 2 mm$^2$ has a breaking stress of $5\times10^{8}$ Pa. Find the largest load that it can carry.

$F = \text{stress}\times A = 5\times10^{8}\times 2\times10^{-6} = 1000$ N.

Example 4 (bulk modulus): The bulk modulus of water is $2.2\times10^{9}$ Pa. Find the fractional change in the volume when the pressure increases by $2\times10^{7}$ Pa.

$\dfrac{\Delta V}{V} = \dfrac{\Delta P}{B} = \dfrac{2\times10^{7}}{2.2\times10^{9}} \approx 9.1\times10^{-3}$, so the volume decreases by about 0.91%.

Example 5 (shear modulus): A tangential force of 1000 N acts on the top face of a cube of side 10 cm ($G = 2.5\times10^{10}$ Pa). Find the displacement of the top face.

$\theta = \dfrac{F_t}{AG} = \dfrac{1000}{0.01\times 2.5\times10^{10}} = 4\times10^{-6}$ rad, so the displacement is $\theta L = 4\times10^{-6}\times 0.1 = 4\times10^{-7}$ m.

Example 6 (elastic energy): Find the energy stored in the wire of Example 1.

$U = \tfrac{1}{2}F\,\Delta L = \tfrac{1}{2}\times 100\times 10^{-3} = 0.05$ J.

Example 7 (energy density): A material with $Y = 2\times10^{11}$ Pa is under a stress of $10^{8}$ Pa. Find the energy per unit volume.

$u = \dfrac{\sigma^{2}}{2Y} = \dfrac{10^{16}}{4\times10^{11}} = 2.5\times10^{4}$ J/m$^{3}$.

Example 8 (a wire that hangs under its own weight): A steel wire of length 10 m hangs from a roof ($\rho = 7800$ kg/m$^3$, $Y = 2\times10^{11}$ Pa). Find the elongation.

$\Delta L = \dfrac{\rho gL^{2}}{2Y} = \dfrac{7800\times10\times100}{4\times10^{11}} \approx 1.95\times10^{-5}$ m, which is about 0.02 mm.

Example 9 (thermal stress): A steel rod is held between rigid walls and heated by 50 K ($\alpha = 1.2\times10^{-5}\ \text{K}^{-1}$, $Y = 2\times10^{11}$ Pa). Find the stress.

$\text{stress} = Y\alpha\,\Delta T = 2\times10^{11}\times1.2\times10^{-5}\times 50 = 1.2\times10^{8}$ Pa.

Example 10 (Poisson’s ratio): For a material, $Y = 2\times10^{11}$ Pa and $G = 0.8\times10^{11}$ Pa. Find Poisson’s ratio.

$Y = 2G(1 + \sigma)$, so $1 + \sigma = \dfrac{2\times10^{11}}{1.6\times10^{11}} = 1.25$, and $\sigma = 0.25$.

Example 11 (a comparison of two wires): Two wires of the same material and length have radii in the ratio $1 : 2$. They carry the same load. Find the ratio of their extensions.

$\Delta L \propto \dfrac{1}{r^{2}}$, so the ratio is $\dfrac{4}{1}$, which is $4 : 1$.

Example 12 (the wire as a spring): Find the force constant of the wire of Example 1, and use it to find the extension for a load of 100 N.

$k = \dfrac{YA}{L} = \dfrac{2\times10^{11}\times10^{-6}}{2} = 10^{5}$ N/m, so $\Delta L = \dfrac{F}{k} = \dfrac{100}{10^{5}} = 10^{-3}$ m, which matches Example 1.

Common Mistakes in Mechanical Properties of Solids Class 11 Problems

How to Study Mechanical Properties of Solids Class 11 Notes

1. Read the NCERT chapter first, and download the textbook from eSaral’s NCERT books for Class 11 page.
2. Learn the definitions of stress, strain and the three moduli, then the stress–strain curve, and then Poisson’s ratio and the energy formulas.
3. Practise one problem each on a stretched wire, the breaking load, the bulk modulus, the shear modulus, the elastic energy, thermal stress and a wire under its own weight.
4. Revise with eSaral’s Class 11 Physics notes, and use the free Physics revision series on YouTube for quick formula revision.
5. Test yourself with eSaral’s JEE Main chapter-wise PYQs and NEET chapter-wise PYQs.