Class 11 — Mechanical Properties of Solids

Chapter Overview & Weightage

Mechanical Properties of Solids carries a steady $4$ - $5$ marks weightage in CBSE Class 11 board exams, usually as one short-answer (3 marks) and one MCQ/very short (1-2 marks). Topics include stress, strain, Hooke’s law, Young’s modulus, bulk modulus, shear modulus, and Poisson’s ratio.

The chapter is application-heavy. Most questions are direct numerical substitutions or two-step problems combining definitions and formulas.

Key Concepts You Must Know

Stress: force per unit area, $\sigma = F/A$ , units $\text{N/m}^2$ or pascal.
Strain: relative deformation, dimensionless. Linear strain $= \Delta L / L_0$ .
Hooke’s law: stress is proportional to strain within the elastic limit.
Young’s modulus ( $Y$ ): for tensile/compressive deformation, $Y = \sigma / \epsilon$ .
Bulk modulus ( $K$ ): for volumetric deformation under pressure, $K = -P / (\Delta V / V)$ .
Shear modulus ( $G$ ): for tangential deformation, $G = (F/A) / \tan\theta$ .
Poisson’s ratio ( $\nu$ ): lateral strain to longitudinal strain ratio, typically $0.2$ to $0.5$ for solids.
Elastic energy stored: $U = \frac{1}{2} \times \text{stress} \times \text{strain} \times V$ .

Important Formulas

$Y = \frac{F L_0}{A \Delta L}$

Use when a wire is stretched or compressed under longitudinal force.

$u = \frac{1}{2} \sigma \epsilon = \frac{1}{2} Y \epsilon^2 = \frac{\sigma^2}{2Y}$

The energy density stored in a stressed body.

$K = -\frac{V \Delta P}{\Delta V}$

The minus sign reflects that increasing pressure decreases volume.

Solved Previous Year Questions

PYQ 1 (CBSE 2023): A wire of length $2$ m and cross-section $1$ mm $^2$ is stretched by $1$ mm under a force of $50$ N. Find Young’s modulus.

Solution: $A = 10^{-6}$ m $^2$ , $\Delta L = 10^{-3}$ m. $Y = (F L_0)/(A \Delta L) = (50 \times 2)/(10^{-6} \times 10^{-3}) = 10^{11}$ N/m $^2$ .

PYQ 2 (CBSE 2022): Define elastic limit and explain stress-strain curve for a ductile material.

Answer outline: Elastic limit is the maximum stress beyond which permanent deformation occurs. Stress-strain curve has linear (Hooke’s law) region up to proportional limit, then a yield region, then plastic flow, then necking, then fracture point.

PYQ 3 (CBSE 2021): A copper wire of length $4$ m and diameter $1$ mm is stretched by a force such that the elongation is $0.5$ mm. Calculate the elastic energy stored. ( $Y_{\text{Cu}} = 1.2 \times 10^{11}$ N/m $^2$ .)

Solution: $A = \pi(0.5 \times 10^{-3})^2 = 7.85 \times 10^{-7}$ m $^2$ . Strain $\epsilon = 0.5 \times 10^{-3}/4 = 1.25 \times 10^{-4}$ . Stress $\sigma = Y\epsilon = 1.5 \times 10^7$ N/m $^2$ . Energy density $u = \frac{1}{2}\sigma\epsilon = 937.5$ J/m $^3$ . Volume $V = AL = 3.14 \times 10^{-6}$ m $^3$ . Total energy $U = uV \approx 2.94 \times 10^{-3}$ J.

Difficulty Distribution

Easy (50%): Direct substitution into $Y = FL/(A\Delta L)$ .
Medium (35%): Energy stored, comparing two materials, ratio problems.
Hard (15%): Multi-step problems combining stress-strain with thermal expansion or rotational systems.

Expert Strategy

Memorise typical Young’s modulus values: steel $\sim 2 \times 10^{11}$ , copper $\sim 1.2 \times 10^{11}$ , rubber $\sim 10^6$ N/m $^2$ . Examiners often ask which material is “stiffer” — that’s whichever has higher $Y$ .

For 3-mark numerical problems, write the formula clearly, substitute with units, and compute step-by-step. Examiners give partial credit even if the final number is wrong, as long as the method is right.

For derivation-type questions (energy stored, work done in stretching), start from $W = \int F \, dx$ where $F = (YA/L)x$ for a Hookean wire. Integrate to get $\frac{1}{2}YA(\Delta L)^2/L$ , the standard result.

Common Traps

Trap 1: Using length in cm or mm instead of metres. Always convert to SI.

Trap 2: Forgetting that strain is dimensionless. If you compute strain as $0.5$ mm and forget to divide by $L_0$ , every subsequent step is wrong.

Trap 3: Confusing Young’s modulus with bulk modulus. $Y$ is for length change under axial force; $K$ is for volume change under pressure.

Trap 4: Using the diameter instead of the radius when computing $A = \pi r^2$ for a wire. CBSE problems often give diameter — halve it before squaring.