PiTheory

PiTheory JEE Physics Practice & Revision Pack · Class 11 · Chapter 8 · JEE Main

Mechanical Properties of Solids

Chapter-wise practice and revision for JEE Main: concise notes on stress and strain, Young's, bulk and shear moduli, thermal stress and elastic energy, concept figures, and 35 selected previous-year JEE Main questions covering the major concepts and question patterns, each with a step-by-step solution.

35 selected questions3 topics3 concept figuresPreview · first 3 of 35 questions

How to study this chapter

  1. Look at the topic's figures, then read its notes (5 to 10 minutes). They give the major equations and methods the questions use.
  2. Solve the topic's questions with a timer: 2 minutes for Easy, 3 to 4 for Medium and Hard.
  3. Check the solution. Read the Point to note line even if you were right: it is the idea you will reuse.
  4. Mark each question Got it or Revise. Come back to the Revise list after two days.

Where the marks come from

Main idea of each of the 95 previous-year JEE questions in the PiTheory bank

Elastic moduli59
Stress and strain27
Thermal stress, energy9

Contents

  1. 1Stress, strain and breaking stressQ1–Q9
  2. 2Young's, bulk and shear moduliQ10–Q28
  3. 3Thermal stress and elastic energyQ29–Q35
  4. ✓Last-day checklist
01

Topic 1 · Q1–Q9

Stress, strain and breaking stress

27 of the 95 JEE PYQs in the bank test this as their main idea

Look first

strainstressproportional limityield pointultimate stressfractureslope = Yplastic region:permanent set
Read the stress–strain curve from left to right: elastic, plastic, broken.

Quick notes

$$\text{stress}=\frac FA,\qquad \text{strain}=\frac{\Delta L}{L}\ \Big(\text{or }\frac{\Delta V}{V},\ \theta\Big),\qquad \text{stress}=E\times\text{strain}$$

Stress has units of pressure (N m$^{-2}$ = Pa); strain has no units. Hooke's law holds up to the proportional limit; $E$ is the modulus of elasticity.

Point on the stress–strain curveMeaning
Proportional limitend of the straight line (Hooke's law)
Elastic limit / yield pointbeyond it the wire does not return fully: permanent set
Ultimate (breaking) stresslargest stress; the wire breaks soon after
Ductile vs brittleductile: long plastic region (copper); brittle: breaks just after the elastic limit (glass)
  • The tension at a point of a hanging wire equals the weight of everything below that point. Upper wires carry more.
  • Breaking: $T_{\max}=\sigma_bA$. Check every wire; the weakest limit decides. Accelerating up, $T=m(g+a)$.
  • Longest wire that can hang under its own weight: $L_{\max}=\dfrac{\sigma_b}{\rho g}$, independent of the area.
  • Same material, scaled up by $n$: weight $\times n^3$, area $\times n^2$, stress $\times n$.
  • Two readings $(T_1,l_1)$, $(T_2,l_2)$: natural length $L=\dfrac{l_1T_2-l_2T_1}{T_2-T_1}$.
Traps. Breaking stress is a material property: a thicker wire carries a larger load, not a larger stress. In an Atwood machine the tension is $\dfrac{2m_1m_2g}{m_1+m_2}$, not either weight.

Practice: first 3 of 9 questions

Q1JEE Main 2023 (6 Apr, Shift 2)Easy

A metal block of mass $m$ is suspended from a rigid support through a metal wire of diameter $14\ \text{mm}$. The tensile stress developed in the wire under equilibrium state is $7\times10^5\ \text{N m}^{-2}$. The value of mass $m$ is ______ kg. (Take $g=9.8\ \text{m s}^{-2}$ and $\pi=\frac{22}{7}$)

Numerical answer type

Answer and solution

Answer: 11

Concept. Tensile stress is the force per unit cross-sectional area: $\sigma=\dfrac FA$. In equilibrium the wire's tension equals the weight it holds, so $mg=\sigma A$.

Step 1. Area of the wire. Radius $=7$ mm $=7\times10^{-3}$ m: $$A=\pi r^2=\frac{22}{7}\times(7\times10^{-3})^2=154\times10^{-6}\ \text{m}^2$$

Step 2. Tension. $F=\sigma A=7\times10^5\times154\times10^{-6}=107.8$ N.

Step 3. Mass. $m=\dfrac{107.8}{9.8}=11$ kg.

Answer. $11$.

Point to note. Tension $=$ weight; $\sigma=F/\pi r^2$.

Q2JEE Main 2026 (6 Apr, Shift 1)Medium

A lift of mass $1600$ kg is supported by thick iron wire. If the maximum stress which the wire can withstand is $4 \times 10^8\ \text{N/m}^2$ and its radius is $4$ mm, then maximum acceleration the lift can take is ______ $\text{m/s}^2$. (take $g = 10\ \text{m/s}^2$ and $\pi = 3.14$)

  1. A$2.56$
  2. B$3.89$
  3. C$4.32$
  4. D$5.16$
Answer and solution

Answer: Option A: $2.56$

Concept. The wire breaks if its stress exceeds the maximum stress, so the largest tension it can take is $T_{\max}=\sigma_{\max}A$. When the lift accelerates upward, Newton's law gives $T-mg=ma$, so the tension is largest then.

Step 1. Largest tension. $$T_{\max}=4\times10^8\times\pi(4\times10^{-3})^2=4\times10^8\times3.14\times16\times10^{-6}=20\,096\ \text{N}$$

Step 2. Newton's law for the lift (upward acceleration $a$): $$T_{\max}=m(g+a)\ \Rightarrow\ g+a=\frac{20\,096}{1600}=12.56$$

Step 3. Acceleration. $a=12.56-10=2.56$ m s$^{-2}$.

Answer. Option A.

Point to note. Largest tension $\sigma A$ when accelerating up: $T=m(g+a)$.

Q3JEE Main 2024 (5 Apr, Shift 1)Easy

The density and breaking stress of a wire are $6\times10^4\ \text{kg/m}^3$ and $1.2\times10^8\ \text{N/m}^2$ respectively. The wire is suspended from a rigid support on a planet where the acceleration due to gravity is $\dfrac13$ of the value on the surface of the earth. The maximum length of the wire without breaking is ______ m (take $g = 10\ \text{m/s}^2$).

Numerical answer type

Answer and solution

Answer: 600

Concept. A hanging wire holds its own weight. The stress is largest at the top, where the whole wire hangs below: $\sigma=\dfrac{\rho ALg}{A}=\rho gL$. It breaks when this reaches the breaking stress. The area cancels.

Step 1. Gravity on the planet. $g'=\dfrac{10}{3}$ m s$^{-2}$.

Step 2. Longest wire. $$L_{\max}=\frac{\sigma_b}{\rho g'}=\frac{1.2\times10^8}{6\times10^4\times\frac{10}{3}}=\frac{1.2\times10^8}{2\times10^5}=600\ \text{m}$$

Answer. $600$ m. The maximum length does not depend on the thickness of the wire.

Point to note. Self-weight: $L_{\max}=\sigma_b/\rho g$; area cancels.

32 more questions and 2 more topics in the full chapter

  1. 2Young's, bulk and shear moduli19 questions with solutions
  2. 3Thermal stress and elastic energy7 questions with solutions
  3. ✓Last-day checklist

The complete Mechanical Properties of Solids chapter is in the Class 11 Pack and the Complete Pack, as a PDF on WhatsApp. Want to see a whole chapter first? Laws of Motion and Current Electricity are free in full.

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