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EAPCET Physics · Intermediate first year

Mechanical Properties of Solids

Mechanical Properties of Solids covers stress, strain, Hooke’s law and the elastic moduli, along with the energy stored in a stretched body.

Subject
Physics
Taught in
Intermediate first year
Physics in the paper
40 of 160 questions

Key formulas and results

  • Stress =F/A= F/A; strain =Δl/l= \Delta l/l
  • Young’s modulus Y=F/AΔl/lY = \dfrac{F/A}{\Delta l/l}
  • Bulk modulus B=−ΔPΔV/VB = -\dfrac{\Delta P}{\Delta V/V}
  • Poisson’s ratio = lateral strain / longitudinal strain
  • Elastic energy per unit volume =12×stress×strain= \tfrac{1}{2} \times \text{stress} \times \text{strain}
  • Elongation of a wire: x=FLAYx = \frac{FL}{AY}; for a hanging mass MM, x=MgLAYx = \frac{MgL}{AY}
  • Elongation of a uniform rod hanging under its own weight: x=MgL2AYx = \frac{MgL}{2AY}
  • Rod on a smooth floor pulled by forces F1F_{1} and F2F_{2} at opposite ends: x=(F1+F2)L2AYx = \frac{(F_{1} + F_{2})L}{2AY}; a single force FF at one end gives FL2AY\frac{FL}{2AY}
  • A stretched wire acts like a spring of force constant k=YALk = \frac{YA}{L}
  • Total elastic energy stored in a stretched wire: U=12×stress×strain×volume=12FxU = \tfrac{1}{2} \times \text{stress} \times \text{strain} \times \text{volume} = \tfrac{1}{2}Fx
  • Rigidity (shear) modulus: η=shear stressshear strain=FAθ\eta = \frac{\text{shear stress}}{\text{shear strain}} = \frac{F}{A\theta}
  • Compressibility is the reciprocal of bulk modulus: K=1/BK = 1/B

What this chapter covers

  • Stress and strain
  • Hooke’s law and the stress–strain curve
  • Young’s, bulk and shear moduli
  • Poisson’s ratio
  • Elastic potential energy

How it is usually asked

  • Find the extension of a wire
  • Find the energy stored in a stretched wire
  • Compare elastic moduli

Understanding Mechanical Properties of Solids

When a deforming force acts on a solid, internal restoring forces develop. Stress is the restoring force per unit area, and strain is the fractional change in size: change in length per unit length, change in volume per unit volume, or the angle of shear. Within the elastic limit, Hooke’s law says stress is proportional to strain, and the constant of proportionality is an elastic modulus.

Young’s modulus describes stretching, the bulk modulus describes uniform compression, and the shear (rigidity) modulus describes change of shape. A modulus is a property of the material alone, so two wires of the same material have the same Y whatever their length or thickness. Rearranging Y = FL/(AΔl) gives the extension Δl = FL/(AY), which is the most common calculation.

Energy stored per unit volume of a stretched wire is ½ × stress × strain, and the total energy is ½FΔl. EAPCET often asks for extensions, energy stored, or comparisons between two wires. For comparisons, write Δl = FL/(AY) and keep only what changes: for wires of the same material under the same load, Δl ∝ L/r², so a wire twice as long and twice as thick stretches half as much.

Worked example

EAPCET-style question

A steel wire of length 2 m and cross-sectional area 1 mm21\ \text{mm}^{2} supports a load of 10 kg. If Young’s modulus of steel is 2×1011 N/m22 \times 10^{11}\ \text{N/m}^{2} and g=10 m/s2g = 10\ \text{m/s}^{2}, the extension of the wire is

  1. (1)1 mm
  2. (2)0.1 mm
  3. (3)10 mm
  4. (4)2 mm

Solution

  1. Force F=mg=10×10=100 NF = mg = 10 \times 10 = 100\ \text{N}; area A=1 mm2=1×10−6 m2A = 1\ \text{mm}^{2} = 1 \times 10^{-6}\ \text{m}^{2}.
  2. Δl=FLAY=100×210−6×2×1011\Delta l = \frac{FL}{AY} = \frac{100 \times 2}{10^{-6} \times 2 \times 10^{11}}.
  3. Δl=2002×105=1×10−3 m\Delta l = \frac{200}{2 \times 10^{5}} = 1 \times 10^{-3}\ \text{m}.
  4. Extension = 1 mm.

Answer: 1 mm

Shortcuts and results to remember

  • On a stress–strain graph the steeper straight line belongs to the material with the larger Young’s modulus.
  • Poisson’s ratio for real materials lies between 0 and 0.5.
  • Ductile materials such as copper have a long plastic region before breaking; brittle materials such as cast iron break soon after the elastic limit.
  • When a rod is pulled with unequal forces at its two ends, use the average of the two forces as the effective tension for elongation.

More practice questions

Practice question 1

A wire 2 m long with cross-section 1 mm21\ \text{mm}^{2} and Young’s modulus 2×1011 N/m22 \times 10^{11}\ \text{N/m}^{2} supports a 10 kg mass (g=10 m/s2g = 10\ \text{m/s}^{2}). Its extension is

  1. (1)0.1 mm
  2. (2)0.5 mm
  3. (3)1 mm
  4. (4)2 mm

Solution

  1. x=MgLAYx = \frac{MgL}{AY}
  2. MgL=10×10×2=200 N mMgL = 10 \times 10 \times 2 = 200\ \text{N m}
  3. AY=10−6×2×1011=2×105 NAY = 10^{-6} \times 2 \times 10^{11} = 2 \times 10^{5}\ \text{N}
  4. x=2002×105=10−3 m=1 mmx = \frac{200}{2 \times 10^{5}} = 10^{-3}\ \text{m} = 1\ \text{mm}

Answer: 1 mm

Common mistakes to avoid

  • Converting mm² to m² as 10⁻³ — the correct factor is 1 mm² = 10⁻⁶ m².
  • Using the diameter in place of the radius when finding the area πr² of a wire.
  • Writing the energy stored as F × Δl — the force builds up from zero, so the energy is ½FΔl.
  • Thinking Young’s modulus changes when a wire is made longer or thicker — it depends only on the material.

How to prepare it

Keep stress in pascals and strain dimensionless — unit slips are the usual source of wrong options here.

Once the formulas feel familiar, test them the way the exam does: practise Mechanical Properties of Solids questions chapter-wise, then attempt a full previous year paper to see whether it holds up under a 3-hour timer.

Related Physics chapters

See every chapter in the EAPCET Physics syllabus.

Practise Mechanical Properties of Solids chapter-wise

Chapter-wise quizzes show exactly which topics in this chapter you still get wrong — before a full paper does.

Frequently asked questions

Is Mechanical Properties of Solids part of the EAPCET Physics syllabus?
Yes. Mechanical Properties of Solids is taught in Intermediate first year Physics, and the EAPCET Engineering syllabus is based on the Intermediate first- and second-year syllabus. Check the official notification each year for any changes.
How should I prepare Mechanical Properties of Solids for EAPCET?
Keep stress in pascals and strain dimensionless — unit slips are the usual source of wrong options here.
How many Physics questions are there in EAPCET?
40 of the 160 questions in the Engineering paper are Physics, spread across 30 chapters.
What mistakes do students make in Mechanical Properties of Solids?
Converting mm² to m² as 10⁻³ — the correct factor is 1 mm² = 10⁻⁶ m². Using the diameter in place of the radius when finding the area πr² of a wire. Writing the energy stored as F × Δl — the force builds up from zero, so the energy is ½FΔl. Thinking Young’s modulus changes when a wire is made longer or thicker — it depends only on the material.