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Mechanical Properties of Solids – Class 11 Physics notes, important questions and MCQs

NCERT Class 11 Physics, chapter 8. Notes, key terms, important questions with answers and MCQs.

Chapter overview

Solids deform under forces. Stress, strain and elastic moduli describe how much, and Hooke's law holds within the elastic limit.

Stress and strain

Stress = force/area (N/m²); strain = change in dimension/original dimension (no unit).
Hooke's law: stress ∝ strain within the elastic limit.
Stress–strain curve: proportional limit, elastic limit (yield point), ultimate strength, fracture point.

Elastic moduli

Young's modulus Y = (F/A)/(ΔL/L); bulk modulus B = −pV/ΔV; shear modulus G = shear stress/shear strain.
Compressibility = 1/B; Poisson's ratio = lateral strain/longitudinal strain.
Elastic PE per unit volume = ½ × stress × strain; steel is more elastic than rubber (larger Y).

Key terms

Elastic limit: maximum stress up to which a body regains its shape
Young's modulus: ratio of longitudinal stress to longitudinal strain
Shear modulus (rigidity): ratio of shearing stress to shearing strain, G = (F/A)/θ

Formulas

Y = FL/(AΔL)
Energy density = ½ stress × strain

Important questions with answers

Why are bridges and buildings designed with materials of high Young's modulus?2 marks

A high Y means less strain for a given stress, so beams bend and stretch less under loads and remain safe within the elastic limit.

Derive the elastic potential energy stored per unit volume in a stretched wire.3 marks

Work done in stretching by ΔL: W = ½ × F × ΔL (average force). Energy per volume = W/(AL) = ½ × (F/A) × (ΔL/L) = ½ × stress × strain.

Describe the stress–strain curve of a ductile metal, marking important points, and explain why rubber is less elastic than steel though it stretches more.5 marks

From O to A, stress ∝ strain (Hooke's law, proportional limit). Up to B (yield point / elastic limit), the wire regains its shape when unloaded. Beyond B, strain increases rapidly with little extra stress (plastic region); the wire gets permanently deformed. D is the ultimate tensile strength (maximum stress); after this the wire thins and breaks at the fracture point E. Elasticity in physics means resistance to deformation measured by Y. Steel needs a much larger stress for the same strain (Y ≈ 2 × 10¹¹ N/m²) than rubber (Y ≈ 10⁷ N/m²), so steel is more elastic.

Quick revision

Strain has no unit.
Steel is more elastic than rubber.
Y depends on material, not on dimensions.

Exam tips

Extension of a wire ΔL = FL/(AY): doubling length doubles ΔL; doubling radius makes it one-fourth.
Liquids and gases have only bulk modulus; solids have all three moduli (Young's, bulk, shear).

Practice MCQs

SI unit of stress is
1. N
2. J
3. N/m²
4. no unit
Show answer
Answer: (c) N/m². Force per area.
Strain has
1. unit N
2. unit N/m
3. unit m
4. no unit
Show answer
Answer: (d) no unit. Ratio.
Hooke's law holds up to the
1. breaking point
2. elastic limit
3. yield point beyond
4. plastic region
Show answer
Answer: (b) elastic limit. Elastic limit.
Which is more elastic
1. steel
2. rubber
3. clay
4. wax
Show answer
Answer: (a) steel. Larger Y.
A steel wire of length 2 m and area 1 mm² stretches by 1 mm under 100 N. Young's modulus is
1. 10¹¹ N/m²
2. 2 × 10⁸ N/m²
3. 2 × 10¹¹ N/m²
4. 2 × 10⁵ N/m²
Show answer
Answer: (c) 2 × 10¹¹ N/m². FL/(AΔL) = 200/(10⁻⁶ × 10⁻³).
If the radius of a wire is doubled (same load and length), its extension becomes
1. double
2. half
3. one-fourth
4. four times
Show answer
Answer: (c) one-fourth. ΔL ∝ 1/r².
Reciprocal of bulk modulus is
1. compressibility
2. Young's modulus
3. rigidity
4. Poisson's ratio
Show answer
Answer: (a) compressibility. 1/B.
Two wires of the same material have length ratio 1 : 2 and radius ratio 2 : 1. Under the same load, the ratio of their extensions is
1. 2 : 1
2. 8 : 1
3. 1 : 4
4. 1 : 8
Show answer
Answer: (d) 1 : 8. ΔL ∝ L/r².

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Written by Kundali IQ from standard, well-established facts to help with revision. This is not an official NCERT publication; always read your textbook too.