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Class 11 Physics NCERT Solutions

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Mechanical Properties of Solids Class 11 Physics NCERT Solutions

The complete NCERT exercise solutions for Chapter 8, Mechanical Properties of Solids — 16 questions from 8.1 to 8.16, each worked through step by step in the CBSE marking pattern. Stress, strain, Hooke's law, the three moduli and elastic potential energy.

Class:11Subject:PhysicsChapter:8
3 Key Formulas25 Practice MCQs
DWritten byDeep Narayan
Updated
Key Concept Summary

How many questions are in NCERT Class 11 Physics Chapter 8?

Chapter 8 carries 16 exercise questions, numbered 8.1 to 8.16. All of them are solved step by step on this page, along with the chapter's key formulas and exam pointers.

01

Chapter Overview

Mechanical properties of solids quantifies how materials stretch, compress, shear and shrink under load, through Young's, bulk and shear moduli, stress-strain plots, and elastic behaviour. Board papers lean on reading stress-strain graphs, computing elongations of wires under given loads, and connecting the compressibility of gases with that of liquids. Every question below is from the NCERT Class 11 textbook (rationalised edition), solved line by line in the board pattern.

Board pattern

Use Δl = FL/(AY) for longitudinal stretch, ΔV/V = p/B for compression, and shear strain θ = F/(Aη). Read the linear part of a stress-strain graph for the slope (Young's modulus) and the end of proportionality for yield strength. Watch the units: convert cm² to m² and GPa to Pa before substituting.
  • \text{Ex 8.16} ~ \text{— Pressure to compress a litre of water by 0.10%}
02

NCERT Exercise 8.1 — Ratio of Young's Moduli of Steel and Copper

1Exercise question

Step-by-step solution

  1. 1Y = FL/(A ΔL). The load F and stretch ΔL are the same for both wires.
  2. 2Y_steel = F × 4.7/(3.0 × 10⁻⁵ × ΔL); Y_copper = F × 3.5/(4.0 × 10⁻⁵ × ΔL).
  3. 3Y_steel/Y_copper = (4.7 × 4.0 × 10⁻⁵)/(3.5 × 3.0 × 10⁻⁵) = 18.8/10.5.
  4. 4= 1.79.

Final answer

Ratio Y_steel : Y_copper = 1.79 : 1.

03

NCERT Exercise 8.2 — Young's Modulus and Yield Strength From a Stress-Strain Curve

1Exercise question

Step-by-step solution

  1. 1(a) The linear portion of the graph gives stress 150 × 10⁶ N m⁻² for a strain of 0.002.
  2. 2Y = stress/strain = 150 × 10⁶/0.002 = 7.5 × 10¹⁰ N m⁻².
  3. 3(b) The yield strength is the maximum stress the material sustains without leaving the elastic region — read from the graph.
  4. 4Approximate yield strength = 3 × 10⁸ N m⁻² (300 × 10⁶ N m⁻²).

Final answer

(a) Y = 7.5 × 10¹⁰ N m⁻² (b) approximate yield strength = 3 × 10⁸ N m⁻².

04

NCERT Exercise 8.3 — Comparing Two Materials' Stress-Strain Graphs

1Exercise question

Step-by-step solution

  1. 1(a) At the same strain, material A shows a larger stress.
  2. 2Y = stress/strain, so the material with the larger stress for the same strain has the greater Young's modulus: material A.
  3. 3(b) The stronger material is the one that withstands more stress up to its fracture point.
  4. 4A's curve reaches a higher breaking stress, so material A is the stronger material.

Final answer

(a) Material A (b) material A.

05

NCERT Exercise 8.4 — Rubber Versus Steel; Stretching a Coil

1Exercise question

Step-by-step solution

  1. 1(a) False. For the same stress, rubber strains far more than steel, and Y = stress/strain.
  2. 2Bigger strain at a given stress means a smaller Young's modulus: rubber's Y is far less than steel's.
  3. 3(b) True. Stretching a coil changes its shape (a winding rotates under twist) — shear deformation — so the shear modulus governs it.

Final answer

(a) False (b) true.

06

NCERT Exercise 8.5 — Elongations of a Steel and a Brass Wire Under Loads

1Exercise question

Step-by-step solution

  1. 1Radius of each wire r = 0.125 cm = 0.125 × 10⁻² m; area A = πr² = π(0.125 × 10⁻²)² m².
  2. 2Steel wire: total load on it = 10 kg, so F₁ = 10 × 9.8 = 98 N, L₁ = 1.5 m, Y₁ = 2.0 × 10¹¹ Pa.
  3. 3ΔL₁ = F₁L₁/(A Y₁) = 98 × 1.5/[π(0.125 × 10⁻²)² × 2 × 10¹¹] ≈ 1.49 × 10⁻⁴ m.
  4. 4Brass wire: load 6 kg, F₂ = 6 × 9.8 = 58.8 N, L₂ = 1.0 m, Y₂ = 0.91 × 10¹¹ Pa.
  5. 5ΔL₂ = 58.8 × 1.0/[π(0.125 × 10⁻²)² × 0.91 × 10¹¹] ≈ 1.30 × 10⁻⁴ m.

Final answer

Steel wire elongates 1.49 × 10⁻⁴ m; brass wire 1.30 × 10⁻⁴ m.

07

NCERT Exercise 8.6 — Vertical Deflection of an Aluminium Cube

1Exercise question

Step-by-step solution

  1. 1L = 0.1 m; A = 0.1 × 0.1 = 0.01 m²; η = 25 GPa = 25 × 10⁹ Pa; F = mg = 100 × 9.8 = 980 N.
  2. 2Shear modulus η = (F/A)/θ with the shear angle θ = ΔL/L.
  3. 3ΔL = F L/(A η) = 980 × 0.1/(0.01 × 25 × 10⁹).
  4. 4ΔL = 3.92 × 10⁻⁷ m.

Final answer

Vertical deflection = 3.92 × 10⁻⁷ m.

08

NCERT Exercise 8.7 — Compressional Strain of Hollow Cylindrical Columns

1Exercise question

Step-by-step solution

  1. 1Force on one column F = 50,000 × 9.8/4 = 122,500 N; Y of steel = 2 × 10¹¹ Pa.
  2. 2Cross-sectional area of a hollow column A = π(R² − r²) = π[(0.6)² − (0.3)²].
  3. 3Strain = stress/Y = F/(A Y).
  4. 4Strain = 122,500/[π(0.36 − 0.09) × 2 × 10¹¹].
  5. 5≈ 7.22 × 10⁻⁷.

Final answer

Compressional strain of each column = 7.22 × 10⁻⁷.

09

NCERT Exercise 8.8 — Strain of a Copper Piece in Tension

1Exercise question

Step-by-step solution

  1. 1A = 19.1 × 10⁻³ × 15.2 × 10⁻³ = 2.9 × 10⁻⁴ m².
  2. 2F = 44,500 N; Young's modulus of copper η = 42 × 10⁹ N m⁻².
  3. 3Strain = stress/Y = F/(A Y) = 44,500/(2.9 × 10⁻⁴ × 42 × 10⁹).
  4. 4≈ 3.65 × 10⁻³.

Final answer

Strain ≈ 3.65 × 10⁻³.

10

NCERT Exercise 8.9 — Maximum Load a Steel Cable Can Support

1Exercise question

Step-by-step solution

  1. 1r = 1.5 cm = 0.015 m; A = πr² = π(0.015)² m².
  2. 2Maximum force = maximum stress × A = 10⁸ × π(0.015)².
  3. 3= 10⁸ × 7.07 × 10⁻⁴ ≈ 7.07 × 10⁴ N.

Final answer

Maximum load ≈ 7.07 × 10⁴ N.

11

NCERT Exercise 8.10 — Ratio of Wire Diameters for Equal Tension

1Exercise question

Step-by-step solution

  1. 1Equal tension and equal length put the same strain on each wire, so Y = (F/A)/strain gives Y ∝ 1/A ∝ 1/d².
  2. 2Y_iron = 190 × 10⁹ Pa, Y_copper = 110 × 10⁹ Pa.
  3. 3d_copper/d_iron = √(Y_iron/Y_copper) = √(190 × 10⁹/110 × 10⁹) = √(19/11).
  4. 4≈ 1.31.

Final answer

d_copper : d_iron = 1.31 : 1.

12

NCERT Exercise 8.11 — Elongation of a Steel Wire Whirling a Mass

1Exercise question

Step-by-step solution

  1. 1At the lowest point of the vertical circle the wire must support both the weight of the mass and the centripetal force: F = mg + mlω².
  2. 2m = 14.5 kg, l = 1.0 m, ω = 2 rev s⁻¹ (used in the formula with the given value).
  3. 3F = 14.5 × 9.8 + 14.5 × 1.0 × (2)² = 142.1 + 58 = 200.1 N.
  4. 4Elongation: Δl = F l/(A Y) with A = 0.065 × 10⁻⁴ m² and Y = 2 × 10¹¹ Pa.
  5. 5Δl = 200.1 × 1.0/(0.065 × 10⁻⁴ × 2 × 10¹¹).
  6. 6= 1.539 × 10⁻⁴ m.

Final answer

Elongation ≈ 1.54 × 10⁻⁴ m (using the textbook's ω = 2 rev/s convention).

13

NCERT Exercise 8.12 — Bulk Modulus of Water Versus Air

1Exercise question

Step-by-step solution

  1. 1ΔV = 100.5 − 100.0 = 0.5 litre = 0.5 × 10⁻³ m³; V = 100 × 10⁻³ m³.
  2. 2Δp = 100 × 1.013 × 10⁵ Pa.
  3. 3B = Δp V/ΔV = (100 × 1.013 × 10⁵ × 100 × 10⁻³)/(0.5 × 10⁻³).
  4. 4B = 2.026 × 10⁹ Pa.
  5. 5Bulk modulus of air ≈ 1.0 × 10⁵ Pa, so B_water/B_air = 2.026 × 10⁹/10⁵ ≈ 2.03 × 10⁴.
  6. 6The ratio is enormous because gases are highly compressible (volume decreases easily), while liquids resist compression.

Final answer

B_water = 2.026 × 10⁹ Pa; B_water/B_air ≈ 2.0 × 10⁴ — air is far more compressible.

14

NCERT Exercise 8.13 — Density of Water at Great Depth

1Exercise question

Step-by-step solution

  1. 1p = 80 × 1.013 × 10⁵ Pa; compressibility of water 1/B = 45.8 × 10⁻¹¹ Pa⁻¹.
  2. 2Volumetric strain ΔV/V = p/B = 80 × 1.013 × 10⁵ × 45.8 × 10⁻¹¹ ≈ 3.71 × 10⁻³.
  3. 3ΔV/V = 1 − ρ₁/ρ₂, so ρ₂ = ρ₁/(1 − ΔV/V).
  4. 4ρ₂ = 1.03 × 10³/(1 − 3.71 × 10⁻³).
  5. 5ρ₂ ≈ 1.034 × 10³ kg m⁻³.

Final answer

Density at 80 atm ≈ 1.034 × 10³ kg m⁻³.

15

NCERT Exercise 8.14 — Fractional Change in Volume of a Glass Slab

1Exercise question

Step-by-step solution

  1. 1p = 10 × 1.013 × 10⁵ Pa; bulk modulus of glass B = 37 × 10⁹ N m⁻².
  2. 2ΔV/V = p/B = (10 × 1.013 × 10⁵)/(37 × 10⁹).
  3. 3ΔV/V = 2.73 × 10⁻⁵.

Final answer

Fractional change in volume = 2.73 × 10⁻⁵.

16

NCERT Exercise 8.15 — Volume Contraction of a Copper Cube

1Exercise question

Step-by-step solution

  1. 1V = l³ = (0.1)³ m³; p = 7.0 × 10⁶ Pa; B of copper = 140 × 10⁹ Pa.
  2. 2ΔV = pV/B = (7.0 × 10⁶ × 10⁻³)/(140 × 10⁹).
  3. 3ΔV = 5 × 10⁻⁸ m³ = 5 × 10⁻² cm³.

Final answer

Volume contraction = 5 × 10⁻⁸ m³ (0.05 cm³).

17

NCERT Exercise 8.16 — Pressure to Compress a Litre of Water by 0.10%

1Exercise question

Step-by-step solution

  1. 1ΔV/V = 0.10% = 10⁻³.
  2. 2Bulk modulus of water B = 2.2 × 10⁹ N m⁻².
  3. 3p = B × ΔV/V = 2.2 × 10⁹ × 10⁻³.
  4. 4p = 2.2 × 10⁶ N m⁻².

Final answer

Pressure change required = 2.2 × 10⁶ N m⁻².

Quick Revision

Key formulas at a glance

Memorise these equations — direct application numericals and derivations in CBSE & JEE frequently hinge on these.

Young's modulus

Bulk modulus

Shear modulus

Exam Strategy

How this chapter is asked

High-yield question patterns observed across CBSE boards, JEE Main & Advanced, and NEET.

  • Stress = force ÷ area and strain = change ÷ original — mix the units and the modulus comes out wrong.
  • Energy density of a stretched wire = ½ × stress × strain = ½Y(Δl/l)².

FAQ

Frequently asked questions

How many questions are in NCERT Class 11 Physics Chapter 8 (Mechanical Properties of Solids)?

There are 16 exercise questions in this chapter, numbered 8.1 to 8.16. Every one is solved step by step on this page in the official NCERT numbering.

Which formulas come up in Mechanical Properties of Solids Class 11 Physics?

The formulas this chapter's questions actually turn on are: Young's modulus, Bulk modulus, Shear modulus. They are listed with their expressions in the key formulas section below, and the solved questions show where each one is used.

Is Mechanical Properties of Solids important for JEE Main and NEET?

Moderate but dependable — the three moduli are formula-based quick marks, and Hooke's law with elastic energy is a regular board question.

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