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Mechanical Properties of Solids — NCERT Class 11 Physics

Complete NCERT Solutions for CBSE Class 11 Physics Chapter 9 — all intext and chapter-end exercise questions on stress, strain, Young's modulus, shear modulus, bulk modulus, Poisson's ratio, elastic potential energy answered clearly.

Updated 11 June 202613 min read

Exercise Solutions Available

Step-by-step answers for all textbook questions in this chapter

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How to Use This Page

Read each NCERT exercise question, attempt it on your own, then compare your method and answer with the step-by-step solution below. Pay close attention to the unit conversions and sign conventions — they are where most students lose marks.

Key Concepts (Quick Recap)

  • Stress = F/A (unit: Pa or N m⁻²)
  • Strain = ΔL/L (dimensionless, no unit)
  • Young's modulus Y = longitudinal stress / longitudinal strain
  • Shear modulus G = shearing stress / shearing strain
  • Bulk modulus B = −ΔP / (ΔV/V)
  • Poisson's ratio σ = −(lateral strain) / (longitudinal strain)
  • Elastic potential energy U = ½ × stress × strain × volume = ½ × Y × strain² × volume
  • Stress-strain curve: proportional limit → elastic limit → yield point → ultimate stress → fracture

Intext Questions

Page 242

Q1. The stress-strain graphs for two materials A and B are shown. The graphs are drawn to the same scale. (a) Which has greater Young's modulus? (b) Which is stronger?
Answer

(a) Material A has a steeper slope in the linear (proportional) region. Since Y = slope of the stress-strain curve, material A has greater Young's modulus.

(b) The strength of a material is indicated by the ultimate stress (highest point on the curve). Material B reaches a higher stress before fracture, so material B is stronger.

Q2. How much should the pressure on a litre of water be increased to compress it by 0.1%? Given bulk modulus of water = 2.2 × 10⁹ Pa.
ΔV/V = 0.1% = 0.1/100 = 10⁻³ B = −ΔP/(ΔV/V) ΔP = B × (ΔV/V) = 2.2 × 10⁹ × 10⁻³ = 2.2 × 10⁶ Pa
Answer

Required increase in pressure = 2.2 × 10⁶ Pa (approximately 22 atmospheres).

Page 246

Q3. A structural steel rod has a radius of 10 mm and a length of 1.0 m. A 100 kN force stretches it along its length. Find (a) stress (b) elongation (c) strain. Given Y = 2 × 10¹¹ Pa.
Answer

(a) Stress = F/A

A = πr² = π × (10 × 10⁻³)² = 3.14 × 10⁻⁴ m² Stress = 100 × 10³ / (3.14 × 10⁻⁴) = 3.18 × 10⁸ N m⁻²

(b) Elongation: Y = stress/strain → strain = stress/Y

Strain = 3.18 × 10⁸ / (2 × 10¹¹) = 1.59 × 10⁻³ ΔL = strain × L = 1.59 × 10⁻³ × 1.0 = 1.59 × 10⁻³ m = 1.59 mm

(c) Strain = 1.59 × 10⁻³ (dimensionless)

Q4. A cable is replaced by another of the same length and material but of diameter twice that of the original. For a given load, the extension of the new cable is what fraction of the original?
Y = (F × L)/(A × ΔL) → ΔL = FL/(AY) = FL/(πr²Y) If diameter doubles, radius doubles, so A becomes 4A. ΔL_new = FL/(4AY) = (1/4) × ΔL_original
Answer

Extension of new cable = 1/4 of the original.

Chapter-End Exercise

Q1. A steel wire of length 4.7 m and cross-sectional area 3.0 × 10⁻⁵ m² stretches by the same amount as a copper wire of length 4.0 m and cross-sectional area 4.0 × 10⁻⁵ m² under a given load. What is the ratio of Young's modulus of steel to that of copper?
ΔL_s = ΔL_c F L_s / (A_s Y_s) = F L_c / (A_c Y_c) Y_s / Y_c = (L_s / L_c) × (A_c / A_s) = (4.7 / 4.0) × (4.0 × 10⁻⁵ / 3.0 × 10⁻⁵) = (4.7 / 4.0) × (4.0 / 3.0) = 4.7 / 3.0 = 1.567
Answer

Ratio Y_steel / Y_copper = 1.57

Q2. Figure shows the strain-stress curve for a given material. What are (a) Young's modulus (b) approximate yield strength?
Answer

(a) Young's modulus = slope of the linear portion of the stress-strain curve. From the proportional region, read off stress/strain:

Y = stress/strain ≈ (300 × 10⁶) / (0.006) = 5 × 10¹⁰ Pa

(b) Yield strength ≈ stress at the yield point ≈ 300 MPa (the point where the curve departs from linearity and plastic deformation begins).

Q3. The stress-strain curves for two materials A and B are shown. Which of the two is stronger?
Answer

The material with the higher ultimate tensile stress (peak of the curve) is stronger. If material B's curve extends to a higher stress before fracture, then B is stronger.

Q4. Two wires of diameter 0.25 cm, one made of steel and the other of copper, are loaded as shown. Determine the elongation of the steel and the copper wires. Given: L_steel = 1.5 m, L_copper = 2.0 m, Y_steel = 2 × 10¹¹ Pa, Y_copper = 1.1 × 10¹¹ Pa, total load = 6 kg each.
r = 0.25/2 × 10⁻² = 1.25 × 10⁻³ m A = πr² = π × (1.25 × 10⁻³)² = 4.91 × 10⁻⁶ m² F = 6 × 9.8 = 58.8 N Steel: ΔL = FL/(AY) = 58.8 × 1.5 / (4.91 × 10⁻⁶ × 2 × 10¹¹) = 88.2 / (9.82 × 10⁵) = 8.97 × 10⁻⁵ m ≈ 0.09 mm Copper: ΔL = 58.8 × 2.0 / (4.91 × 10⁻⁶ × 1.1 × 10¹¹) = 117.6 / (5.40 × 10⁵) = 2.18 × 10⁻⁴ m ≈ 0.22 mm
Q5. The edge of an aluminium cube is 10 cm. One face is rigidly fixed and a shearing force of 100 kN is applied to the opposite face. Find the shear modulus of aluminium if the edge moves by 0.01 cm.
Shearing stress = F/A = 100 × 10³ / (0.1 × 0.1) = 10⁷ N m⁻² Shearing strain = Δx/L = 0.01/10 = 10⁻³ G = shearing stress / shearing strain = 10⁷ / 10⁻³ = 10¹⁰ Pa
Answer

Shear modulus G = 1 × 10¹⁰ Pa

Q6. A mild steel wire of length 1.0 m and cross-sectional area 0.50 × 10⁻⁶ m² is stretched, within its elastic limit, by a force of 50 N. Calculate: (a) the stress (b) the strain (c) the energy stored in the wire.
Answer

(a) Stress = F/A = 50 / (0.50 × 10⁻⁶) = 1.0 × 10⁸ Pa

(b) Given Y = 2 × 10¹¹ Pa (mild steel)

Strain = stress/Y = 1.0 × 10⁸ / (2 × 10¹¹) = 5 × 10⁻⁴

(c) Energy = ½ × stress × strain × volume

Volume = A × L = 0.50 × 10⁻⁶ × 1.0 = 5.0 × 10⁻⁷ m³ U = ½ × 1.0 × 10⁸ × 5 × 10⁻⁴ × 5.0 × 10⁻⁷ = 1.25 × 10⁻² J

U = 0.0125 J

Q7. A steel cable, 1.5 cm in diameter, is used to lift a load of 3500 kg. Find the maximum load it can carry if the breaking stress for steel is 6 × 10⁸ N m⁻².
A = π(d/2)² = π × (0.75 × 10⁻²)² = 1.767 × 10⁻⁴ m² Max force = breaking stress × A = 6 × 10⁸ × 1.767 × 10⁻⁴ = 1.06 × 10⁵ N Max load = 1.06 × 10⁵ / 9.8 = 1.08 × 10⁴ kg ≈ 10800 kg
Answer

Maximum load ≈ 10800 kg (the cable can safely lift 3500 kg since 10800 kg > 3500 kg).

Q8. A 14.5 kg mass, fastened to the end of a steel wire of unstretched length 1.0 m, is whirled in a vertical circle with an angular velocity of 2 rev/s. The cross-sectional area of the wire is 4.0 × 10⁻⁶ m². Find the elongation. Y_steel = 2 × 10¹¹ Pa.
At the lowest point: T = mg + mω²r ω = 2π × 2 = 4π rad/s, r = 1.0 m T = 14.5 × 9.8 + 14.5 × (4π)² × 1.0 = 142.1 + 14.5 × 157.91 = 142.1 + 2289.7 = 2431.8 N Stress = T/A = 2431.8 / (4.0 × 10⁻⁶) = 6.08 × 10⁸ Pa Strain = stress/Y = 6.08 × 10⁸ / (2 × 10¹¹) = 3.04 × 10⁻³ ΔL = strain × L = 3.04 × 10⁻³ × 1.0 = 3.04 × 10⁻³ m ≈ 3.0 mm
Q9. Compute the bulk modulus of water from the following data: Initial volume = 100.0 litre, pressure increase = 100.0 atm (1 atm = 1.013 × 10⁵ Pa), final volume = 100.5 litre. Compare with the standard value.
ΔV = 100.5 − 100.0 = 0.5 litre; ΔV/V = 0.5/100 = 5 × 10⁻³ ΔP = 100 × 1.013 × 10⁵ = 1.013 × 10⁷ Pa B = ΔP / (ΔV/V) = 1.013 × 10⁷ / (5 × 10⁻³) = 2.026 × 10⁹ Pa
Answer

Standard value = 2.2 × 10⁹ Pa. Our computed value is close but slightly lower because 100 atm is well beyond the linear range.

Q10. What is the density of water at a depth where pressure is 80.0 atm, given that its density at the surface is 1.03 × 10³ kg m⁻³? Compressibility = 45.8 × 10⁻¹² Pa⁻¹.
ΔP = 80 atm = 80 × 1.013 × 10⁵ = 8.104 × 10⁶ Pa Compressibility K = 1/B, so B = 1/K = 1/(45.8 × 10⁻¹²) = 2.183 × 10¹⁰ Pa ΔV/V = ΔP/B = 8.104 × 10⁶ / (2.183 × 10¹⁰) = 3.713 × 10⁻⁴ New density = ρ₀ / (1 − ΔV/V) = 1.03 × 10³ / (1 − 3.713 × 10⁻⁴) = 1.03 × 10³ / 0.999629 ≈ 1030.38 kg m⁻³
Answer

Density at 80 atm ≈ 1030.4 kg m⁻³

Q11. A rigid bar of mass 15 kg is supported symmetrically by three wires each 2.0 m long. The two outer wires are made of steel and the inner one of aluminium. They are stretched by the same amount. Determine the ratio of their tensions.
Answer

Since all three wires have the same elongation:

T_s/(A_s Y_s) = T_al/(A_al Y_al) (same ΔL and L) T_s/T_al = (A_s Y_s)/(A_al Y_al) Given same cross-section area: T_s/T_al = Y_s/Y_al = 2 × 10¹¹ / 7 × 10¹⁰ = 20/7 ≈ 2.86

Ratio T_steel : T_aluminium = 20 : 7

Q12. A 4.0 m long copper wire of cross-sectional area 1.2 cm² is stretched by a force of 500 N. Find (a) stress (b) strain (c) elongation. Y_copper = 1.2 × 10¹¹ Pa.
Answer

(a) Stress = F/A = 500 / (1.2 × 10⁻⁴) = 4.17 × 10⁶ Pa

(b) Strain = stress/Y = 4.17 × 10⁶ / (1.2 × 10¹¹) = 3.47 × 10⁻⁵

(c) ΔL = strain × L = 3.47 × 10⁻⁵ × 4.0 = 1.39 × 10⁻⁴ m ≈ 0.14 mm

Q13. A steel rod of diameter 2 cm and length 2 m is stretched by a load of 2000 kg. Find the work done in stretching it. Y = 2 × 10¹¹ Pa.
A = π × (1 × 10⁻²)² = 3.14 × 10⁻⁴ m² F = 2000 × 9.8 = 19600 N Stress = 19600 / (3.14 × 10⁻⁴) = 6.24 × 10⁷ Pa Strain = stress/Y = 6.24 × 10⁷ / (2 × 10¹¹) = 3.12 × 10⁻⁴ Volume = A × L = 3.14 × 10⁻⁴ × 2 = 6.28 × 10⁻⁴ m³ U = ½ × stress × strain × volume = ½ × 6.24 × 10⁷ × 3.12 × 10⁻⁴ × 6.28 × 10⁻⁴ = ½ × 6.24 × 3.12 × 6.28 × 10⁻¹ = 6.11 J
Answer

Work done = 6.11 J

Q14. Define Poisson's ratio. What are its theoretical limits?
Answer

Poisson's ratio σ = −(lateral strain)/(longitudinal strain). The negative sign ensures σ is positive for most materials. Theoretical limits: −1 ≤ σ ≤ 0.5. For most metals, σ ≈ 0.25–0.35.

Q15. What is the elastic potential energy of a body strained within elastic limit, per unit volume?
Answer

Per unit volume, U = ½ × stress × strain = ½ × Y × strain². This is the area under the stress-strain curve (triangular region within the elastic limit).

Q16. Why are springs made of steel and not of copper?
Answer

Steel has a higher Young's modulus and a much larger elastic range than copper. A steel spring can undergo larger reversible deformations and return to its original shape. Copper would permanently deform (plastic deformation) more easily, making it unsuitable for spring action.

Q17. Explain why: (a) steel is more elastic than rubber (b) the stretching of a coil spring is determined by its shear modulus.
Answer

(a) Elasticity is measured by Young's modulus, not by how much a material stretches. Y_steel (~2 × 10¹¹ Pa) is far greater than Y_rubber (~10⁶ Pa). For the same stress, rubber shows more strain — that means it is less elastic, not more.

(b) When a helical spring is stretched, the wire twists. This is a shearing deformation, not a tensile one. Hence the restoring force depends on the shear modulus (G), not Young's modulus.

Q18. How does the stress-strain curve differ for ductile and brittle materials?
Answer
  • Ductile materials (e.g., copper, mild steel): The curve shows a large plastic region between the yield point and the fracture point. The material can be drawn into wires.
  • Brittle materials (e.g., glass, cast iron): The curve shows almost no plastic region — the material fractures soon after the elastic limit.
Q19. Derive the expression for elastic potential energy of a stretched wire.
Answer

Work done in stretching a wire from 0 to ΔL:

dW = F d(ΔL), where F = (YA/L) × ΔL (Hooke's law) W = ∫₀^ΔL (YA/L) x dx = (YA/2L)(ΔL)² = ½ × (YAΔL/L) × ΔL = ½ × stress × strain × volume
Q20. A wire elongates by l mm when a load W is hanged from it. If the wire goes over a pulley and two weights W each are hung at the two ends, what is the elongation?
Answer

With two weights W on either side of a frictionless pulley, the tension in the wire = W (not 2W). The elongation is the same as before: l mm.

Q21. A truck is moving with a speed of 54 km/h on a straight road. The driver applies brakes to produce a uniform retardation of 0.5 m/s². What is the shortest distance in which the truck can be stopped?
v² = u² + 2as → 0 = (15)² + 2(−0.5)s s = 225/1 = 225 m
Answer

Stopping distance = 225 m

Q22. State Hooke's law.
Answer

Within the elastic limit, stress is directly proportional to strain: stress ∝ strain, or stress = Y × strain, where Y is the modulus of elasticity.

Internal Links

  • [Chapter 9 Guide — Mechanical Properties of Solids](/school/cbse/class/11/physics/cbse-mechanical-properties-of-solids)
  • [All Class 11 Physics Solutions](/school/cbse/class/11/physics/solutions)
  • [Class 11 Physics Subject Guide](/school/cbse/class/11/physics)

Frequently Asked Questions

Stress is the restoring force per unit area (SI unit Pa), while strain is the ratio of change in dimension to the original dimension — it has no unit.