Mechanical Properties of Solids — Class 11 Physics NCERT Solutions (Free)
Free step-by-step NCERT solutions for Class 11 Physics chapter "Mechanical Properties of Solids" — 8 important questions with detailed answers for CBSE board exam preparation.
TL;DR: Free step-by-step NCERT solutions for Class 11 Physics chapter "Mechanical Properties of Solids" — 8 important questions with detailed answers for CBS…
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NCERT Solutions for Class 11 Physics — Mechanical Properties of Solids. Step-by-step answers to all 8 textbook questions from this chapter, written for CBSE board preparation and free to use.
Mechanical Properties of Solids — All 8 Questions Solved
Q1. A steel wire of diameter 2 mm and length 1.5 m is suspended vertically with a load of 20 kg attached to its lower end. Calculate the extension in the wire. Given: Young's modulus of steel Y = 2.0 × 10¹¹ Pa, g = 10 m/s².
Step 1: Calculate the stress in the wire.
Force F = mg = 20 × 10 = 200 N
Area A = π r² = π (1 × 10⁻³)² = π × 10⁻⁶ m²
Stress σ = F/A = 200/(π × 10⁻⁶) = 200 × 10⁶/π = 6.37 × 10⁷ Pa
Step 2: Use Young's modulus formula Y = stress/strain
Strain ε = stress/Y = (6.37 × 10⁷)/(2.0 × 10¹¹) = 3.19 × 10⁻⁴
Step 3: Calculate extension.
Extension ΔL = strain × L = 3.19 × 10⁻⁴ × 1.5 = 4.78 × 10⁻⁴ m = 0.478 mm
Final Answer: Extension = 0.48 mm (approximately)
Q2. Two wires of same material and same length are subjected to the same tensile force. Wire A has diameter 1 mm and wire B has diameter 2 mm. Find the ratio of their extensions.
Step 1: Young's modulus Y = (F/A)/(ΔL/L)
Rearranging: ΔL = (F × L)/(Y × A)
Step 2: For wire A: ΔL_A = (F × L)/(Y × π r_A²) = (F × L)/(Y × π (0.5 × 10⁻³)²)
For wire B: ΔL_B = (F × L)/(Y × π r_B²) = (F × L)/(Y × π (1 × 10⁻³)²)
Step 3: Ratio ΔL_A/ΔL_B = r_B²/r_A² = (1)²/(0.5)² = 4/1
Final Answer: ΔL_A : ΔL_B = 4 : 1
Q3. A rubber ball of mass 50 g is thrown vertically upward with velocity 10 m/s. During collision with the ground, it rebounds with 60% of the incident velocity. Calculate the coefficient of restitution.
Step 1: Velocity just before collision with ground (downward).
Using v² = u² + 2as for downward motion:
v² = 10² + 2 × 10 × (maximum height)
At maximum height, v = 0, so h_max = 10²/(2 × 10) = 5 m
Velocity just before impact = √(2 × 10 × 5) = 10 m/s (downward)
Step 2: Rebound velocity = 60% of 10 = 6 m/s (upward)
Step 3: Coefficient of restitution e = velocity of separation/velocity of approach
e = 6/10 = 0.6
Final Answer: e = 0.6
Q4. A copper wire of length 1 m is stretched by 0.1%. Calculate the tensile stress if Young's modulus for copper is 1.3 × 10¹¹ Pa.
Step 1: Given strain ε = 0.1% = 0.1/100 = 0.001 = 1 × 10⁻³
Step 2: Using Young's modulus Y = stress/strain
Stress σ = Y × strain
σ = 1.3 × 10¹¹ × 1 × 10⁻³
σ = 1.3 × 10⁸ Pa = 1.3 × 10⁸ N/m²
Final Answer: Tensile stress = 1.3 × 10⁸ Pa or 130 MPa
Q5. A cylindrical rod of length L and radius r is twisted by an angle θ. The torque required is τ. Express the rigidity modulus in terms of these quantities.
Step 1: The torsional rigidity or modulus of rigidity is defined as:
G = torque × original length / (angle of twist × second moment of area)
Step 2: Second moment of area for a cylinder I_p = π r⁴/2
Step 3: Rigidity modulus G = τ × L / (θ × I_p)
G = τ × L / (θ × π r⁴/2)
G = 2τL / (π r⁴ θ)
Final Answer: Modulus of rigidity G = 2τL/(πr⁴θ)
Q6. A wire of uniform cross-section is bent into a circle of radius 1 m. If Young's modulus is 2 × 10¹¹ Pa and the bending moment is 10 N·m, find the tensile stress at the top of the bent wire.
Step 1: For bending of a wire, the bending moment M relates to radius of curvature R by:
M = E × I / R
where I is the second moment of area about neutral axis.
Step 2: For a circular cross-section of radius r: I = π r⁴/4
Bending stress σ = M × y / I = M/R
where y is distance from neutral axis.
Step 3: At the top of bent wire, maximum stress occurs at outer fiber.
M = E × I / R → 10 = 2 × 10¹¹ × I / 1
I = 10/(2 × 10¹¹) = 5 × 10⁻¹² m⁴
Step 4: Bending stress at outermost fiber:
σ = M × r_outer / I
Without specific wire radius, stress = E/R = (2 × 10¹¹)/1 = 2 × 10¹¹ Pa
Final Answer: Maximum bending stress = 2 × 10¹¹ Pa or as M/R = 10 Pa (depending on moment arm)
Q7. Three identical wires of length L and cross-sectional area A are joined in parallel. A total load W is applied. Calculate the extension of the combination. Given: Young's modulus Y.
Step 1: When wires are in parallel, load distributes equally among them.
Load on each wire = W/3
Step 2: Extension of each wire using Y = (F/A)/(ΔL/L):
ΔL = (F × L)/(Y × A)
ΔL = (W/3) × L / (Y × A)
Step 3: Since all wires are in parallel with same attachment points, the extension of the combination equals extension of each wire.
ΔL_total = W × L / (3 × Y × A)
Final Answer: Extension of parallel combination = WL/(3YA)
Q8. An iron rod of length 1 m and radius 5 mm is subjected to a tensile force of 100 kN. Calculate the percentage change in volume. Given: Y = 2 × 10¹¹ Pa, Poisson's ratio μ = 0.3.
Step 1: Calculate longitudinal strain ε_l
Force F = 100 × 10³ N = 10⁵ N
Area A = π (5 × 10⁻³)² = 25π × 10⁻⁶ m²
Longitudinal stress σ = F/A = 10⁵/(25π × 10⁻⁶) = 4 × 10⁹/π Pa
ε_l = σ/Y = (4 × 10⁹/π)/(2 × 10¹¹) = 2 × 10⁻² / π ≈ 6.37 × 10⁻³
Step 2: Lateral strain ε_t = -μ × ε_l = -0.3 × 6.37 × 10⁻³ = -1.91 × 10⁻³
Step 3: Volumetric strain = ε_l + 2ε_t = 6.37 × 10⁻³ + 2(-1.91 × 10⁻³)
Volumetric strain = 6.37 × 10⁻³ - 3.82 × 10⁻³ = 2.55 × 10⁻³
Step 4: Percentage change = 2.55 × 10⁻³ × 100 = 0.255%
Final Answer: Percentage change in volume = 0.26% (approximately)
Mechanical Properties of Solids — Practice MCQs with Answers
Attempt these 20 multiple-choice questions after working through the solutions above. Each carries the correct option and the reasoning behind it, so a wrong answer tells you which idea to revisit.
MCQ 1. Which of the following is the correct SI unit for stress?
- A. Newton
- B. Joule
- C. Pascal
- D. Meter
Answer: C. Pascal — Stress is defined as force per unit area. The SI unit for force is Newton and for area is square meter, so the unit of stress is N/m², which is also known as Pascal.
MCQ 2. What is the unit of strain?
- A. Newton/meter
- B. Joule
- C. Pascal
- D. Dimensionless
Answer: D. Dimensionless — Strain is defined as the ratio of change in dimension to the original dimension. Since it is a ratio of two similar quantities, it has no units and is dimensionless.
MCQ 3. According to Hooke's Law, within the elastic limit, what is directly proportional to stress?
- A. Area
- B. Volume
- C. Strain
- D. Temperature
Answer: C. Strain — Hooke's Law states that for small deformations, the stress in a body is directly proportional to the strain produced in it, within the elastic limit.
MCQ 4. The SI unit of Young's Modulus is the same as that of which quantity?
- A. Force
- B. Energy
- C. Pressure
- D. Strain
Answer: C. Pressure — Young's Modulus is defined as the ratio of stress to longitudinal strain. Since strain is dimensionless, the unit of Young's Modulus is the same as the unit of stress, which is Pascal (or N/m²), the unit of pressure.
MCQ 5. Up to which point on the stress-strain curve does a material regain its original shape after the removal of the deforming force?
- A. Fracture point
- B. Ultimate tensile strength
- C. Yield point
- D. Elastic limit
Answer: D. Elastic limit — The elastic limit is the maximum stress a material can withstand without undergoing permanent deformation. If the stress is removed before this point, the material returns to its original dimensions.
MCQ 6. A material that breaks without significant plastic deformation is called a/an __________ material.
- A. Ductile
- B. Elastic
- C. Brittle
- D. Malleable
Answer: C. Brittle — Brittle materials exhibit very little plastic deformation and fracture relatively easily under stress, like glass or cast iron. Ductile materials undergo significant plastic deformation before fracturing.
MCQ 7. The property of a body by virtue of which it tends to regain its original size and shape after the removal of the deforming force is called:
- A. Plasticity
- B. Ductility
- C. Elasticity
- D. Malleability
Answer: C. Elasticity — Elasticity is the property that allows a material to return to its original form once the deforming force is removed.
MCQ 8. In the stress-strain curve, the region beyond the yield point but before the ultimate tensile strength is characterized by:
- A. Material behaving purely elastically
- B. Material undergoing permanent deformation
- C. Material completely losing its strength
- D. Material returning to its original shape
Answer: B. Material undergoing permanent deformation — Beyond the yield point, the material enters the plastic region where it experiences permanent deformation even after the removal of the stress.
MCQ 9. A wire of length L and cross-sectional area A is stretched by a force F. If the Young's modulus of the material is Y, the change in length is:
- A. FL/YA
- B. YA/FL
- C. FA/YL
- D. YL/FA
Answer: A. FL/YA — Young's Modulus Y = (Stress / Strain) = (F/A) / (ΔL/L). Rearranging this for ΔL gives ΔL = FL / YA.
MCQ 10. If a solid sphere of radius R is subjected to a uniform pressure P, the fractional change in its volume (ΔV/V) is given by (where B is the bulk modulus):
- A. P/B
- B. B/P
- C. PB
- D. B/(P*V)
Answer: A. P/B — Bulk modulus B is defined as the ratio of volumetric stress (pressure P) to volumetric strain (ΔV/V). So, B = P / (ΔV/V), which implies ΔV/V = P/B.
MCQ 11. For a material, if the Young's modulus is Y and the Poisson's ratio is σ, the modulus of rigidity (G) is given by:
- A. Y / (2(1+σ))
- B. Y (1+σ) / 2
- C. 2Y / (1+σ)
- D. Y / (1-2σ)
Answer: A. Y / (2(1+σ)) — There is a fundamental relationship between the elastic moduli: Y = 2G(1+σ). Rearranging this to find G gives G = Y / (2(1+σ)).
MCQ 12. The potential energy stored per unit volume in a stretched wire is:
- A. (1/2) * Stress * Strain
- B. Stress * Strain
- C. (1/2) * Stress / Strain
- D. Stress / Strain
Answer: A. (1/2) * Stress * Strain — The elastic potential energy stored per unit volume in a material is given by the area under the stress-strain curve in the elastic region, which is (1/2) * stress * strain.
MCQ 13. Which property describes the inability of a material to return to its original shape after deformation?
- A. Elasticity
- B. Ductility
- C. Plasticity
- D. Malleability
Answer: C. Plasticity — Plasticity is the property of a material to undergo permanent deformation without fracture. A plastically deformed material does not regain its original shape.
MCQ 14. Generally, with an increase in temperature, the elastic properties of a material:
- A. Increase
- B. Decrease
- C. Remain unchanged
- D. First increase then decrease
Answer: B. Decrease — As temperature increases, the atomic vibrations become more vigorous, weakening the interatomic forces, which generally leads to a decrease in the elastic properties (e.g., Young's modulus) of materials.
MCQ 15. Which of the following materials is expected to have the highest Young's Modulus?
- A. Rubber
- B. Steel
- C. Copper
- D. Aluminum
Answer: B. Steel — Steel is known for its high stiffness and resistance to deformation, which is reflected in its high Young's Modulus compared to other common engineering materials like copper, aluminum, or rubber.
MCQ 16. Shear modulus (Modulus of Rigidity) is relevant for which type of deformation?
- A. Change in length
- B. Change in volume
- C. Change in shape
- D. Change in density
Answer: C. Change in shape — Shear modulus describes a material's resistance to shear deformation, which involves a change in shape without a change in volume.
MCQ 17. Why are bridges designed with a certain factor of safety, considering the elastic properties of materials?
- A. To make them lighter
- B. To ensure they operate within the elastic limit and can withstand unexpected loads
- C. To increase their cost
- D. To make them aesthetically pleasing
Answer: B. To ensure they operate within the elastic limit and can withstand unexpected loads — A factor of safety ensures that the maximum stress experienced by the bridge under expected and unexpected loads remains well below the elastic limit and ultimate tensile strength, preventing permanent deformation or catastrophic failure.
MCQ 18. For most common materials, the value of Poisson's ratio lies between:
- A. -1 and 0.5
- B. 0 and 1
- C. 0 and 0.5
- D. -0.5 and 0.5
Answer: C. 0 and 0.5 — Theoretically, Poisson's ratio can range from -1 to 0.5. However, for most common isotropic materials, it lies between 0 and 0.5, with 0.5 representing an incompressible material.
MCQ 19. Materials like rubber that can be stretched to many times their original length and still regain their original shape are called:
- A. Brittle materials
- B. Ductile materials
- C. Elastomers
- D. Plastic materials
Answer: C. Elastomers — Elastomers are materials that exhibit high elasticity, meaning they can undergo very large reversible elastic deformations. Rubber is a classic example.
MCQ 20. If a wire of length L and cross-sectional area A is stretched by an extension 'x' under a force 'F', the elastic potential energy stored in the wire is:
- A. FL
- B. (1/2)Fx
- C. Fx
- D. (1/2)FL
Answer: B. (1/2)Fx — The work done in stretching the wire, which is stored as elastic potential energy, is the average force multiplied by the extension. Since the force increases linearly from 0 to F, the average force is F/2, so energy = (1/2)Fx.
How to Use These Solutions
Attempt each question yourself first and only then compare with the worked answer. Marks in the CBSE board exam are awarded for the METHOD as much as the final result, so reproduce the steps rather than memorising the last line. Where a solution states a law, a formula or a definition, learn that wording — examiners look for it.
Related practice for this chapter: chapter MCQs, previous-year questions, revision notes and sample papers.