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.

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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…

Written & reviewed by the Syllab.in Academic Team (CBSE/NCERT subject experts) · Updated Jul 23, 2026

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Key Questions Covered:

A steel wire of diameter 2 mm and length 1.5 m is suspended vertically with a…
Two wires of same material and same length are subjected to the same tensile …
A rubber ball of mass 50 g is thrown vertically upward with velocity 10 m/s. …
A copper wire of length 1 m is stretched by 0.1%. Calculate the tensile stres…
A cylindrical rod of length L and radius r is twisted by an angle θ. The torq…
A wire of uniform cross-section is bent into a circle of radius 1 m. If Young…
+ 2 more questions in the full chapter

Solutions Summary:

Question	Status
A steel wire of diameter 2 mm and length 1.5 m is suspend…	✓ Solved
Two wires of same material and same length are subjected …	✓ Solved
A rubber ball of mass 50 g is thrown vertically upward wi…	✓ Solved
A copper wire of length 1 m is stretched by 0.1%. Calcula…	✓ Solved
A cylindrical rod of length L and radius r is twisted by …	✓ Solved
A wire of uniform cross-section is bent into a circle of …	✓ Solved

Showing 6 of 8 questions

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 /...

Showing 6 of 8 questions. Visit the full page for complete solutions.