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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Key Questions Covered:

  1. A steel wire of diameter 2 mm and length 1.5 m is suspended vertically with a…
  2. Two wires of same material and same length are subjected to the same tensile …
  3. A rubber ball of mass 50 g is thrown vertically upward with velocity 10 m/s. …
  4. + 5 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

Showing 3 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 Ste...

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...

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 =...

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