Questions & explanations
1. Two rods of equal length L, Young's moduli Y₁ and Y₂, cross-sectional areas A₁ and A₂, and coefficients of linear expansion α₁ and α₂ are placed side by side between rigid walls. The temperature is raised by ΔT. What is the thermal stress developed in each rod?
- Y₁α₁ΔT and Y₂α₂ΔT
- (Y₁α₁ + Y₂α₂)ΔT / 2
- (Y₁α₁A₁ + Y₂α₂A₂)ΔT / (A₁ + A₂)
- (Y₁α₁A₂ + Y₂α₂A₁)ΔT / (A₁ + A₂)
Answer: (Y₁α₁A₁ + Y₂α₂A₂)ΔT / (A₁ + A₂)
Both rods are constrained between rigid walls, so total length change is zero. Let common strain be ε. Each rod's net stress = Y(αΔT - ε). Force balance: Y₁A₁(α₁ΔT - ε) + Y₂A₂(α₂ΔT - ε) = 0. Solve for ε, then stress = Y(αΔT - ε) yields same value (Y₁α₁A₁ + Y₂α₂A₂)ΔT/(A₁+A₂).
2. Why is steel preferred over copper for making crane ropes?
- Steel has higher Young's modulus and lower ultimate strength.
- Steel has lower Young's modulus and higher ultimate strength.
- Steel has lower Young's modulus and lower ultimate strength.
- Steel has higher Young's modulus and higher ultimate strength.
Answer: Steel has higher Young's modulus and higher ultimate strength.
For crane ropes, we need small extension under load and high breaking strength. Steel has Young's modulus ≈ 2×10¹¹ Pa (higher than copper's 1.2×10¹¹ Pa) and ultimate strength ≈ 500 MPa (higher than copper's ~200 MPa). Thus steel stretches less and is stronger.
3. A 2 kg mass is gently hung from a steel wire of length 1 m and cross-section 1 mm². Young's modulus of steel is 2×10¹¹ N/m². If the same mass is dropped from a height of 0.1 m onto the wire, the maximum extension is:
- 6.5 mm
- 0.4 mm
- 0.2 mm
- 0.1 mm
Answer: 6.5 mm
Static extension = mgL/(AY) = 0.1 mm, so spring constant k = AY/L = 2×10⁵ N/m. Energy conservation: mg(h+ΔL) = ½ k ΔL². Substituting m=2 kg, g=10 m/s², h=0.1 m, k=2×10⁵ N/m gives 20(0.1+ΔL) = 10⁵ ΔL². Solving quadratic: ΔL ≈ 6.5×10⁻³ m = 6.5 mm.
4. A steel rod of length 1 m is fixed between two rigid walls. It is heated by 50°C. What is the stress developed in the rod? (α = 1.2×10⁻⁵ /°C, Y = 2×10¹¹ N/m²)
- 1.2×10⁸ N/m² tensile
- 1.2×10⁸ N/m² compressive
- 2.4×10⁸ N/m² compressive
- 2.4×10⁸ N/m² tensile
Answer: 1.2×10⁸ N/m² compressive
Since the rod is fixed, total strain is zero. Thermal strain = αΔT = 1.2×10⁻⁵ × 50 = 6×10⁻⁴. Mechanical strain = σ/Y. Setting αΔT + σ/Y = 0 gives σ = -YαΔT = -2×10¹¹ × 6×10⁻⁴ = -1.2×10⁸ N/m². Negative sign indicates compressive stress.
5. A 10 kg mass is attached to a steel wire of length 2 m and cross-sectional area 1 mm² (Y = 2×10¹¹ N/m², g = 10 m/s²). If the mass is released suddenly from the unstretched position, the maximum extension of the wire is:
- 2 mm
- 1 mm
- 0.5 mm
- 4 mm
Answer: 2 mm
For sudden release, the mass oscillates. Maximum extension occurs when kinetic energy is zero. By energy conservation: mgΔL = (1/2)k(ΔL)², giving ΔL = 2mg/k. k = AY/L = (1e-6)(2e11)/2 = 1e5 N/m. So ΔL = 2*10*10/1e5 = 0.002 m = 2 mm.
6. Which of the following statements about Young's modulus is correct?
- It is defined as the ratio of longitudinal stress to longitudinal strain.
- It is defined as the ratio of longitudinal strain to longitudinal stress.
- It has the same unit as strain, which is dimensionless.
- It is the same for all materials.
Answer: It is defined as the ratio of longitudinal stress to longitudinal strain.
Young's modulus Y = (F/A) / (ΔL/L) = longitudinal stress / longitudinal strain. Its SI unit is pascal (Pa), same as stress. It is a material property and varies widely (e.g., steel ~2×10¹¹ Pa, rubber ~10⁶ Pa).
7. Two rods of same length 1 m are placed side by side. Steel rod: Y=2×10¹¹ N/m², area 2 cm². Copper rod: Y=1×10¹¹ N/m², area 4 cm². A total force of 8000 N is applied. What is the elongation?
- 1 mm
- 0.2 mm
- 0.5 mm
- 0.1 mm
Answer: 0.1 mm
Convert area: 2 cm² = 2×10⁻⁴ m², 4 cm² = 4×10⁻⁴ m². Stiffness: k1 = Y1A1/L = 2e11×2e-4/1 = 4×10⁷ N/m, k2 = 1e11×4e-4/1 = 4×10⁷ N/m. Total k = 8×10⁷ N/m. Elongation ΔL = F/k = 8000/(8×10⁷) = 1×10⁻⁴ m = 0.1 mm.
8. Which of the following statements is true about the stress-strain curve of a brittle material like glass?
- It has a long yield plateau.
- It shows significant strain hardening.
- It has a nearly linear region up to fracture with very little plastic deformation.
- It exhibits necking before fracture.
Answer: It has a nearly linear region up to fracture with very little plastic deformation.
Brittle materials like glass have a stress-strain curve that is almost linear up to the fracture point. They undergo negligible plastic deformation and break suddenly without a yield plateau or necking.
9. A vertical rod of length L, density ρ, Young's modulus Y, and coefficient of linear expansion α is clamped at both ends. If the temperature is increased by ΔT, the net stress at the top of the rod is:
- ρgL/2 - YαΔT
- ρgL - YαΔT
- ρgL + YαΔT
- ρgL/2 + YαΔT
Answer: ρgL - YαΔT
At the top, the rod supports the entire weight below, giving tensile stress ρgL. Thermal expansion prevented by clamps induces compressive stress YαΔT. Net stress = tensile - compressive = ρgL - YαΔT.
10. A uniform steel rod of length 2 m, density 8000 kg/m³, and Young's modulus 2×10¹¹ Pa hangs vertically from a ceiling. Find its elongation due to its own weight. (Take g = 10 m/s²)
- 16 μm
- 4 μm
- 8 μm
- 32 μm
Answer: 8 μm
The elongation of a uniform rod under its own weight is ΔL = ρgL²/(2Y). Substituting ρ=8000 kg/m³, g=10 m/s², L=2 m, Y=2×10¹¹ Pa gives ΔL = (8000×10×4)/(2×2×10¹¹) = 320000/(4×10¹¹) = 8×10⁻⁶ m = 8 μm.
11. A rubber ball is squeezed uniformly from all sides. The type of stress developed inside is:
- volumetric stress
- compressive stress
- shear stress
- tensile stress
Answer: volumetric stress
When a body is subjected to uniform pressure from all sides, the internal restoring force per unit area is called volumetric or hydrostatic stress. This is the case for a ball squeezed uniformly.
12. The area inside a loading-unloading hysteresis loop in the stress-strain plane represents:
- total work done per unit volume per cycle
- energy stored per unit volume per cycle
- energy lost per unit area per cycle
- energy dissipated per unit volume per cycle
Answer: energy dissipated per unit volume per cycle
The hysteresis loop area equals the energy dissipated as heat per unit volume per cycle. Stress has units of Pa (J/m³) and strain is dimensionless, so the product gives energy per unit volume.