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Question 55
Question Statement:
A 1 m long rod of 1 cm × 1 cm cross section is subjected to an axial tensile force
of 35 kN. The Young's modulus of the material is 70 GPa. The cross-section of
the deformed rod is 0.998 cm × 0.998 cm. The Poisson's ratio of the material is
________ (rounded off to one decimal place).
Step-by-Step Solution & Explanation
### **Detailed Step-by-Step Solution (GATE AE 2025 Q.55)**
**1. Intuitive Physical Setup & Given Parameters**:
- Tensile bar: length $L = 1\text{ m}$.
- Initial square cross-section: $b_0 = 1\text{ cm} = 0.01\text{ m}$, $A_0 = 1\text{ cm}^2 = 10^{-4}\text{ m}^2$.
- Axial tensile force: $P = 35\text{ kN} = 35,000\text{ N}$.
- Young's modulus: $E = 70\text{ GPa} = 70 \times 10^9\text{ N/m}^2$.
- Deformed cross-section: $b = 0.998\text{ cm}$.
We need to determine Poisson's ratio $\nu$ (rounded off to one decimal place).
**2. Step-by-Step Mechanics of Materials Derivation**:
1. **Longitudinal (Axial) Strain ($\varepsilon_{\text{long}}$)**:
Axial tensile stress:
$\sigma = \frac{P}{A_0} = \frac{35,000\text{ N}}{10^{-4}\text{ m}^2} = 350 \times 10^6\text{ Pa} = 350\text{ MPa}$
By Hooke's law:
$\varepsilon_{\text{long}} = \frac{\sigma}{E} = \frac{350 \times 10^6\text{ Pa}}{70 \times 10^9\text{ Pa}} = 5 \times 10^{-3} = 0.005$
2. **Lateral Strain ($\varepsilon_{\text{lat}}$)**:
$\Delta b = b - b_0 = 0.998\text{ cm} - 1.000\text{ cm} = -0.002\text{ cm}$
$\varepsilon_{\text{lat}} = \frac{\Delta b}{b_0} = \frac{-0.002}{1.000} = -0.002$
3. **Definition of Poisson's Ratio ($\nu$)**:
$\nu = -\frac{\varepsilon_{\text{lat}}}{\varepsilon_{\text{long}}} = -\frac{-0.002}{0.005} = \frac{2}{5} = 0.4$
**3. Comprehensive Options & Distractor Analysis**:
- **Evaluation Criteria**: Numerical Answer Type (NAT).
- **Exact Analytical Result**: $\nu = 0.4$.
- **Rounding Requirement**: One decimal place $\implies \mathbf{0.4}$.
**4. Official Master Key & Exam Takeaway**:
- **Official Answer Key**: **0.4** (Acceptable Range: **0.4 to 0.4**)
- **Exam Takeaway**: $\nu = |\varepsilon_{\text{lat}}| / \varepsilon_{\text{long}} = \frac{|\Delta b / b_0|}{\sigma / E} = \frac{0.002}{0.005} = 0.4$.
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