# Solution to Q.4 (c) Let's solve each part step by step: --- ## **(i) Cross-sectional area of the rod in mm²** The cross-section is an **equilateral triangle** with side = 50 mm. The area \( A \) of an equilateral triangle with side \( a \) is: \[ A = \frac{\sqrt{3}}{4} a^2 \] Substituting \( a = 50 \) mm: \[ A = \frac{\sqrt{3}}{4} \times (50)^2 \] \[ A = \frac{\sqrt{3}}{4} \times 250 \] \[ A = .433 \times 250 \] \[ A = 1082.5~\text{mm}^2 \] --- ## **(ii) Stress of the rod in N/mm²** Stress (\( \sigma \)) is: \[ \sigma = \frac{F}{A} \] Given force \( F = 325~\text{kN} = 325,000~\text{N} \), and \( A = 1082.5~\text{mm}^2 \): \[ \sigma = \frac{325,000}{1082.5} \] \[ \sigma \approx 300.3~\text{N/mm}^2 \] --- ## **(iii) Strain of the rod** Strain (\( \epsilon \)) is: \[ \epsilon = \frac{\sigma}{E} \] Given \( E = 300,000~\text{N/mm}^2 \): \[ \epsilon = \frac{300.3}{300,000} \] \[ \epsilon = .001001 \] --- ## **(iv) Deformation of length (extension) to the rod in mm** Extension (\( \Delta L \)) is: \[ \Delta L = \epsilon \times L \] Given \( L = 4~\text{m} = 400~\text{mm} \): \[ \Delta L = .001001 \times 400 \] \[ \Delta L = 4.004~\text{mm} \] --- ## **Summary Table** | Part | Quantity | Formula/Calculation | Value | |------|-------------------------------|-------------------------------|-------------------| | (i) | Cross-sectional area (\( A \)) | \( \frac{\sqrt{3}}{4} a^2 \) | \( 1082.5~\text{mm}^2 \) | | (ii) | Stress (\( \sigma \)) | \( \frac{F}{A} \) | \( 300.3~\text{N/mm}^2 \) | | (iii)| Strain (\( \epsilon \)) | \( \frac{\sigma}{E} \) | \( .001001 \) | | (iv) | Extension (\( \Delta L \)) | \( \epsilon \times L \) | \( 4.004~\text{mm} \) | --- ### **Final Answers** 1. **Cross-sectional area:** 1082.5 mm² 2. **Stress:** 300.3 N/mm² 3. **Strain:** .001001 4. **Deformation (extension):** 4.004 mm