Let's solve each question one by one: --- ## **Question 1** **Find the value of shape factor F1 for cylindrical cavity of diameter 10 cm and height 15 cm:** For a cylindrical cavity (vertical), the **shape factor (F1)**, also known as the view factor, from the bottom surface (1) to the side surface (2) is given by: \[ F_{1-2} = \frac{1}{1 + \frac{d}{4h}} \] Where: - \( d = 10 \) cm (diameter) - \( h = 15 \) cm (height) \[ F_{1-2} = \frac{1}{1 + \frac{10}{4 \times 15}} \] \[ = \frac{1}{1 + \frac{10}{60}} \] \[ = \frac{1}{1 + 0.1667} \] \[ = \frac{1}{1.1667} \approx 0.857 \] **Correct option:** - **0.857** --- ## **Question 2** **Given:** - Length \( l = 1.5 \) m - Impact energy \( U = 50 \) Nm - Stress \( \sigma = 200 \) MPa \( = 200 \times 10^6 \) Pa - Elastic modulus \( E = 95 \) GPa \( = 95 \times 10^9 \) Pa ### **Step 1: Strain energy in bar** Strain energy stored in a rod: \[ U = \frac{\sigma^2}{2E} \times \text{Volume} \] The volume of the rod: \[ \text{Volume} = A \times l \] So, \[ U = \frac{\sigma^2}{2E} \times A \times l \] \[ A = \frac{U \times 2E}{\sigma^2 \times l} \] ### **Step 2: Substitute values** \[ A = \frac{50 \times 2 \times 95 \times 10^9}{(200 \times 10^6)^2 \times 1.5} \] \[ A = \frac{9500 \times 10^9}{(40000 \times 10^{12}) \times 1.5} \] \[ A = \frac{9500 \times 10^9}{60000 \times 10^{12}} \] \[ A = \frac{9500}{60000} \times 10^{-3} \] \[ A = 0.1583 \times 10^{-3} = 1.583 \times 10^{-4} \text{ m}^2 \] ### **Step 3: Find diameter** \[ A = \frac{\pi}{4} d^2 \] \[ d^2 = \frac{4A}{\pi} \] \[ d^2 = \frac{4 \times 1.583 \times 10^{-4}}{3.1416} \] \[ d^2 = \frac{6.332 \times 10^{-4}}{3.1416} \] \[ d^2 = 2.016 \times 10^{-4} \] \[ d = \sqrt{2.016 \times 10^{-4}} \] \[ d \approx 0.0142 \text{ m} = 14.2 \text{ mm} \] **Correct option:** - **14.20 mm** --- ### **Summary of Answers:** - **Question 1:** **0.857** - **Question 2:** **14.20 mm** Let me know if you need detailed steps or explanations!