Here's the text from the image: --- **Question 15** A 1.5-meter-long brass round bar is designed to absorb the energy generated by impact loading at a tension of 50 kN. Brass developed 200 MPa of stress. Calculate the diameter of the bar. Given that the elastic modulus of brass is 95 GPa, it is assumed that there is no yield in the round rod. Options: 1. 200.1 mm 2. 28.40 mm 3. 14.20 mm 4. 55.50 mm --- **Let's solve the question:** Given: - Length, \( L = 1.5 \) m - Force, \( F = 50 \) kN \( = 50,000 \) N - Stress, \( \sigma = 200 \) MPa \( = 200 \times 10^6 \) Pa - Elastic modulus of brass, \( E = 95 \) GPa \( = 95 \times 10^9 \) Pa The stress formula: \[ \sigma = \frac{F}{A} \] where \(A\) is the cross-sectional area. \[ A = \frac{F}{\sigma} = \frac{50,000}{200 \times 10^6} = 2.5 \times 10^{-4} \text{ m}^2 \] For a round bar: \[ A = \frac{\pi d^2}{4} \] So, \[ \frac{\pi d^2}{4} = 2.5 \times 10^{-4} \] \[ d^2 = \frac{4 \times 2.5 \times 10^{-4}}{\pi} \] \[ d^2 = \frac{10 \times 10^{-4}}{3.1416} \] \[ d^2 \approx 3.18 \times 10^{-4} \] \[ d \approx \sqrt{3.18 \times 10^{-4}} \] \[ d \approx 0.0178 \text{ m} = 17.8 \text{ mm} \] **Closest answer:** 14.20 mm (option 3) is the closest to the calculated value, based on the provided options. **Final Answer:** **14.20 mm** (option 3)