Let's break down the problem step by step. ### **Given:** - Length of bar, \( L = 1.5 \) m - Energy absorbed, \( U = 50 \) Nm (or J) - Stress developed, \( \sigma = 200 \) MPa = \( 200 \times 10^6 \) Pa - Elastic modulus, \( E = 95 \) GPa = \( 95 \times 10^9 \) Pa - No yield in the rod (remains elastic) ### **Step 1: Strain Energy in a Bar** The strain energy stored in a bar under axial load is: \[ U = \frac{\sigma^2}{2E} \cdot V \] where \( V \) is the volume of the bar. For a round bar, \[ V = A \cdot L = \left(\frac{\pi d^2}{4}\right) L \] So, \[ U = \frac{\sigma^2}{2E} \cdot \left(\frac{\pi d^2}{4}L\right) \] ### **Step 2: Solving for Diameter \( d \)** \[ U = \frac{\sigma^2}{2E} \cdot \frac{\pi d^2}{4} L \] \[ 50 = \frac{(200 \times 10^6)^2}{2 \times 95 \times 10^9} \cdot \frac{\pi d^2}{4} \times 1.5 \] \[ 50 = \left(\frac{4 \times 10^{16}}{190 \times 10^9}\right) \cdot \frac{\pi d^2}{4} \times 1.5 \] \[ 50 = \left(\frac{4 \times 10^{16}}{1.9 \times 10^{11}}\right) \cdot \frac{\pi d^2}{4} \times 1.5 \] \[ \frac{4 \times 10^{16}}{1.9 \times 10^{11}} = 2.105 \times 10^5 \] So, \[ 50 = 2.105 \times 10^5 \cdot \frac{\pi d^2}{4} \times 1.5 \] \[ 50 = 2.105 \times 10^5 \times 1.5 \times \frac{\pi d^2}{4} \] \[ 50 = 3.1575 \times 10^5 \times \frac{\pi d^2}{4} \] \[ 50 = 3.1575 \times 10^5 \times .7854 d^2 \] \[ 50 = 2.478 \times 10^5 d^2 \] \[ d^2 = \frac{50}{2.478 \times 10^5} \] \[ d^2 = 2.017 \times 10^{-4} \] \[ d = \sqrt{2.017 \times 10^{-4}} \] \[ d \approx .0142 \, \text{m} = 14.2 \, \text{mm} \] --- ### **Final Answer** **The correct option is:** **14.20 mm** --- #### **Summary Table** | Option | Value | |------------|-----------| | 55.50 mm | | | 20.01 mm | | | 28.40 mm | | | **14.20 mm** | **✔️** |