A tensile test of a certain material provided the following data. Calculate the percentage reduction in area. Initial diameter \( =12 \mathrm{~mm} \) Initial gauge length \( =50.0 \mathrm{~mm} \) Final diameter \( =10.5 \mathrm{~mm} \) Final gauge length \( =60.2 \mathrm{~mm} \) A) \( 28.3 \% \) (B) \( 23.4 \% \) (C) \( 36.2 \% \) (D) \( 40.6 \% \)

To calculate the percentage reduction in area, we first need to find the initial and final cross-sectional areas. The area can be computed using the formula for the area of a circle: \( A = \pi \times \left(\frac{d}{2}\right)^2 \). 1. **Initial area** \( A_i \): \[ A_i = \pi \times \left(\frac{12\, \text{mm}}{2}\right)^2 = \pi \times (6\, \text{mm})^2 = 36\pi \, \text{mm}^2 \] 2. **Final area** \( A_f \): \[ A_f = \pi \times \left(\frac{10.5\, \text{mm}}{2}\right)^2 = \pi \times (5.25\, \text{mm})^2 = 27.5625\pi \, \text{mm}^2 \] 3. **Percentage reduction in area** is calculated using the formula: \[ \text{Percentage Reduction} = \left(\frac{A_i - A_f}{A_i}\right) \times 100\% \] \[ = \left(\frac{36\pi - 27.5625\pi}{36\pi}\right) \times 100\% = \left(\frac{8.4375\pi}{36\pi}\right) \times 100\% = \left(\frac{8.4375}{36}\right) \times 100\% \approx 23.4\% \] So, the correct answer is (B) \( 23.4\% \).