To solve this problem, need to understand the relationship between the stress, the shear stress on the slip plane, and the resulting shear stress on the slip plane. ### Given: - Tensile stress, \( \sigma = \, \text{MPa} \) - The slip occurs on the plane \( (111) \) in the \( [110] \) direction ### Step 1: Find the shear stress on the slip plane The critical resolved shear stress \( \tau_{cr} \) is given by: \[ \tau_{cr} = \sigma \times \cos \phi \times \cos \lambda \] where: - \( \phi \) is the angle between the normal to the slip plane and the applied stress direction - \( \lambda \) is the angle between the slip direction and the applied stress direction ### Step 2: Determine the angles - For the plane \( (111) \), the normal is \( [111] \) - For the slip direction \( [110] \) - The applied stress direction is along \( [010] \) Calculate the angles: \[ \cos \phi = \frac{ \textbf{normal} \cdot \textbf{stress direction} }{ |\textbf{normal}| |\textbf{stress direction}| } = \frac{ [111] \cdot [010] }{ \sqrt{1^2 + 1^2 + 1^2} \times \sqrt{^2 + 1^2 + ^2} } = \frac{ \times 1 + 1 \times + 1 \times }{ \sqrt{3} \times 1 } = \] Since \( \cos \phi = \), the shear stress on the slip plane is maximized when the plane is oriented at the critical angle. But, in terms of resolved shear stress, the maximum shear occurs when angles \( \phi \) and \( \lambda \) satisfy the following: \[ \tau = \sigma \ \cos \phi \times \cos \lambda \] Using the known orientations, the resolved shear stress is: \[ \tau = \sigma \times \frac{1}{\sqrt{6}} \approx \frac{10}{\sqrt{6}} \approx 4.082\, \text{MPa} \] ### **Answer:** **Correct option:** \[ \boxed{4.083\, \text{MPa}} \] which matches the first option.