Step 1: The maximum shear stress in a two-dimensional stress state can be calculated using the following formula: \[ \tau_{{max}} = \frac{1}{2} \sqrt{ (\sigma_{xx} - \sigma_{yy})^2 + 4\sigma_{xy}^2 } \] where \( \sigma_{xx} \), \( \sigma_{yy} \), and \( \sigma_{xy} \) are the normal and shear stresses.
Step 2: Substituting the given values into the formula: \[ \tau_{{max}} = \frac{1}{2} \sqrt{ (800 - 0)^2 + 4(300)^2 } \] \[ \tau_{{max}} = \frac{1}{2} \sqrt{ 640000 + 360000 } \] \[ \tau_{{max}} = \frac{1}{2} \sqrt{ 1000000 } \] \[ \tau_{{max}} = \frac{1}{2} \times 1000 = 500 \, {MPa} \] Step 3: Therefore, the maximum shear stress is \( 500 \, {MPa} \), which corresponds to option (A).
A force of \( P = 100 \, {N} \) is applied at the ends of the pliers as shown in the figure. Neglecting friction, the force exerted by the upper jaw on the workpiece is ........... N (in integer).
Consider a beam with a square box cross-section as shown in the figure. The outer square has a length of 10 mm. The thickness of the section is 1 mm. The area moment of inertia about the x-axis is ........... mm\(^4\) (in integer). 
The value of the determinant 
is: