The state of stress at a point, for a body in plane stress, is shown in the figure below. If the minimum principal stress is 10 kPa, then the normal stress σ_y (in kPa) is      

The correct option is 37.78.

Given Data from the stress element in the figure:
Normal stress in the x-direction, ${\sigma }_x=100\text{kPa}$ (tensile)
Shear stress, ${\tau }_{xy}=50\text{kPa}$
Minimum principal stress, ${\sigma }_2=10\text{kPa}$

Formula:
The expression for the minimum principal stress (${\sigma }_2$) in plane stress is given by:
${\sigma }_2=\frac{{\sigma }_x+{\sigma }_y}{2}-\sqrt{{\left(\frac{{\sigma }_x-{\sigma }_y}{2}\right)}^2+{\tau }_{xy}^2}$

Step-by-step Calculation:
Substitute the given values into the formula:
$10=\frac{100+{\sigma }_y}{2}-\sqrt{{\left(\frac{100-{\sigma }_y}{2}\right)}^2+{50}^2}$
Rearrange the equation to isolate the radical term:
$\sqrt{{\left(\frac{100-{\sigma }_y}{2}\right)}^2+2500}=\frac{100+{\sigma }_y}{2}-10$
Simplify the right-hand side:
$\sqrt{{\left(50-0.5{\sigma }_y\right)}^2+2500}=40+0.5{\sigma }_y$
Square both sides of the equation:
${\left(50-0.5{\sigma }_y\right)}^2+2500={\left(40+0.5{\sigma }_y\right)}^2$
Expand both sides:
$2500-50{\sigma }_y+0.25{\sigma }_y^2+2500=1600+40{\sigma }_y+0.25{\sigma }_y^2$
Subtract $0.25{\sigma }_y^2$ from both sides:
$5000-50{\sigma }_y=1600+40{\sigma }_y$
Rearrange the terms to solve for ${\sigma }_y$:
$5000-1600=40{\sigma }_y+50{\sigma }_y$
$3400=90{\sigma }_y$
${\sigma }_y=\frac{3400}{90}\approx 37.78\text{kPa}$

Therefore, the normal stress ${\sigma }_y$ is 37.78 kPa.