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.

Step-by-step Derivation:

From the stress state shown in the given image, we can identify the following plane stress parameters:
- Normal stress in the x-direction: ${\sigma }_x=100\text{kPa}$
- Shear stress: ${\tau }_{xy}=50\text{kPa}$
- Minimum principal stress: ${\sigma }_{\text{min}}=10\text{kPa}$

The formula for the minimum principal stress in plane stress is given by:

$${\sigma }_{\text{min}}=\frac{{\sigma }_x+{\sigma }_y}{2}-\sqrt{{\left(\frac{{\sigma }_x-{\sigma }_y}{2}\right)}^2+{\tau }_{xy}^2}$$
Substitute the known values into the equation:

$$10=\frac{100+{\sigma }_y}{2}-\sqrt{{\left(\frac{100-{\sigma }_y}{2}\right)}^2+{50}^2}$$
Rearrange the equation to isolate the square root term:

$$\sqrt{{\left(\frac{100-{\sigma }_y}{2}\right)}^2+2500}=\frac{100+{\sigma }_y}{2}-10$$
Simplify the right-hand side:

$$\frac{100+{\sigma }_y}{2}-10=50+\frac{{\sigma }_y}{2}-10=40+\frac{{\sigma }_y}{2}$$
This gives:

$$\sqrt{{\left(\frac{100-{\sigma }_y}{2}\right)}^2+2500}=40+\frac{{\sigma }_y}{2}$$
Square both sides of the equation to eliminate the radical:

$${\left(\frac{100-{\sigma }_y}{2}\right)}^2+2500={(40+\frac{{\sigma }_y}{2})}^2$$
Expand both sides of the equation:

- Left-hand side (LHS):

$${(50-\frac{{\sigma }_y}{2})}^2+2500=2500-50{\sigma }_y+\frac{{\sigma }_y^2}{4}+2500=5000-50{\sigma }_y+\frac{{\sigma }_y^2}{4}$$
- Right-hand side (RHS):

$${(40+\frac{{\sigma }_y}{2})}^2=1600+40{\sigma }_y+\frac{{\sigma }_y^2}{4}$$
Set the expanded expressions equal to each other:

$$5000-50{\sigma }_y+\frac{{\sigma }_y^2}{4}=1600+40{\sigma }_y+\frac{{\sigma }_y^2}{4}$$
Subtract $\frac{{\sigma }_y^2}{4}$ from both sides:

$$5000-50{\sigma }_y=1600+40{\sigma }_y$$
Rearrange to solve for ${\sigma }_y$:

$$5000-1600=40{\sigma }_y+50{\sigma }_y$$
$$3400=90{\sigma }_y$$
$${\sigma }_y=\frac{3400}{90}\approx 37.78\text{kPa}$$