A metallic square plate is subjected to a uniform hydrostatic pressure (P). Choose the correct Mohr’s circle representing the state of stress at any point in the plate from the options given below. For the Mohr’s circle, normal stress is positive towards the right and shear stress is positive in the upward direction.

Correct Answer:

Step-by-Step Explanation:

1. Understanding the Hydrostatic Stress State:
When a body (such as a metallic square plate) is subjected to a uniform hydrostatic pressure $P$, equal compressive stresses act on it from all directions in two mutually perpendicular axes (2D plane stress state).
According to standard sign convention, tensile stresses are considered positive and compressive stresses (like hydrostatic pressure) are considered negative.

Thus, the principal stresses acting at any point are:

${\sigma }_x=-P$

${\sigma }_y=-P$

There is no shear stress acting on these mutually perpendicular planes, so:

${\tau }_{xy}=0$

2. Center and Radius of Mohr's Circle:
The center of Mohr's circle along the normal stress axis ($\sigma$-axis) is given by the formula:

$\text{Center}=\left(\frac{{\sigma }_x+{\sigma }_y}{2},0\right)=\left(\frac{-P-P}{2},0\right)=\left(-P,0\right)$

The radius $R$ of Mohr's circle is given by:

$R=\sqrt{{\left(\frac{{\sigma }_x-{\sigma }_y}{2}\right)}^2+{\tau }_{xy}^2}=\sqrt{{\left(\frac{-P-(-P)}{2}\right)}^2+0^2}=0$

3. Conclusion:
Since the radius of the Mohr's circle is $R=0$, the Mohr's circle reduces to a single point located on the negative normal stress axis at $\sigma =-P$ and $\tau =0$.
This point is clearly represented in the second image option, showing a single point labelled $-P$ on the horizontal normal stress ($\sigma$) axis to the left of the origin $\text{O}$.