Correct Answer - Option 3 : 25 MPa

__Concept:__

\(Stress\;(\sigma) = \frac{Load}{Area\; of\; cross\; section }\)

Here, the Area of the cross-section is perpendicular to the applied load.

__Calculation:__

__Given:__

Load = 100 N, Area of rectangular plate = 4 mm2.

\(Stress\;(\sigma) = \frac{Load}{Area\; of\; cross\; section }\)

\(\sigma = \frac{100}{4}\) = 25 N/mm^{2} or MPa

∴ The stress developed in the plate is **25 MPa**