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DSSSB JE ME 2019 Official Paper Shift - 2 (Held on 06 Nov 2019)

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**