The bar is subjected to equal and opposite forces as shown in the figure PQ is plane making angle $\mathrm{\theta }$ with the cross-section of the bar. If the area of cross-section be 'a', then what is the tensile stress on PQ ? 

From the figure,

${\mathrm{F}}_{\perp } = \mathrm{Fsin\theta } \mathrm{and} {\mathrm{F}}_{\parallel }= \mathrm{Fsin\theta }$

On the plane PQ, produces ${\mathrm{F}}_{\parallel }$ shear stress and produces ${\mathrm{F}}_{\perp }$ tensile stress. Let A' = the plane PQ's surface area

$\mathrm{cos\theta } =\frac{\mathrm{a}}{\mathrm{A}\text{'}}$

$\mathrm{A}\text{'}=\frac{\mathrm{a}}{\mathrm{cos\theta }}$

the normal force $\mathrm{Fcos\theta }$ acting on the specified plane. As a result, $\frac{\mathrm{a}}{\mathrm{cos\theta }}$ gives the area along which the force acts. As a result, the plane PQ's tensile strength will be:

$\mathrm{Tensile} \mathrm{stress}=\frac{\mathrm{Normal} \mathrm{Force}}{\mathrm{Area}}$

$\mathrm{Tensile} \mathrm{stress}=\frac{\mathrm{Fcos\theta }}{{\displaystyle \frac{\mathrm{a}}{\mathrm{cos\theta }}}}$

$\mathrm{Tensile} \mathrm{stress}=\frac{{\mathrm{Fcos}}^2\mathrm{\theta }}{\mathrm{a}}$

Hence the correct answer is $\frac{{\mathrm{Fcos}}^2\mathrm{\theta }}{\mathrm{a}}.$