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(A) Consider a bar of cross-sectional area $A$ subjected to a tensile force $F$. Let the cross-section $PQRS$ be inclined at an angle $	heta$ with th

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(A) Consider a bar of cross-sectional area $A$ subjected to a tensile force $F$. Let the cross-section $PQRS$ be inclined at an angle $	heta$ with the plane perpendicular to the force $F$. The area of the inclined cross-section $PQRS$ is $A' = \frac{A}{\cos 	heta}$. The component of force $F$ normal to the cross-section $PQRS$ is $F_n = F \cos 	heta$. The tensile stress $\sigma$ on the cross-section $PQRS$ is given by: $\sigma = \frac{F_n}{A'} = \frac{F \cos 	heta}{A / \cos 	heta} = \frac{F}{A} \cos^2 	heta$. For the tensile stress $\sigma$ to be maximum,$\cos^2 	heta$ must be maximum. The maximum value of $\cos^2 	heta$ is $1$,which occurs when $	heta = 0^o$.

Column $I$	Column $II$
$A$. Hooke's law	$1$. Tangential strain
$B$. Shearing strain	$2$. Temporary loss of elastic property
$C$. Bulk strain	$3$. Elastic limit
$D$. Elastic fatigue	$4$. $3$ times the linear strain

For the tensile stress on the cross-section $PQRS$ to be maximum,$\theta =$ ......... $^o$

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