A steel rod of length ell,cross-sectional are | vedclass

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(B) The thermal expansion of the rod is $\Delta \ell = \ell \alpha t$.The force $F$ required to prevent this expansion (or the force exerted by the ro

Vedclass · Accepted Answer

$\frac{1}{2}(YA\alpha t) 	imes (\ell\alpha t)$

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(B) The thermal expansion of the rod is $\Delta \ell = \ell \alpha t$.The force $F$ required to prevent this expansion (or the force exerted by the rod) is given by Hooke's Law: $F = Y A \frac{\Delta \ell}{\ell} = Y A \frac{\ell \alpha t}{\ell} = Y A \alpha t$.The work done by the rod during this expansion is equivalent to the elastic potential energy stored in the rod,which is given by $W = \frac{1}{2} 	imes 	ext{Force} 	imes 	ext{Extension}$.Substituting the values: $W = \frac{1}{2} 	imes (Y A \alpha t) 	imes (\ell \alpha t)$.

$A$ steel rod of length $\ell$,cross-sectional area $A$,Young's modulus of elasticity $Y$,and thermal coefficient of linear expansion $\alpha$ is heated so that its temperature increases by $t\,^\circ C$. The work that can be done by the rod on heating is:

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