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(C) The volume of the wire remains constant during stretching. Let the initial length be $l$ and initial area be $A$. The initial volume is $V = A \cd

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$2Mg/A$

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(C) The volume of the wire remains constant during stretching. Let the initial length be $l$ and initial area be $A$. The initial volume is $V = A \cdot l$.When the length becomes double $(l' = 2l)$,the new area $A'$ must satisfy $A' \cdot l' = A \cdot l$.Substituting $l' = 2l$,we get $A' \cdot (2l) = A \cdot l$,which implies $A' = A/2$.Stress is defined as the force applied per unit area: $	ext{Stress} = \frac{F}{A'}$.Substituting the values $F = Mg$ and $A' = A/2$,we get $	ext{Stress} = \frac{Mg}{A/2} = \frac{2Mg}{A}$.

$A$ wire of length $l$ and cross-sectional area $A$ is stretched by a weight $Mg$ such that its length becomes double. What is the stress in the wire?

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