On suspending a weight Mg, the length l of elastic wire and area of cross-section A its length becomes double the initial length. The instantaneous stress action on the wire is

The correct option is 2Mg/A.

Let the initial length of the wire be $l$ and its initial cross-sectional area be $A$.
When a weight of mass $M$ is suspended, the stretching force acting on the wire is:
$F=Mg$

According to the problem, the length of the wire doubles, which means the final length $l_f$ becomes:
$l_f=2l$

Assuming the volume of the elastic wire remains constant during elongation:
$\text{Initial Volume}=\text{Final Volume}$
$A\times l=A^{\prime }\times \left(2l\right)$

where $A^{\prime }$ is the new instantaneous cross-sectional area of the wire. Solving for $A^{\prime }$ yields:
$A^{\prime }=\frac{A}{2}$

The instantaneous stress on the wire is defined as the force divided by the instantaneous cross-sectional area:
$\text{Instantaneous Stress}=\frac{F}{A^{\prime }}$

Substituting $F=Mg$ and $A^{\prime }=\frac{A}{2}$ into the equation:
$\text{Instantaneous Stress}=\frac{Mg}{\left(\frac{A}{2}\right)}$
$\text{Instantaneous Stress}=\frac{2Mg}{A}$