A metallic wire of length L metres extends by l metres when stretched by suspending a weight Mg to it. The mechanical energy stored in the wire is

To find the mechanical energy stored in a stretched wire, we can analyze the work done by the stretching force.

Let the initial length of the metallic wire be L. When a weight of mass M is suspended from it, it experiences a gravitational force (or tension) of:
$F=Mg$
This force causes the wire to extend by a length l.

The stretching of a wire behaves like a spring obeying Hooke's Law, where the restoring force is proportional to the extension. The tension in the wire increases linearly from 0 (at zero extension) to its maximum value Mg (at extension l).

The average force acting on the wire during this extension is given by:
$F_{\text{avg}}=\frac{0+Mg}{2}=\frac{Mg}{2}$

The work done (W) in stretching the wire through a distance l is the product of the average force and the extension:
$W=F_{\text{avg}}\times l$
$W=\frac{Mg}{2}\times l=\frac{Mgl}{2}$

This work done is stored in the wire as potential energy (mechanical energy). Therefore, the mechanical energy stored in the wire is:
$U=\frac{Mgl}{2}$

The correct option is Mgl/2.