Derive an expression for elastic potential energy per unit volume stored for the wire is
`1/2 xx` stress `xx` strain.

Explanation

Young modulus `Y= ( F//A)/(x//L)`
where x = length of wire increased
`therefore F = ((YA)/(L ))x`
Now, work done to be done to increae length dx,
`dW = Fdx`
`= ((YA)/(L)) x dx`
`therefore `Work to be done to increase the length l against restoring force,
`W = int _(0) ^(l) ((YA)/( L ))d dx`
`= (AY)/(2L) l ^(2)`
Which is the elastic potential energy stored in wire.
`therefore U= (AY)/(2L) l ^(2)`
`therefore ` Energy stored per unit volume `= (U )/(LA)`
`=1/2 (Yl ^(2))/(L ^(2))`
`=1/2 xx (("stress")/("strain")) xx ("strain") ^(2)`
`=1/2 xx ` stress `xx` strain