Elastic potential energy 'u' stored in a streched wire is

Question

Accepted Answer

The correct option is (2):
When a wire is stretched, work is done against the internal restoring forces, and this work is stored as elastic potential energy in the wire. To derive the formula, consider a wire of length $L$ and cross-sectional area $A$. When a force $F$ is applied, it causes an extension $\Delta L$. The elastic potential energy (U) stored per unit volume (energy density) is given by the area under the stress-strain curve, which for elastic deformation is a triangle.
Thus, energy density ($u$) = $\frac{1}{2} 	imes 	ext{stress} 	imes 	ext{strain}$.
Since the total elastic potential energy (U) stored in the wire is the energy density multiplied by the volume of the wire ($V = A 	imes L$), we have:
$U = u 	imes V = \frac{1}{2} 	imes 	ext{stress} 	imes 	ext{strain} 	imes 	ext{volume of wire}$.
This formula directly relates the stored energy to stress, strain, and the volume of the material.

The incorrect option is (1):
$\frac{1}{2}$ Young's modulus $	imes$ stress $	imes$ volume of wire. This option is incorrect because Young's modulus (Y) is defined as stress/strain. Substituting this into the energy density formula would give $u = \frac{1}{2} 	imes Y 	imes 	ext{strain}^2$, not involving stress in this manner.

The incorrect option is (3):
Stress $	imes$ strain$^2$ $	imes$ volume of wire. This option is incorrect as it misses the factor of $\frac{1}{2}$ and incorrectly squares the strain while omitting the stress term from its proper place in the energy density calculation.

The incorrect option is (4):
Stress $	imes$ strain $	imes$ volume of wire. This option is incorrect because the elastic potential energy stored, when stress and strain are proportional (within the elastic limit), is represented by the area of a triangle under the stress-strain curve, which requires a factor of $\frac{1}{2}$.