Elastic Potential Energy

As stated in NCERT Class 11, Chapter 9 (Mechanical Properties of Solids), the elastic potential energy stored in a stretched wire within the elastic limit is $U=\frac{1}{2}\times \text{stress}\times \text{strain}\times \text{volume}$, which for a given wire reduces to $U=\frac{1}{2}k(\Delta L{)}^2$, behaving like a spring with effective stiffness $k$. The energy is therefore proportional to the square of the extension. Doubling the extension multiplies the stored energy by $2^2=4$. The option 'two times' wrongly assumes a linear dependence on extension. The option 'half' inverts the relationship. The option 'eight times' applies a cubic power that does not appear in the energy expression. This is the same quadratic relationship found in a compressed or stretched spring, where the stored energy is $\frac{1}{2}k{x}^2$, highlighting that elastic solids store energy just like ideal springs within their elastic limit. As a sanity check, since force grows linearly with extension and the stored energy is the triangular area under the force-extension graph, doubling the extension both doubles the peak force and doubles the displacement, giving a four-fold area. This confirms the result and explains why slightly overstretching a catapult or bowstring greatly increases the energy it can release.