Elastic potential energy scales inversely with cross-sectional area

Steps:

• Elastic PE $U = \frac{1}{2} \frac{F^2}{k}$, where $k$ is the effective spring constant.
• For wires, $k = \frac{Y A}{L}$, with $Y$ (Young's modulus) and $L$ (length) the same for both.
• Diameter of Y is twice that of X, so radius doubles and area $A_Y = 4 A_X$.
• Thus, $k_Y = 4 k_X$, so $U_Y = \frac{1}{2} \frac{F^2}{4 k_X} = \frac{1}{4} U_X = 0.25E$.

Why A is correct:

• $U \propto 1/k$ and $k \propto A$, so quadrupling area quarters the energy for the same force.

Why the others are wrong:

• B: Assumes area doubles (halving energy), but area quadruples.
• C: Assumes stiffness halves (doubling energy), opposite of actual increase.
• D: Assumes stiffness quarters (quadrupling energy), opposite of actual increase.

Final answer: A