Question

If a wire is stretched by applying force at one of its ends, then the elastic potential energy density in terms of Young's modulus Y and linear strain $\alpha$ will be

• A. $\frac{\alpha Y^2}{2}$
• B. $\frac{Y\alpha }{2}$
• C. $\frac{{\alpha }^{2}Y}{2}$
• D. $2{\alpha }^{2}Y$

Answer 1

Answer: (C)

$\frac{{\alpha }^{2}Y}{2}$

$PotentialEnergyDensity=\frac{1}{2}\times Stress\times Strain=\frac{1}{2}\times \frac{xY}{l}\times \frac{x}{l}(\because Stress=Y\times Strain)or,\frac{x^{2}Y}{2L^2}Since{\left(\frac{x}{L}\right)}^2=Strai{n}^2={\alpha }^{2}Thusgiving,\frac{{\alpha }^{2}Y}{2}$