Question

A metallic wire is stretched by suspended weight to it. If $\alpha$ is the longitudinal strain and $Y$ its Young’s modulus of elasticity, elastic potential energy per unit volume is

• A. $ Y \alpha^2$
• B. $\frac{1}{2} Y \alpha^2$
• C. $\frac{3}{2} Y \alpha^2$
• D. $ \frac{Y}{\alpha^2}$

Answer 1

Answer: (B)

Elastic energy per unit volume
$ = \frac{1}{2} \times\, stress \times \, strain$
Here, strain $= \alpha ,$
$= $and $ \, Y = \frac{stress}{strain} \, or \, stress= Y \times\, strain$
$\therefore \, stress = Y \times \alpha$
Elastic energy per unit volume
$ = \frac{1}{2} \times (Y \times \alpha ) \times \alpha = \frac{1}{2} Y \alpha^2$