MCQ
The elastic energy stored in a wire of Young's modulus $Y$ is
  • A
    $Y \times \frac{{{\rm{Strai}}{{\rm{n}}^{\rm{2}}}}}{{{\rm{Volume}}}}$
  • B
    Stress $ \times $ Strain $ \times $ Volume
  • $\frac{{{\rm{Stres}}{{\rm{s}}^{\rm{2}}} \times {\rm{Volume}}}}{{2Y}}$
  • D
    $\frac{1}{2}Y \times $ Stress $ \times $ Strain $ \times $ Volume

Answer

Correct option: C.
$\frac{{{\rm{Stres}}{{\rm{s}}^{\rm{2}}} \times {\rm{Volume}}}}{{2Y}}$
c
When a wire is stretched work is done against the interatomic forces. This work is stored in the wire in the form of elastic potential energy.

$W=\frac{1}{2} \times$ stress $\times$ strain $\times$ volume of wire

Also, when strain in small, ratio of longitudinal stress to corresponding longitudinal strain is called Young's modulus of material of body.

$Y=\frac{\text { longitudinal stress }}{\text { longitudinal strain }}$

$W=\frac{1}{2} \times \text { stress } \times \frac{\text { stress }}{Y} \times \text { volume }$

$W=\frac{(\text {stress)}^2 \times \text { volume }}{2 Y}$

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