Question

if p longitudinal strain is produced in a wire of young modulus Q then energy stored in the material of the wire per unit volume is​

Answer 1

Answer:

Explanation:

Energy=1/2×stress×strain

E=1/2×4×strain^2=stress/strain

E=1/2p^2q

Answer 2

Answer:

$\frac{Qp^2}{2}$

Explanation:

Given longitudinal strain = p

The Young's modulus is Q

We know that

$\boxed{Q=\frac{\text{Stress}}{\text{Strain}}}$

Therefore,

$Y=\frac{\text{Stress}}{p}$

$\implies \text{Stress}=Q\times p$

The energy stored per unit volume is given by

$E=\frac{1}{2}\times \text{Stress}\times\text{Strain}$

$E=\frac{1}{2}\times Qp\times p$

$\implies E=\frac{1}{2}Qp^2$

or, $E=\frac{Qp^2}{2}$

Hope this helps.