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If in a wire of Young's modulus $Y$, longitudinal strain $X$ is produced then the potential energy stored in its unit volume will be
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Verified Answer
The correct answer is:
$0.5 Y X^{2}$
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. Elastic potential energy per unit volume is
$$
U=\frac{1}{2} \times \text { stress } \times \text { strain }
$$
From definition of Young's modulus of wire
$$
\begin{aligned}
Y &=\frac{\text { stress }}{\text { strain }} \\
\Rightarrow \quad \text { Stress } &=Y \times \text { strain }
\end{aligned}
$$
Given, $\quad$ strain $=X$
Therefore, $U=\frac{1}{2} Y X^{2}$
$$
\Rightarrow \quad U=0.5 Y X^{2}
$$
$$
U=\frac{1}{2} \times \text { stress } \times \text { strain }
$$
From definition of Young's modulus of wire
$$
\begin{aligned}
Y &=\frac{\text { stress }}{\text { strain }} \\
\Rightarrow \quad \text { Stress } &=Y \times \text { strain }
\end{aligned}
$$
Given, $\quad$ strain $=X$
Therefore, $U=\frac{1}{2} Y X^{2}$
$$
\Rightarrow \quad U=0.5 Y X^{2}
$$
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