
Strain energy per unit volume in a stretched string is
$A.\dfrac{1}{2}\times Stress\times Strain$
$B.Stress\times Strain$
$C.{{\left( Stress\times Strain \right)}^{2}}$
$D.\dfrac{Stress}{Strain}$
Answer
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Hint: We know that the restoring force per unit area is called stress and strain is defined as the measure of deformation. Strain energy is the energy stored by a system when a system is undergoing deformation. We have to apply these concepts to find the strain energy per unit volume in relation to stress and strain.
Formula Used:
We will use the following formula to find the strain energy per unit volume:-
$F=\dfrac{YAx}{L}$.
Complete step-by-step solution:
Let Young’s modulus be $Y$, wire of length be $L$, Area of cross-section be $A$, the load is $F$, the extension is $x$, stress is $\sigma $ and strain be $\varepsilon $ and work done be $W$ and $V$ is the volume.
$\Rightarrow Y=\dfrac{FL}{Ax}$………………. $(i)$
From $(i)$ we can write
$F=\dfrac{YAx}{L}$……………… $(ii)$
$\Rightarrow W=F.dx$……………… $(iii)$
Where $dx$ is the displacement
Putting $(ii)$ in $(iii)$ we get
$W=YA\left( \dfrac{x}{L} \right)dx$……………….$(iv)$
We have to integrate it to find the work done in elongation $x$
$W=\dfrac{YA}{L}\int{xdx}$………………….. $(v)$
Integrating $(v)$ we get
$\Rightarrow W=\dfrac{1}{2}\left( \dfrac{YA{{x}^{2}}}{L} \right)$
This can be written as
$\Rightarrow W=\dfrac{1}{2}\left( \dfrac{YAx}{L} \right)x$
Using $(ii)$ we get
$\Rightarrow W=\dfrac{1}{2}\times F\times x$……………… $(iv)$
Work done is equal to strain energy, $E$ can be written as
$\Rightarrow E=\dfrac{1}{2}\times F\times x$
Therefore, strain energy per unit volume can be written as
$\dfrac{1\times F\times x}{2\times V}$
$\Rightarrow \dfrac{1\times F\times x}{2\times A\times L}$………….. $(v)$
This can be rearranged into
$\Rightarrow \dfrac{1}{2}\times \sigma \times \varepsilon $
Because $\sigma =\dfrac{F}{A}$and $\varepsilon =\dfrac{x}{L}$.
Hence, option $(A)$ is the correct option.
Additional Information:
We should also have knowledge of Hooke’s law whenever we analyze stress, strain, and strain energy. Hooke’s law states that within proportional limit stress is directly proportional to strain produced. Mathematically it can be written as
$\sigma \propto \varepsilon $
$\Rightarrow \sigma = E\varepsilon $
$\Rightarrow \dfrac{\sigma }{\varepsilon }= E$
Where $E$ is proportionality constant.
Note: We should not get confused over the fact that strain and strain energies are different terms and their formulae are also different. We should also read the question carefully as it is said to find strain energy per unit volume not only strain energy. Strain energy stored in a material can be represented as work done on it.
Formula Used:
We will use the following formula to find the strain energy per unit volume:-
$F=\dfrac{YAx}{L}$.
Complete step-by-step solution:
Let Young’s modulus be $Y$, wire of length be $L$, Area of cross-section be $A$, the load is $F$, the extension is $x$, stress is $\sigma $ and strain be $\varepsilon $ and work done be $W$ and $V$ is the volume.
$\Rightarrow Y=\dfrac{FL}{Ax}$………………. $(i)$
From $(i)$ we can write
$F=\dfrac{YAx}{L}$……………… $(ii)$
$\Rightarrow W=F.dx$……………… $(iii)$
Where $dx$ is the displacement
Putting $(ii)$ in $(iii)$ we get
$W=YA\left( \dfrac{x}{L} \right)dx$……………….$(iv)$
We have to integrate it to find the work done in elongation $x$
$W=\dfrac{YA}{L}\int{xdx}$………………….. $(v)$
Integrating $(v)$ we get
$\Rightarrow W=\dfrac{1}{2}\left( \dfrac{YA{{x}^{2}}}{L} \right)$
This can be written as
$\Rightarrow W=\dfrac{1}{2}\left( \dfrac{YAx}{L} \right)x$
Using $(ii)$ we get
$\Rightarrow W=\dfrac{1}{2}\times F\times x$……………… $(iv)$
Work done is equal to strain energy, $E$ can be written as
$\Rightarrow E=\dfrac{1}{2}\times F\times x$
Therefore, strain energy per unit volume can be written as
$\dfrac{1\times F\times x}{2\times V}$
$\Rightarrow \dfrac{1\times F\times x}{2\times A\times L}$………….. $(v)$
This can be rearranged into
$\Rightarrow \dfrac{1}{2}\times \sigma \times \varepsilon $
Because $\sigma =\dfrac{F}{A}$and $\varepsilon =\dfrac{x}{L}$.
Hence, option $(A)$ is the correct option.
Additional Information:
We should also have knowledge of Hooke’s law whenever we analyze stress, strain, and strain energy. Hooke’s law states that within proportional limit stress is directly proportional to strain produced. Mathematically it can be written as
$\sigma \propto \varepsilon $
$\Rightarrow \sigma = E\varepsilon $
$\Rightarrow \dfrac{\sigma }{\varepsilon }= E$
Where $E$ is proportionality constant.
Note: We should not get confused over the fact that strain and strain energies are different terms and their formulae are also different. We should also read the question carefully as it is said to find strain energy per unit volume not only strain energy. Strain energy stored in a material can be represented as work done on it.
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