
In a wire stretched by hanging a weight from its end, the elastic potential energy per unit volume in terms of the longitudinal strain $\sigma $ and modulus of elasticity Y is
(A) $\dfrac{Y{{\sigma }^{2}}}{2}$
(B) $\dfrac{Y\sigma }{2}$
(C) $\dfrac{2Y{{\sigma }^{2}}}{2}$
(D) $\dfrac{{{Y}^{2}}\sigma }{2}$
Answer
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Hint: The elastic potential energy per unit volume of a wire stretched by hanging a weight from its end is given by half stress on the wire due to the weight ( i.e. the Young’s modulus times strain) multiplied by the strain on the wire due to its weight.
Complete step by step answer:
Let us consider a stretchable wire attached to surface and a mass suspended from its end as shown in the figure below,
The wire will naturally stretch under the weight of the mass attached to it since it is a stretchable wire. The wire experiences stress and strain, stress is the force acting on the unit area of a material and strain is the measure of the deformation of an object.
The elastic potential energy per unit volume of the wire is given by,
$\Rightarrow \dfrac{1}{2}\times $ strain $\times $ stress
Here, stress is Young’s modulus Y times strain, so,
$\Rightarrow \dfrac{1}{2}\times \sigma \times Y\sigma $
$\Rightarrow \dfrac{1}{2}\times Y{{\sigma }^{2}}$
$\Rightarrow \dfrac{Y{{\sigma }^{2}}}{2}$
Hence, the correct answer is option (A).
Additional Information:
The stress (i.e. force acting on the unit area of a material) and strain (i.e. the measure of the deformation of an object) are proportional to each other. This is given by Hooke's law which states that the strain on the object is proportional to the stress applied on it.
$F=-k\cdot x$
Where, F is force applied, x is length of spring and k is spring constant.
Note:
Students should remember the units of stress and strain. S.I. the unit of stress is N/${{m}^{2}}$, but strain is a dimensionless quantity. The formulae of elastic potential energy per unit volume ($\dfrac{Y{{\sigma }^{2}}}{2}$), stress ($\sigma =\dfrac{F}{A}$ ) and strain ( $\dfrac{\partial l}{L}$ ) should be memorized to solve such questions.
Complete step by step answer:
Let us consider a stretchable wire attached to surface and a mass suspended from its end as shown in the figure below,
The wire will naturally stretch under the weight of the mass attached to it since it is a stretchable wire. The wire experiences stress and strain, stress is the force acting on the unit area of a material and strain is the measure of the deformation of an object.
The elastic potential energy per unit volume of the wire is given by,
$\Rightarrow \dfrac{1}{2}\times $ strain $\times $ stress
Here, stress is Young’s modulus Y times strain, so,
$\Rightarrow \dfrac{1}{2}\times \sigma \times Y\sigma $
$\Rightarrow \dfrac{1}{2}\times Y{{\sigma }^{2}}$
$\Rightarrow \dfrac{Y{{\sigma }^{2}}}{2}$
Hence, the correct answer is option (A).
Additional Information:
The stress (i.e. force acting on the unit area of a material) and strain (i.e. the measure of the deformation of an object) are proportional to each other. This is given by Hooke's law which states that the strain on the object is proportional to the stress applied on it.
$F=-k\cdot x$
Where, F is force applied, x is length of spring and k is spring constant.
Note:
Students should remember the units of stress and strain. S.I. the unit of stress is N/${{m}^{2}}$, but strain is a dimensionless quantity. The formulae of elastic potential energy per unit volume ($\dfrac{Y{{\sigma }^{2}}}{2}$), stress ($\sigma =\dfrac{F}{A}$ ) and strain ( $\dfrac{\partial l}{L}$ ) should be memorized to solve such questions.
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