A wire of length L and cross-sectional area A is made of a material of Young's modulus Y. If the wire is stretched by the amount x. the work done is:
(A) $ \dfrac{{Y{\text{ }}A{x^2}}}{{2L}} $
(B) $ \dfrac{{Y{\text{ }}A{x^2}}}{L} $
(C) $ Y{\text{ a}}{{\text{x}}^2}L $
(D) $ \dfrac{{Y{\text{ }}Ax}}{{2L}} $
Answer
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Hint :In order to answer this question, we need to understand that Young's modulus is a mechanical property of a solid material that measures the stiffness of that material. It can be defined as the relationship between stress (force per unit area) and strain (proportional deformation) in a material in the linear elasticity regime of a uniaxial deformation.
$ Work Done = \dfrac{1}{2} \times Stress \times Strain \times Volume $
Complete Step By Step Answer:
Young’s modulus describes the relationship between stress and strain.
Stress i.e. force per unit area and
strain i.e. proportional deformation in an object.
Here, we have to find the work done.
$ Work Done = \dfrac{1}{2} \times Stress \times Strain \times Volume $
We know that,
$ \because Stress = Y \times Strain $
Now substituting the formula for stress in the work done we get,
$ \Rightarrow Workdone = \dfrac{1}{2} \times Y \times {\left( {Strain} \right)^2} \times Volume $
Here, we know that
$ Strain = \dfrac{x}{L} $
$ Volume = AL $
were,
L is the length of wire
A is the cross-sectional area
Y is the Young's modulus
x is the amount of wire stretched
$ \Rightarrow Workdone = \dfrac{1}{2} \times Y \times {\left( {\dfrac{x}{L}} \right)^2} \times AL $
$ \Rightarrow W = \dfrac{{Y{x^2}A}}{{2L}} $
Hence, the correct option is (B).
Note :
Students need to first write the given information given in the question and use the suitable formula next define the physical quantities being discussed properly and then use the understanding and mathematical expression of these quantities such as stress, strain and Young’s modulus to solve the question. Additionally, students try to skip steps when solving these types of questions that can result in silly mistakes. So avoid skipping steps. The Young’s modulus is named after Thomas Young who was a British scientist.
$ Work Done = \dfrac{1}{2} \times Stress \times Strain \times Volume $
Complete Step By Step Answer:
Young’s modulus describes the relationship between stress and strain.
Stress i.e. force per unit area and
strain i.e. proportional deformation in an object.
Here, we have to find the work done.
$ Work Done = \dfrac{1}{2} \times Stress \times Strain \times Volume $
We know that,
$ \because Stress = Y \times Strain $
Now substituting the formula for stress in the work done we get,
$ \Rightarrow Workdone = \dfrac{1}{2} \times Y \times {\left( {Strain} \right)^2} \times Volume $
Here, we know that
$ Strain = \dfrac{x}{L} $
$ Volume = AL $
were,
L is the length of wire
A is the cross-sectional area
Y is the Young's modulus
x is the amount of wire stretched
$ \Rightarrow Workdone = \dfrac{1}{2} \times Y \times {\left( {\dfrac{x}{L}} \right)^2} \times AL $
$ \Rightarrow W = \dfrac{{Y{x^2}A}}{{2L}} $
Hence, the correct option is (B).
Note :
Students need to first write the given information given in the question and use the suitable formula next define the physical quantities being discussed properly and then use the understanding and mathematical expression of these quantities such as stress, strain and Young’s modulus to solve the question. Additionally, students try to skip steps when solving these types of questions that can result in silly mistakes. So avoid skipping steps. The Young’s modulus is named after Thomas Young who was a British scientist.
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