
Dimensional formula of Young’s Modulus is:
A. \[[{{M}^{1}}{{L}^{-1}}{{T}^{-2}}]\]
B. \[[{{M}^{1}}{{L}^{1}}{{T}^{1}}]\]
C. \[[{{M}^{-1}}{{L}^{2}}{{T}^{-1}}]\]
D. \[[{{M}^{-1}}{{L}^{3}}{{T}^{-2}}]\]
Answer
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Hint: When a strain is small, the ratio of the longitudinal stress to the corresponding longitudinal strain is the ‘Young’s Modulus’ of the material of the body.
Complete step by step solution:
Young Modulus Y of a material is given by:
\[Y=\dfrac{\text{longitudinal stress}}{\text{longitudinal strain}}\]
Now,
Longitudinal stress \[=\dfrac{\text{Force}}{\text{area}}\] and longitudinal strain \[=\dfrac{\text{increase in length}}{\text{original length}}\]
Thus, \[Y=\dfrac{\text{Force/area}}{\text{increase in length/original length}}\]
The S.I. unit of force is newton (N), area is metre square (\[{{\text{m}}^{2}}\]), and that of length is metre (m)
So, the S.I. unit of Young modulus is:
\[Y=\dfrac{\text{N/}{{\text{m}}^{2}}}{\text{m/m}}=\text{N/}{{\text{m}}^{2}}\]
Now,
\[\dfrac{\text{newton}}{\text{metr}{{\text{e}}^{\text{2}}}}=\dfrac{\text{kg}\times \text{metre}\times \text{secon}{{\text{d}}^{-2}}}{\text{metr}{{\text{e}}^{\text{2}}}}=\text{kg metr}{{\text{e}}^{-1}}\text{secon}{{\text{d}}^{-2}}\]
So, the dimensional formula of Young Modulus is \[[{{M}^{1}}{{L}^{-1}}{{T}^{-2}}]\]
Hence, option A is the correct answer.
Additional information:
Young Modulus is determined only for solids and is the characteristic of the material of a solid.
By Hooke’s law, if strain is small, then the stress produced in a body is proportional to the strain. The ratio of stress to strain is a constant for the material of the given body and is called the ‘modulus of elasticity’ E.
Thus, E = stress/strain
The value of the modulus of elasticity of a material depends upon the type of stress and strain produced. For longitudinal strain, the modulus of elasticity is called ‘Young’s modulus’. If strain is in volume then it is ‘Bulk modulus’ and if strain is in shape then it is called ‘modulus of rigidity’.
The unit of measurement and dimension of Young Modulus, Bulk Modulus, and Modulus of Rigidity is same, that is, \[\text{N/}{{\text{m}}^{2}}\] and \[[{{M}^{1}}{{L}^{-1}}{{T}^{-2}}]\] respectively.
Note: Young Modulus is a modulus of elasticity and so has the same unit of measurement and dimension as modulus of elasticity.
Complete step by step solution:
Young Modulus Y of a material is given by:
\[Y=\dfrac{\text{longitudinal stress}}{\text{longitudinal strain}}\]
Now,
Longitudinal stress \[=\dfrac{\text{Force}}{\text{area}}\] and longitudinal strain \[=\dfrac{\text{increase in length}}{\text{original length}}\]
Thus, \[Y=\dfrac{\text{Force/area}}{\text{increase in length/original length}}\]
The S.I. unit of force is newton (N), area is metre square (\[{{\text{m}}^{2}}\]), and that of length is metre (m)
So, the S.I. unit of Young modulus is:
\[Y=\dfrac{\text{N/}{{\text{m}}^{2}}}{\text{m/m}}=\text{N/}{{\text{m}}^{2}}\]
Now,
\[\dfrac{\text{newton}}{\text{metr}{{\text{e}}^{\text{2}}}}=\dfrac{\text{kg}\times \text{metre}\times \text{secon}{{\text{d}}^{-2}}}{\text{metr}{{\text{e}}^{\text{2}}}}=\text{kg metr}{{\text{e}}^{-1}}\text{secon}{{\text{d}}^{-2}}\]
So, the dimensional formula of Young Modulus is \[[{{M}^{1}}{{L}^{-1}}{{T}^{-2}}]\]
Hence, option A is the correct answer.
Additional information:
Young Modulus is determined only for solids and is the characteristic of the material of a solid.
By Hooke’s law, if strain is small, then the stress produced in a body is proportional to the strain. The ratio of stress to strain is a constant for the material of the given body and is called the ‘modulus of elasticity’ E.
Thus, E = stress/strain
The value of the modulus of elasticity of a material depends upon the type of stress and strain produced. For longitudinal strain, the modulus of elasticity is called ‘Young’s modulus’. If strain is in volume then it is ‘Bulk modulus’ and if strain is in shape then it is called ‘modulus of rigidity’.
The unit of measurement and dimension of Young Modulus, Bulk Modulus, and Modulus of Rigidity is same, that is, \[\text{N/}{{\text{m}}^{2}}\] and \[[{{M}^{1}}{{L}^{-1}}{{T}^{-2}}]\] respectively.
Note: Young Modulus is a modulus of elasticity and so has the same unit of measurement and dimension as modulus of elasticity.
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