Question

M.K.S. unit of Young’s modulus is
A. \[M/m\]
B. \[M/{m^2}\]
C. \[M \times m\]
D. \[M \times {m^2}\]

Answer 1

Hint:Young’s modulus is given by the division of normal stress by longitudinal strain. It is the property that measures the tensile stiffness of a material. The normal stress depends upon force and area and the longitudinal strain deals with the original length and the length that increases or decreases due to elasticity.

Formula Used:
The formula for Young’s modulus is \[Y = \dfrac{{{\sigma _n}}}{{{\varepsilon _l}}}\]
where, \[{\sigma _n}\] is the normal stress and\[{\varepsilon _l}\]is the longitudinal strain.

Complete step by step answer:
The ratio of the normal stress to the longitudinal strain is called Young's modulus. It is generally denoted as \[Y\]. And, the formula of Young’s modulus is given by
\[Y = \dfrac{{{\sigma _n}}}{{{\varepsilon _l}}}\]
where, \[{\sigma _n}\] is the normal stress and\[{\varepsilon _l}\] is the longitudinal strain.

But, \[{\sigma _n} = \dfrac{F}{a}\], where \[F\]is the stretching force and \[a\] is the area of the cross section. Also, \[{\varepsilon _l} = \dfrac{l}{L}\], where \[l\] is the change in length and \[L\] is the original length.
\[Y = \dfrac{{\dfrac{F}{a}}}{{\dfrac{l}{L}}}\\
\Rightarrow Y= \dfrac{{FL}}{{al}}\]
Therefore, the units of Young’s modulus is:
\[\therefore\dfrac{{N.m}}{{{m^3}}} = \dfrac{N}{{{m^2}}}\]

Hence, option B is the correct answer.

Note:The dimensions of Young’s modulus can also be given by modulus of elasticity. The modulus of elasticity is defined as the ratio of stress to strain. Since, strain is a unit less quantity, therefore, the unit of modulus of elasticity is the same as that of stress. Hence, the modulus of elasticity is also expressed as Newton per meter square.