Question

The unit of Young's Modulus is
(A) $N{m^{ - 1}}$
(B) $Nm$
(C) $N{m^{ - 2}}$
(D) $N{m^2}$

Answer 1

Hint : In order to solve this question we need to understand that 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.
${\text{The formula for Young's modulus is Y = }}\dfrac{{{\sigma _n}}}{{{\varepsilon _l}}}$
${\text{were, }}{\sigma _n}{\text{ is the normal stress and }}{\varepsilon _l}{\text{ is the longitudinal strain}}{\text{.}}$

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
${\text{Y = }}\dfrac{{{\sigma _n}}}{{{\varepsilon _l}}}$
${\text{were, }}{\sigma _n}{\text{ is the normal stress and }}{\varepsilon _l}{\text{ is the longitudinal strain}}{\text{.}}$
${\text{But, }}{\sigma _n} = \dfrac{F}{a},{\text{ where F is the stretching force}}$
${\text{and }}a{\text{ is the cross section}}{\text{.}}$
${\varepsilon _l} = \dfrac{l}{L},{\text{ where }}l{\text{ is the change in length }}$
${\text{and L is the original length}}{\text{.}}$
$\Rightarrow {\text{Y = }}\dfrac{{\dfrac{F}{a}}}{{\dfrac{l}{L}}}$
$\Rightarrow Y = \dfrac{{FL}}{{al}}$
Therefore, the units of Young’s modulus is:
$\therefore \dfrac{{N.m}}{{{m^3}}} = \dfrac{N}{{{m^2}}}$
Hence, the correct option is (C).

Note :
It should be remembered that the dimensions of Young’s modulus can also be given by the 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.