Question

The dimensional formula for Young’s modulus is:
$\text{A}\text{. }\left[ M{{L}^{-1}}{{T}^{-2}} \right]$
$\text{B}\text{. }\left[ {{M}^{0}}L{{T}^{-2}} \right]$
$\text{C}\text{. }\left[ ML{{T}^{-2}} \right]$
$\text{D}\text{. }\left[ M{{L}^{2}}{{T}^{-2}} \right]$

Answer 1

Hint: Young’s modulus is the ratio of the longitudinal stress to the longitudinal strain produced in a body. The formula of stress and strain are $\sigma =\dfrac{F}{A}$ and $\mu =\dfrac{\Delta l}{l}$ respectively. Find the ratio of the dimensional formula of stress to the dimensional formula of strain to find the dimensional formula of Young’s modulus.

Complete step by step answer:
To compress or stretch a body when a force is applied on a body, it applies an equal force opposite to the direction of the applied force. The force applied per unit area is defined as stress produced in the body. Suppose a force F is applied on the surface of the body of area A, then the stress in the body will be equal to the ratio of force to the area of the surface i.e.$\sigma =\dfrac{F}{A}$ …… (i).
If the applied force is perpendicular to the surface of the body, then the stress is called longitudinal stress.
When the body is compressed or stretched, we say that there is a strain produced in the body. The ratio of the change in the dimension of the body to the original dimension is defined as strain. When the change occurs in the length of the body, the strain is called longitudinal strain. If the change in length is $\Delta $l and the original length is l, then the strain is equal to $\mu =\dfrac{\Delta l}{l}$ ….. (ii).
Therefore, when there is a longitudinal stress produced in the body, there is longitudinal strain in the body.
The ratio of the longitudinal stress to the longitudinal strain is defined as Young’s modulus (Y).
Therefore, $Y=\dfrac{\sigma }{\mu }$ ….. (iii).
If you observe equation (ii), strain is the ratio of the same dimensions. Hence, strain is a dimensionless quantity.
Therefore, Young’s modulus has the same dimensions that stress has.
Therefore, the dimensional formula of Young’s modulus is $\left[ Y \right]=\left[ \sigma \right]=\left[ \dfrac{F}{A} \right]=\dfrac{\left[ F \right]}{\left[ A \right]}$.
The dimensional formula of force is $\left[ ML{{T}^{-2}} \right]$.
The dimensional formula of area is $\left[ {{L}^{2}} \right]$.
Therefore, $\left[ Y \right]=\dfrac{\left[ F \right]}{\left[ A \right]}=\dfrac{\left[ ML{{T}^{-2}} \right]}{\left[ {{L}^{2}} \right]}=\left[ M{{L}^{-1}}{{T}^{-2}} \right]$.
Therefore, the dimensional formula of Young’s modulus is $\left[ M{{L}^{-1}}{{T}^{-2}} \right]$.
Hence, the correct option is A.

Note: Do not confuse between Young’s modulus and modulus of rigidity.
When the applied force is parallel to the surface of the body, the stress produced in the body is called shear stress and the strain caused due to this stress is called shear strain.
Modulus of rigidity is the ratio of shear stress to the shear strain.
Therefore, Young’s modulus and modulus of rigidity are conceptually different but both have the same dimensions.