Question

As per given $M{{L}^{-1}}{{T}^{-2}}$ is the dimensional formula of
(A). Force
(B). Coefficient of friction
(C). Modulus of elasticity
(D). Energy

Answer 1

According to the Hooke's law, without exceeding the elastic limit of the material, the stress or deformation produced is directly proportional to the force applied. We can calculate the dimensional units of stress and strain by using its definition and use them to determine the dimensional unit of modulus of elasticity.

Formulas used:
$\text{M=}\dfrac{\text{stress}}{\text{strain}}$
$\text{stress=}\dfrac{\text{applied}\,\text{force}}{\text{area}}$
$F=ma$

Complete step by step solution:
According to Hooke's law, stress is directly proportional to strain within the elastic unit of the body. Mathematically,
$\begin{aligned} & \text{stress}\propto \text{strain} \\\ & \Rightarrow \text{stress=(M)strain} \\\ \end{aligned}$
Here, $M$ is the constant of proportionality and is also known as the modulus of elasticity. Therefore, from the above equation, we get,
$\text{M=}\dfrac{\text{stress}}{\text{strain}}$ - (1)
Hence, we can define the modulus of elasticity for a material as the ratio of stress to strain.
The restoring force that is developed per unit cross-sectional area in a body when deforming force is applied on it is known as stress. Thus,
$\text{stress=}\dfrac{\text{applied}\,\text{force}}{\text{area}}$
Force is the product of mass and acceleration of a body. Therefore,
$F=ma$
Here, $F$ is the force applied
$m$ is the mass of the body
$a$ is acceleration
Therefore, from the above equation, the dimensions of force will be-
$\begin{aligned} & [F]=[M][L{{T}^{-2}}] \\\ & \Rightarrow [F]=[ML{{T}^{-2}}] \\\ \end{aligned}$
Therefore, the dimensions of stress will be-
$[S]=\dfrac{[ML{{T}^{-2}}]}{[{{L}^{2}}]}$
$\Rightarrow [S]=[M{{L}^{-1}}{{T}^{-2}}]$ - (2)
Therefore, the dimensions of stress are $[M{{L}^{-1}}{{T}^{-2}}]$.
Strain is a unitless quantity because it is the ratio of two same quantities; it does not have any dimensions.
From eq (1) and eq (2), the dimensions of modulus of elasticity will be-
$[M']=[M{{L}^{-1}}{{T}^{-2}}]$
Therefore, the dimensions of modulus are $M{{L}^{-1}}{{T}^{-2}}$. Hence, the correct option is (C).

Note: The dimensional unit of a physical quantity is an equation which represents the unit in terms of seven fundamental units. The modulus of elasticity of a material can also be stated as the slope of the stress-strain curve in the elastic deformation region. The stress- strain curve gives a relationship between stress and strain of a body when a load is applied to it and then gradually removed.