Question

$M{L}^{-1}{T}^{-2}$ is the dimensional formula of?

• A. Coefficient of friction
• B. Energy
• C. Modulus of Elasticity
• D. Force

Answer 1

The correct option is C

Modulus of Elasticity

Dimensional formula

The dimensional formula is the expression of the physical quantity in terms of the fundamental units.

The explanation for the correct answer:

$$\begin{gathered} ModulusofElasticity=\frac{ShearStress}{ShearStrain} \\ ModulusofElasticity=\frac{F/A}{ShearStrain} \end{gathered}$$

Dimensional formula of $shearstress(F/A)=\left[M^1{L}^{-1}{T}^{-2}\right]$ while the dimensional formula of $shearstrain=\left[M^0{L}^0{T}^0\right]$

Now, $$\begin{gathered} ModulusofElasticity=\frac{\left[M^1{L}^{-1}{T}^{-2}\right]}{\left[M^0{L}^0{T}^0\right]} \\ ModulusofElasticity=\left[M^1{L}^{-1}{T}^{-2}\right] \end{gathered}$$

Thus, $M{L}^{-1}{T}^{-2}$ is the dimensional formula of Modulus of Elasticity.

The explanation for the incorrect answer:

For option A

1. $\mu =F/N$
2. here, $\mu$ is the coefficient of friction, $F$ is the frictional force and $N$ is the normal force.
3. Thus, the dimensional formula of coefficient of friction is $M^0{L}^0{T}^0$.

For option B

1. The dimensional formula of Energy is $M^1{L}^2{T}^{-2}$.

For option D

1. $F=m.a$
2. Here, F is the force, m is the mass of the object and a is the acceleration.
3. Thus, the dimensional formula of force is $M^1{L}^1{T}^{-2}$.

Hence, Option C is correct.