Question

The dimensional formula of modulus of rigidity $(\eta )$ is

• A. $\left[M{L}^{-2}{T}^{-2}\right]$
• B. $\left[M{L}^{-3}{T}^{-2}\right]$
• C. $\left[M{L}^2{T}^{-2}\right]$
• D. $\left[M{L}^{-1}{T}^{-2}\right]$

Answer 1

The correct option is D

$\left[M{L}^{-1}{T}^{-2}\right]$

Step 1: Given Data

Modulus of rigidity is the ratio of shear stress to the shear strain. So, the modulus of rigidity, $\left(\eta \right)=\frac{stress}{strain}$

The ratio of the body's changed dimensions to its original dimensions is called strain. So, the dimension of $\eta$ will be equal to the dimension of stress.

Step 2: Formula Used

As we know,

$strain=\frac{\delta l}{l}$

Where $\delta l=$change in length and $l=$actual length. Since both of them are lengths there is no dimensional formula for strain.

As for force,

$\left(F\right)Force=\left(m\right)mass\times \left(a\right)acceleration$

Step 3: Calculation

$F=ma$, which stands for Newton's second law, states that a force acting on an object is equal to that object's mass times its acceleration (a).

$\left(F\right)Force=\left(m\right)mass\times \left(a\right)acceleration$

i.e., $F=\left[M\right]\times \left[L{T}^{-2}\right]=\left[ML{T}^{-2}\right]$.

It is understood that,

$stress=\frac{Force\left(F\right)}{Area\left(A\right)}$

So, the dimension of stress would be, $\frac{\left[ML{T}^{-2}\right]}{\left[L^2\right]}=\left[M{L}^{-1}{T}^{-2}\right]$.

Therefore the dimension of the modulus of rigidity will $\left[M{L}^{-1}{T}^{-2}\right]$ because of stress and modulus of rigidity have the same dimension.

Hence, the correct option is (D).