Question

If force , velocity and area is considered as a fundamental physical quantities then find the dimensional formula of Young modulus of elasticity :

Answer 1

Given :-

◉ $\quad$Force, velocity, and area is considered as a fundamental physical quantities.

To Find :-

◉ $\quad$ Dimensional formula of Young modulus of elasticity.

Solution :-

We know, Young modulus of elasticity has the dimensional formula of Pressure which is M¹L⁻¹T⁻²

Its S.I Unit is pascal

Now, It is given that Force, Velocity, and Area is considered as fundamental physical quantities.

It can be solved in many ways but I would be using the simpler one.

Let the dimensional formula of Young modulus be equal to the dimensional formula in terms of Force (F) , Velocity(V) and Area(A) and the powers raised to each quantity be x, y and z respectively.

We have,

• Force, F = MLT⁻²
• Velocity, V = LT⁻¹
• Area, A = L²

According to our assumption, we have

$\Rightarrow \sf \quad \big[M^{1}L^{-1}T^{-2} \big] = \big[ F^{x} V^{y} A^{z} \big]$

Substituting dimensional formula of F , V and A in the given equation,

$\Rightarrow \sf \quad \big[M^{1}L^{-1}T^{-2} \big] =\big[MLT^{-2}\big]^{x} \big[LT^{-1}\big]^{y} \big[L^{2}\big]^{z} \\ \\ \Rightarrow \sf \quad \big[M^{1}L^{-1}T^{-2} \big] = \big[M^{x}L^{x}T^{-2x}\big]\big[L^{y}T^{-y}\big]\big[L^{2z}\big] \\ \\ \Rightarrow \sf \quad M^{1}L^{-1}T^{-2} =M^{x}L^{(x + y + 2z)}T^{( - 2x - y)} \qquad \big[ \because {a}^{m} \times {a}^{n} = {a}^{(m + n)} \big]$

On Comparing the powers of each physical quantities, we have

Comparing M

⇒ $\quad$1 = x ⇒ x = 1 $\quad$$\quad$ ...(1)

Comparing L

⇒ $\quad$ -1 = x + y + 2z

⇒$\quad$ -1 = 1 + y + 2z $\quad$$\quad$ [ from (1) ]

⇒ $\quad$ 2z + y = -2 $\quad$$\quad$ ...(2)

Comparing T

⇒ $\quad$ -2 = -2x - y

⇒$\quad$ -2 = -2 - y

⇒$\quad$ y = 0 $\quad$$\quad$ ...(3)

Putting y = 0 in (2)

⇒$\quad$ 2z + 0 = -2

⇒$\quad$ 2z = -2

⇒$\quad$ z = -1

Hence, Young modulus of elasticity would have dimension $\displaystyle \big[ F^{1} V^{0} A^{-1} \big]$