Question

If the velocity (V), acceleration (A), and force (F) are taken as fundamental quantities instead of mass (M), length (L) and time (T), the dimensions of Young's modulus (Y) would be

• A. $F{A}^2{V}^{-4}$
• B. $F{A}^2{V}^{-5}$
• C. $F{A}^2{V}^{-3}$
• D. $F{A}^2{V}^{-2}$

Answer 1

The correct option is A $F{A}^2{V}^{-4}$

$\text{Step 1: Dimension of Young's Modulus Y, Force F, Area A and Velocity V}$

We know that,

Velocity $V=\frac{Distance}{time}=L{T}^{-1}$

Acceleration $A=\frac{Changeinvelocity}{time}=L{T}^{-2}$

Force $F=mass\times acceleration=ML{T}^{-2}$

Youngs Modulus $Y=\frac{F/A}{\Delta l/l}=\frac{\frac{ML{T}^{-2}}{L^2}}{\frac{L}{L}}=M{L}^{-1}{T}^{-2}$

$\text{Step 2: Finding Y in terms of F, A and V}$

If Force F, Area A and velocity V are taken as fundamental quantities, so Youngs Modulus Y can be expressed in terms of above as follows:

Let,

$Y=F^a{A}^b{V}^c$ $....\left(1\right)$

Putting their dimensions in above equation

$\Rightarrow M{L}^{-1}{T}^{-2}=\left[ML{T}^{-2}{]}^a\right[L{T}^{-2}{]}^b[L{T}^{-1}{]}^c$

$\Rightarrow M{L}^{-1}{T}^{-2}=M^a{L}^{(a+b+c)}{T}^{(-2a-2b-c)}$

Comparing powers of M, L & T in LHS and RHS in above equation, we get:

$a=1$,

$-1=a+b+c$ $\Rightarrow -1=1+b+c$ $\Rightarrow b+c=-2$ $....\left(2\right)$

$-2=-2a-2b-c$ $\Rightarrow -2=-2-2b-c$ $\Rightarrow c=-2b$

So, Equation $\left(2\right)$ $\Rightarrow b-2b=-2\Rightarrow b=2$

$\Rightarrow c=-2b=-4$

Therefore, putting values of a, b and c in Equation $\left(1\right)$, we get:

$Y=F{A}^2{V}^{-4}$

Hence Option $A$ is correct