Question

If speed V, area A, and force F are chosen as fundamental units then the dimension of Young’s modulus will be:

• A. $F{A}^2{V}^{-3}$
• B. $F{A}^2{V}^{-2}$
• C. $F{A}^{-1}{V}^0$
• D. $F{A}^2{V}^{-1}$

Answer 1

The correct option is C

$F{A}^{-1}{V}^0$

Step 1: Given data

Speed is V

Area is A

Force is F

Step 2: Formula Used

$[YoungModulus]=\left[\frac{Force}{Area}\right]$

Step 3: Solution

Denote young’s modulus as a function of force, speed, and area.

$Y=F^x{A}^{\nu }{V}^z$

Now calculate the dimension of the words in the expression

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

$\left[F\right]=ML{T}^{-2}$

$\left[A\right]=L^2$

$\left[V\right]=L{T}^{-1}$

Rewriting the equation $Y=F^x{A}^{\nu }{V}^z$ in terms of $M,L,$ and $T.$

$M^1{L}^{-1}{T}^{-2}={\left[ML{T}^{-2}\right]}^x{\left[L^2\right]}^y{\left[L{T}^{-1}\right]}^z$

$M^1{L}^{-1}{T}^{-2}={\left[M\right]}^x{\left[L\right]}^{x+2y+z}{\left[T\right]}^{-2x-z}$

Compare the powers on both sides of the equation.

$x=1$

$x+2y+z=-1$

$-2x-z=-2$

Therefore,

$x=1$

$y=-1$

$z=0$

$Y=F{A}^{-1}{V}^0$

Hence, for speed V, area A and force F are chosen as fundamental units $Y=F{A}^{-1}{V}^0$