Question

If velocity $(v),$ acceleration $(a)$ and force $(F)$ are taken as fundamental quantities, the dimensions of Young's modulus $(Y)$ would be

• A. $\left[F{a}^2{v}^{-2}\right]$
• B. $\left[F{a}^2{v}^{-3}\right]$
• C. $\left[F{a}^2{v}^{-4}\right]$
• D. $\left[F{a}^2{v}^{-5}\right]$

Answer 1

Answer: (C)

Let $Y=k{v}^x{a}^y{F}^2$ where $k$ is a dimensionless constant.
$\therefore \left[M{L}^{-1}{T}^{-2}\right]={\left[L{T}^{-1}\right]}^x{\left[L{T}^{-2}\right]}^y{\left[ML{T}^{-2}\right]}^z=\left[M^z{L}^{x+y+z}{T}^{-x-2y-2z}\right]$
Equating the powers of $M,L$ and $T$, we get
$z=1,x+y+z=-1;-x-2y-2z=-2$
Solving, we get, $x=-4,y=2,z=1$
$\therefore Y=v^{-4}{a}^2{F}^1$
or $[Y]=\left[F{a}^2{v}^{-4}\right]$