Question

If velocity (v), force (F) and acceleration (a) are taken as fundamental quantities instead of mass (M), length (L) and time (T), Find the dimensions of Young's modulus of elasticity.

Answer 1

$Young\text{'}sModulus\left(Y\right)=\frac{Stress}{Strain}=\frac{{\displaystyle \raisebox{1ex}{$F$}\!\left/ \!\raisebox{-1ex}{$A$}\right.}}{{\displaystyle \raisebox{1ex}{$l$}\!\left/ \!\raisebox{-1ex}{$L$}\right.}}$
F is a fundamental quantity. Dimension of length in new fundamental units,
$L=\frac{v^2}{a}$
Therefore, A has a dimension,
$A=\frac{v^4}{a^2}$
Strain is dimensionless.
Dimension of Young's Modulus,
$Y=\frac{F}{{\displaystyle \raisebox{1ex}{$v^4$}\!\left/ \!\raisebox{-1ex}{$a^2$}\right.}}=v^{-4}F{a}^2$