If the velocity $v$, acceleration $a$, and force $F$ are chosen as fundamental quantities instead of mass, length, and time, what is the dimensional formula of Young's Modulus $Y$?

Solution

1. Young's Modulus $Y$ has dimensions $[M{L}^{-1}{T}^{-2}]$.
2. Let $$\begin{array}{rl}Y& =F^x{v}^y{a}^z\\ & =[ML{T}^{-2}{]}^x[L{T}^{-1}{]}^y[L{T}^{-2}{]}^z\end{array}$$.
3. Equating powers: $x=1$, $x+y+z=-1$, and $-2x-y-2z=-2$.
4. Solving the system: $$\begin{array}{rl}1+y+z& =-1⟹y+z\\ & =-2\end{array}$$ and $$\begin{array}{rl}-2-y-2z& =-2⟹y+2z\\ & =0\end{array}$$.
5. Subtracting gives $z=2$ and $y=-4$. Hence the answer is (B).