If the velocity $v$, acceleration $a$, and force $F$ are chosen as fundamental quantities, what is the dimension of Young's modulus $Y$ in terms of $v$, $a$, and $F$?

Solution

1. The dimension of Young's modulus is $[Y]=[M][L{]}^{-1}[T{]}^{-2}$.
2. The dimensions of the fundamental quantities are $[v]=[L][T{]}^{-1}$, $[a]=[L][T{]}^{-2}$, and $[F]=[M][L][T{]}^{-2}$.
3. Expressing $[Y]$ as $[F{]}^x[v{]}^y[a{]}^z$, we have $[M][L{]}^{-1}[T{]}^{-2}=([M][L][T{]}^{-2}{)}^x([L][T{]}^{-1}{)}^y([L][T{]}^{-2}{)}^z$.
4. Solving the system of equations for exponents gives $$\begin{array}{rl}x& =1,y\\ & =-4,z\\ & =2\end{array}$$.
5. Hence the answer is (C).