If speed \((V)\), acceleration \((A)\) and force \((F)\) are considered as fundamental units, the dimension of Young’s modulus will be

\(\begin{array}{l}
\frac{F}{A} = y \cdot \frac{{\Delta \ell }}{\ell }\,\,;\,\,\left[ Y \right] = \frac{F}{A}\\
Now\,from\,\dim ension\\
\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,F = \frac{{ML}}{{{T^2}}}\,\,;\,\,L = \frac{F}{M} \cdot {T^2}\\
\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,{L^2} = \frac{{{F^2}}}{{{M^2}}}{\left( {\frac{V}{A}} \right)^4}\,\,\,\,\,\,T = \frac{V}{A}\\
\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,{L^2} = \frac{{{F^2}}}{{{M^2}{A^2}}}\frac{{{V^4}}}{{{A^2}}}\,\,\,\,\,\,F = MA\\
\,\,\,\,\,\,\,\,\,\,\,\,\,\,{L^2} = \frac{{{V^4}}}{{{A^2}}}\\
\left[ Y \right] = \frac{{\left[ F \right]}}{{\left[ A \right]}} = {F^1}{V^{ - 4}}{A^2}
\end{array}\)