If speed $(V)$, acceleration $(A)$ and force $(F)$ are considered as fundamental units, the dimension of Young’s modulus will be
JEE MAIN 2019, Diffcult
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$\frac{F}{A}=y \cdot \frac{\Delta \ell}{\ell} ;[Y]=\frac{F}{A}$
Now from dimension
$F =\frac{M L}{T^2} ; L=\frac{F}{M} \cdot T^2$
$L^2 =\frac{F^2}{M^2}\left(\frac{V}{A}\right)^4 T=\frac{V}{A}$
$L^2 =\frac{F^2}{M^2 A^2} \frac{V^4}{A^2} F=M A$
$L^2 =\frac{V^4}{A^2}$
${[Y]=\frac{[F]}{[A]} } =F^1 V^{-4} A^2$
Now from dimension
$F =\frac{M L}{T^2} ; L=\frac{F}{M} \cdot T^2$
$L^2 =\frac{F^2}{M^2}\left(\frac{V}{A}\right)^4 T=\frac{V}{A}$
$L^2 =\frac{F^2}{M^2 A^2} \frac{V^4}{A^2} F=M A$
$L^2 =\frac{V^4}{A^2}$
${[Y]=\frac{[F]}{[A]} } =F^1 V^{-4} A^2$
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