Question:
If speed $\mathrm{V}$, area $\mathrm{A}$ and force $\mathrm{F}$ are chosen as fundamental units, then the dimension of Young's modulus will be :
Correct Option: 1
Solution:
$\mathrm{Y}=\mathrm{Fx} \mathrm{A}^{\mathrm{y}} \mathrm{V} \mathrm{z}$
$\mathrm{M}^{1} \mathrm{~L}^{-1} \mathrm{~T}^{-2}=\left[\mathrm{MLT}^{-2}\right]^{x}\left[\mathrm{~L}^{2}\right]^{y}\left[\mathrm{LT}^{-1}\right]^{z}$
$\mathrm{M}^{1} \mathrm{~L}^{1} \mathrm{~T}^{-2}=[\mathrm{M}]^{x}[\mathrm{~L}]^{x+2 y+z}[\mathrm{~T}]^{-2 x-z}$
comparing power of $\mathrm{ML}$ and $\mathrm{T}$
$x=1 \ldots(1)$
$x+2 y+z=-1$ ..................(2)
$-2 x-z=-2$ ............(3)
after solving
$x=1$
$y=-1$
$z=0$
$\mathrm{Y}=\mathrm{FA}^{-1} \mathrm{~V}^{0}$