Question

If velocity $(v),$ acceleration $(a)$ and force $(F)$ are taken as fundamental quantities, the dimensions of Young's modulus $(Y)$ would be

• A. $\left[F a^{2} v^{-2}\right]$
• B. $\left[F a^{2} v^{-3}\right]$
• C. $\left[F a^{2} v^{-4}\right]$
• D. $\left[F a^{2} v^{-5}\right]$

Answer 1

Answer: (C)

Let $Y=k v^{x} a^{y} F^{2}$ where $k$ is a dimensionless constant.
$\therefore \left[ ML ^{-1} T ^{-2}\right]=\left[ LT ^{-1}\right]^{x}\left[ LT ^{-2}\right]^{y}\left[ MLT ^{-2}\right]^{z}=\left[ M ^{z} L^{x+y+z} T ^{-x-2 y-2 z}\right]$
Equating the powers of $M , L$ and $T$, we get
$z=1, x+y+z=-1 ;-x-2 y-2 z=-2$
Solving, we get, $x=-4, y=2, z=1$
$\therefore \quad Y=v^{-4} a^{2} F^{1}$
or $[Y]=\left[F a^{2} v^{-4}\right]$