Question

If the velocity $\left(V\right)$, Acceleration $\left(A\right)$ and force $\left(F\right)$ are taken as fundamental quantities instead of mass $\left(M\right)$, length $\left(L\right)$ and Time $\left(T\right)$. The dimensions of young's modulus $\left(Y\right)$ would be :

• A. $F{A}^2{V}^{-4}$
• B. $F{A}^2{V}^{-5}$
• C. $F{A}^2{V}^{-3}$
• D. $F{A}^2{V}^{-2}$

Answer 1

The correct option is A $F{A}^2{V}^{-4}$

Since $\left[Y\right]=\left[M{L}^{-1}{T}^{-2}\right]$

Here,

$\left[Y\right]\alpha V^a{A}^b{F}^c$

$\therefore \left[M^1{L}^{-1}{T}^{-2}\right]=\left[L{T}^{-1}{]}^a\right[L{T}^{-2}{]}^b[ML{T}^{-2}{]}^c$

Hence on Comparing,

$a=-4$

$b=2$

$c=1$

Hence, $Y=F{A}^2{V}^{-4}$