Question

If the velocity $V$ acceleration $A$ and force $F$ are taken as fundamental quantities instead of mass $M$, length $L$ and time $T$, the dimensions of Young's modulus $Y$ 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,

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

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

Hence on comparing;

$a=-4$

$b=2$

$c=1$

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