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 $F{A}^2{V}^{-4}$

Young's modulus $Y=\left[V^a\right]\left[A{]}^b\right[F{]}^c$
$\Rightarrow M{L}^{-1}{T}^{-2}-\left[L{T}^{-1}{]}^a\right[L{T}^{-2}{]}^b[ML{T}^{-2}{]}^c$
$\Rightarrow M{L}^{-1}{T}^{-2}=M^c{L}^{a+b+c}{T}^{-a-2b-2c}$
On comparing both sides,
c = 1, a + b + c = -1, a + 2b + 2c = 2
a + b = -2 and a + 2b = 0
$\therefore$ b = 2 and a = -4
$\therefore Y=V^{-4}{A}^{2}F=F{A}^2{V}^{-4}$