Question

If speed $V$, area $A$ and force $F$ are chosen as fundamental units, then the dimension of Young's modulus will be :

• A. $F{A}^{-1}{V}^0$
• B. $F{A}^2{V}^{-1}$
• C. $F{A}^2{V}^{-3}$
• D. $F{A}^2{V}^{-2}$

Answer 1

Answer: (A)

$Y=F^x{A}^y{V}^z$
$M^1{L}^{-1}{T}^{-2}={\left[ML{T}^{-2}\right]}^x{\left[L^2\right]}^y{\left[L{T}^{-1}\right]}^z$
$M^1{L}^1{T}^{-2}=[M{]}^x[L{]}^{x+2y+z}[T{]}^{-2x-z}$
comparing power of ML and T
$x=1\dots (1)$
$x+2y+z=-1\dots ..(2)$
$-2x-z=-2\dots (3)$
after solving
$x=1$
$y=-1$
$z=0$
$Y=F{A}^{-1}{V}^0$