Question

If Planck's constant $\left(h\right)$, surface tension $\left(s\right)$ and moment of inertia $\left(I\right)$ are taken as fundamental quantities, then dimensions of linear momentum $\left(p\right)$ will be given as:

• A. $h{s}^2{I}^{1/2}$
• B. $h^{1/2}{s}^{1/2}I$
• C. $h^0{s}^{1/2}{I}^{1/2}$
• D. None of these

Answer 1

The correct option is C $h^0{s}^{1/2}{I}^{1/2}$

$p\equiv h^x{T}^y{I}^z$
$ML{T}^{-1}=\left(M{L}^2{T}^{-1}{)}^x\right(M{T}^{-2}{)}^y(M{I}^2{)}^z$
$ML{T}^{-1}=M^{x+y+z}\cdot L^{2x+2z}\times T^{-x-2y}$
$\begin{array}{c}x+y+z=1\\ x+z=\frac{1}{2}\end{array}\}\Rightarrow y=\frac{1}{2}$
$x+2y=1\Rightarrow x+1=1\Rightarrow x=0$
$P\equiv h^0{s}^{1/2}{I}^{1/2}$