Question

Time $\left(T\right)$ , velocity $\left(C\right)$ and angular momentum $\left(h\right)$ are chosen as fundamental quantities instead of mass, length and time. In terms of these, the dimensions of mass would be:

• A. $\left[M\right]=\left[T^{- 1} C^{- 2} h\right]$
• B. $\left[\right.M\left]\right.=\left[\right.T^{- 1}C^{2}h\left]\right.$
• C. $\left[M\right]=\left[T^{- 1} C^{- 2} h^{- 1}\right]$
• D. $\left[\right.M\left]\right.=\left[\right.TC^{- 2}h\left]\right.$

Answer 1

Answer: (A)

$M \propto T^{x}C^{y}h^{z}$
$M^{1}L^{0}T^{0}=\left(T^{1}\right)^{x}\left(L^{1} T^{- 1}\right)^{y}\left(M^{1} L^{2} T^{- 1}\right)^{z}$
$\left[\right.M^{1} \, L^{0}T^{0}\left]\right.=[M^{z} \, L^{y + 2 z}T^{x - y - z}\left]\right.$ ]
$\Longrightarrow z=1$
$y+2z=0 \, ,x-y-z=0$
solving
$y=-2$
and $x=-1$
$M\Rightarrow \left[ T^{- 1} \, C^{- 2} \, h^{1}\right]$