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

Question

Time (T) , velocity (C) and angular momentum (h) are chosen as fundamental quantities instead of mass, length and time. In terms of these, the dimensions of mass would be: (A) [M]=[T- 1 C- 2 h] (B) [.M].=[.T- 1C2h]. (C) [M]=[T- 1 C- 2 h- 1] (D) [.M].=[.TC- 2h].

Tardigrade Admin · Accepted Answer

M propto TxCyhz M1L0T0=(T1)x(L1 T- 1)y(M1 L2 T- 1)z [.M1 L0T0].=[Mz Ly + 2 zTx - y - z]. ] Longrightarrow z=1 y+2z=0 ,x-y-z=0 solving y=-2 and x=-1 M⇒ [ T- 1 C- 2 h1]

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

$[M] = [T^{- 1} C^{- 2} h]$

$[M] = [T^{- 1} C^{2} h]$

$[M] = [T^{- 1} C^{- 2} h^{- 1}]$

$[M] = [T C^{- 2} h]$

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

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

[M]=[T−1C−2h]

[M]=[T−1C2h]

[M]=[T−1C−2h−1]

[M]=[TC−2h]