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-1 C2 h ] (C) [ M ]=[ T-1 C-2 h-1 ] (D) [ M ]=[ T C-2 h ]

Tardigrade Admin · Accepted Answer

M propto Tx vy hz M' L0T0 = (T')x (L1T-1)y (M1L2T-1)z M1L0T0 = Mz Ly+2z + Tx-y-z z = 1 x - y - z = 0 y + 2z = 0 x + 2 - 1 = 0 y = - 2 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 :

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

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

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

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

$M \propto T^{x} v^{y} h^{z}$
$M^{'} L^{0} T^{0} = (T^{'})^{x} (L^{1} T^{- 1})^{y} (M^{1} L^{2} T^{- 1})^{z}$
$M^{1} L^{0} T^{0} = M^{z} L^{y} + 2^{z} + T^{x - y - z}$
$z = 1 x - y - z = 0$
$y + 2 z = 0 x + 2 - 1 = 0$
$y = - 2 x = - 1$
$M \Rightarrow T^{- 1} C^{- 2} h^{1}$

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]

M∝Txvyhz M′L0T0=(T′)x(L1T−1)y(M1L2T−1)z M1L0T0=MzLy+2z+Tx−y−z z=1x−y−z=0 y+2z=0x+2−1=0 y=−2x=−1 M⇒T−1C−2h1

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

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

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

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

$M \propto T^{x} v^{y} h^{z}$
$M^{'} L^{0} T^{0} = (T^{'})^{x} (L^{1} T^{- 1})^{y} (M^{1} L^{2} T^{- 1})^{z}$
$M^{1} L^{0} T^{0} = M^{z} L^{y} + 2^{z} + T^{x - y - z}$
$z = 1 x - y - z = 0$
$y + 2 z = 0 x + 2 - 1 = 0$
$y = - 2 x = - 1$
$M \Rightarrow T^{- 1} C^{- 2} h^{1}$