Time (T), velocity (3) 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

Options

A. `[M]=[T^(-1)C^(-2)h]`
B. `[M]=[T^(-1)C^(2)h]`
C. `[M]=[T^(-1)C^(-2)h^(-1)]`
D. `[M]=[TC^(-2)h]`

Answer

A

Explanation

Let us express the dimensions of the mass in terms of the time T, velocity C and angular momentum h
`[M^(1)l^(0)T^(0)]= [T^(x)C^(y)h^(z)]`
`[M^(1)L^(0)T^(0)]= [T^(x)][(LT^(-1))^(y)][(M^(1)L^(2)T^(-1))^(z)]`
`[M^(1)L^(0)T^(0)]= [M^(z)L^(y+2z)T^(x-y-z)]`
`z= 1" "".........."(i)`
`y+2z=0 " ""........."(ii)`
`x-y-z=0 " ""......."(iii)`
Solving equations (i), (ii), and (iii)
`x= -1, y= -2, z= 1`
Hence, the correct dimensional formula is given by :
`[M]=[T^(-1)C^(-2)h^(1)]`