Dimensional formula for, h=[ML2T−1]
Dimensional formula for, G=[M−1L3T−2]
Dimensional formula for, C=[LT−1]
Let mass be represented as: m=haGbCc
∴ [M]= [ML2T−1]a [M−1L3T−2]b [LT−1]c
[M]= [Ma−bL2a+3b+cT−(a+2b+c)]
On comparing we get:
a−b=1 .........(1)
2a+3b+c=0 .........(2)
a+2b+c=0 .........(3)
Subtracting (3) from (2) we get:
a+b=0
⟹b=−a ...........(4)
Solving (1) and (4):
a=12 b=−12
From (3),
12+2×−12+c=0
⟹c=12
⟹ [m]=[h1/2G−1/2C1/2]