Correct Answer - C
Let `dproprho^(x)S^(y)f^(z)` or `d=krho^(x)S^(y)f^(z)`
where k is a dimensionless constant and x,y and z are the exponents.
Writings dimenions on both sides, we get
`[M^(0)LT^(0)]=[ML^(-3)T^(0)]^(x)[ML^(0)T^(-3)]^(y)[M^(0)L^(0)T^(-1)]^(z)`
`[M^(0)LT^(0)]=M^(x+y)L^(-3x)T^(-3y-z)]`
applying the principle of homogeneity of dimensions,
we get ltBrgt `x+y=0` . . . (i)
`-3x=1` . . . .(ii)
`-3-z=0` . . . (iii)
Solving eqns. (i), (ii) and (iii), we get
`x=-(1)/(3),y=(1)/(3),z=-1,` as `dpropS^(1//3)`
`thereforen=3`