The dimensional form of velocity can be written as

$\left[v\right]=\left[{\lambda }^a{g}^b{\rho }^c\right]$

$\Rightarrow \left[M^{0}L{T}^{-1}\right]={\left[L\right]}^a{\left[L{T}^{-2}\right]}^b{\left[M{L}^{-3}\right]}^c$

$$Comparing the exponents for $M,L,T$,

$$\begin{gathered} \mathrm{For} M, \\ {M}^c=M^0 \\ \mathrm{For} L, \\ {L}^{a+b-3c}=L^1 \\ \mathrm{For} T, \\ {T}^{-2b}=T^{-1} \end{gathered}$$

So,

$$\begin{gathered} c=0 ...\left(i\right) \\ a+b-3c=1 ...\left(ii\right) \\ -2b=-1 ...\left(iii\right) \end{gathered}$$

From equations (i),(ii),(iii),

$a=\frac{1}{2}, b=\frac{1}{2},c=0$