Explanation

Let $\ \ \ \ \ \ \ \ \ \ \ \ v \propto \sigma^{a}\rho^{b}\lambda^{c}$ Equating dimensions on both sides. $\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \left\lbrack M^{0}LT^{- 1} \right\rbrack \propto \left\lbrack MT^{- 2} \right\rbrack^{a}\left\lbrack ML^{- 3} \right\rbrack^{b}\lbrack L\rbrack^{c}$ $\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \propto \lbrack M\rbrack^{a + b}\lbrack L\rbrack^{- 3b + c}\lbrack T\rbrack^{- 2a}$ Equating the powers of M, L, T on both sides, we get $\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ a + b = 0$ $\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ - 3b + c = 1$ $\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ - 2a = - 1\$ Solving, we get $\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ a = \frac{1}{2},b = - \frac{1}{2},\ c = - \frac{1}{2}$ $\therefore\ \ \ \ \ \ \ \ \ \ \ \ \ v \propto \sigma^{1/2}\rho^{- 1/2}\lambda^{- 1/2}$ $\therefore\ \ \ \ \ \ \ \ \ \ \ v^{2} \propto \frac{\sigma}{\rho\lambda}$