Physics - Unit & Dimension Question with Solution | TestHub

Here, we have $\nu =r^a{\rho }^b{s}^c$

Equating the dimensions of each quantity on both sides

$$\begin{gathered} \left[T^{-1}\right]={\left[L^1\right]}^a{\left[M^1{L}^{-3}\right]}^b{\left[\frac{ML{T}^{-2}}{L}\right]}^c \\ \Rightarrow T^{-1}=M^{b+c}{L}^{a-3b}{T}^{-2c} \end{gathered}$$

Comparing the dimensions, we get

$c=\frac{1}{2}, b=-\frac{1}{2}, a-3b=0$

$\Rightarrow a+\frac{3}{2}=0\Rightarrow a=-\frac{3}{2}$

Thus, the value of $a$, $b$ and $c$ are $\left(-\frac{3}{2},-\frac{1}{2}, \frac{1}{2}\right)$.