Explanation

According to question $F = \propto m^{a}v^{b}r^{c}$ $F = km^{a}v^{b}r^{c}$ $k,$ being a dimensionless constant. From homogeneity of dimensions, LHS=RHS $\left\lbrack MLT^{- 2} \right\rbrack = \lbrack M\rbrack^{a}\left\lbrack LT^{- 1} \right\rbrack^{b}\lbrack L\rbrack^{c}$ Or $\left\lbrack MLT^{- 2} \right\rbrack = \lbrack M^{a}L^{b + c}T^{- b}\rbrack$ Comparing the powers, we obtain $a = 1$ $b + c = 1$ $- b = - 2 \Rightarrow b = 2$ $\therefore\ \ 2 + c = 1$ $\Rightarrow \ c = - 1$ Therefore, $F = kmv^{2}r^{- 1} = \frac{kmv^{2}}{r}$ The experimental value of $k$ is found to be 1 here $\therefore\ F = \frac{mv^{2}}{r}$