The expression for centripetal force (F) depends upon mass of body (m), speed (v) of the body and the radius (r) of circular path will be expression for centripetal force

According to question,

$F\propto m^a{v}^b{r}^c$

$F=k{m}^a{v}^b{r}^c$

k, being a dimensionless constant.

From homogeneity of dimensions,

LHS = RHS

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

or $\left[ML{T}^{-2}\right]=\left[M^a{L}^{b+c}{T}^{-b}\right]$

Comparing the powers, we obtain

a =1

b + c =1

$-b=2\Rightarrow b=2$

$\therefore 2+c=1$

$\Rightarrow c=-1$

Therefore, $F=km{v}^2{r}^{-1}=\frac{km{v}^2}{r}$

The experimental value of k is found to be here

$\therefore$  $F=\frac{m{v}^2}{r}$