If the radius of curvature of the path of two particles of same mass are in the ratio 3:4, then in order to have constan…

\(\frac{{r}_{1}}{{r}_{2}}=\frac{3}{4}\)

As centripetal force F \(=\frac{m{v}^{2}}{r}\)
In order to have constant (same in this question) centripetal force

$$\begin{aligned}& F_1=F_2 \\& \frac{m_1 v_1^2}{r_1}=\frac{m_2 v_2^2}{r_2} \\& \Rightarrow \frac{v_1}{v_2}=\sqrt{\frac{r_1}{r_2}}=\frac{\sqrt{3}}{2}\end{aligned}$$