Given $\mathrm{m}_1=\mathrm{m}_2$
$\text { and } \frac{\mathrm{r}_1}{\mathrm{r}_2}=\frac{3}{4}$
As centripetal force $F=\frac{\mathrm{mv}^2}{\mathrm{r}}$
In order to have constant (same in this question) centripetal force
$ \mathrm{F}_1=\mathrm{F}_2 $
$ \frac{\mathrm{m}_1 \mathrm{v}_1^2}{\mathrm{r}_1}=\frac{\mathrm{m}_2 \mathrm{v}_2^2}{\mathrm{r}_2} $
$ \Rightarrow \frac{\mathrm{v}_1}{\mathrm{v}_2}=\sqrt{\frac{\mathrm{r}_1}{\mathrm{r}_2}}=\frac{\sqrt{3}}{2}$