Two beads each of mass m are fixed at the ends of two light rigid rods each of length l as shown in the…

Distance of centre of mass from pivot, ${\mathrm{r}}_{\mathrm{cm}}=\frac{\mathrm{m}l+\mathrm{m}\left(2l\right)}{\mathrm{m}+\mathrm{m}}=\frac{3}{2}l$
Moment of inertia of system about centre of
mass, ${\mathrm{I}}_{\mathrm{cm}}=\mathrm{m}{\left(\frac{l}{2}\right)}^2+\mathrm{m}{\left(\frac{l}{2}\right)}^2=\frac{\mathrm{m}{l}^2}{2}$
Angular velocity of $rod,\omega =\frac{\mathrm{v}}{2l}$

Velocity of centre of mass,
$v_{\mathrm{cm}}=\omega \left(\frac{3}{2}l\right)=\left(\frac{\mathrm{v}}{2l}\right)\left(\frac{3}{2}l\right)=\frac{3}{4}\mathrm{v}$
Translational kinetic energy of system,

$K_{\text{translational }}=\frac{1}{2}{M}_{\text{treal }}{v}_{\text{cin }}^2$

$=\frac{1}{2}\left(2m\right){\left(\frac{3}{4}v\right)}^2=\frac{9}{16}m{v}^2$

Rotational kinetic energy of system, 

${\mathrm{K}}_{\text{rotational }}=\frac{1}{2}{\mathrm{I}}_{\mathrm{cm}}{\omega }^2$

$=\frac{1}{2}\left(\frac{\mathrm{m}{l}^2}{2}\right){\left(\frac{\mathrm{v}}{2l}\right)}^2=\frac{1}{16}{\mathrm{mv}}^2$

$\therefore \frac{{\mathrm{K}}_1}{{\mathrm{K}}_{\mathrm{r}}}=\frac{\frac{9{\mathrm{mv}}^2}{16}}{\frac{{\mathrm{mv}}^2}{16}}=9$