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d
$\frac{\mathrm{\upsilon }}{6\mathrm{L}}$

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detailed solution

Correct option is C

Here, the two masses crash with the rod and become locked together; as a result, the masses enter the system and the imposed external torque is therefore required to be zero. Therefore, the system's angular momentum must be conserved. We are aware that a body's centre of mass is determined by,

$R=\frac{\sum {m}_{i}{r}_{i}}{\sum {m}_{i}}$

After the collision, the system's centre of mass will be located around point O.

$\mathrm{R}=\frac{8\mathrm{m}×0+2\mathrm{m}×\frac{\mathrm{L}}{6}-\mathrm{m}×\frac{\mathrm{L}}{3}}{8\mathrm{m}+2\mathrm{m}+\mathrm{m}}=0$
We also know that a body's angular momentum around a point is determined by,

$\mathrm{L}=\mathrm{mvr}=\mathrm{I\omega }$

Now, the system's angular momentum before impact is,

${\mathrm{L}}_{\mathrm{i}}=\mathrm{m}×2\mathrm{v}×\frac{\mathrm{L}}{3}+2\mathrm{m}×\mathrm{v}×\frac{\mathrm{L}}{6}$

${\mathrm{L}}_{\mathrm{i}}=\frac{2\mathrm{mvL}}{3}+\frac{\mathrm{mvL}}{3}=\mathrm{mvL}$

Let's assume that the system's angular speed upon collision is,
Thus, the system's angular momentum around O after the impact is,

${\mathrm{L}}_{\mathrm{f}}=\left({\mathrm{I}}_{\mathrm{rod}}+{\mathrm{I}}_{\mathrm{m}}+{\mathrm{I}}_{2\mathrm{m}}\right)\mathrm{\omega }$

Now, a rod's moment of inertia around its centre of mass is given by,

$\mathrm{I}=\frac{{\mathrm{ML}}^{2}}{12}$

${\mathrm{I}}_{\mathrm{rod}}=\frac{8{\mathrm{mL}}^{2}}{12}$

The mass m's moment of inertia is ${I}_{m}$$=\mathrm{m}{\left(\frac{\mathrm{L}}{3}\right)}^{2}=\frac{{\mathrm{mL}}^{2}}{9}$

The mass 2m moment of inertia is  ${\mathrm{I}}_{2\mathrm{m}}=2\mathrm{m}{\left(\frac{\mathrm{L}}{6}\right)}^{2}=\frac{{\mathrm{mL}}^{2}}{18}$

thus we get,

${\mathrm{L}}_{\mathrm{f}}=\left({\mathrm{I}}_{\mathrm{rod}}+{\mathrm{I}}_{\mathrm{m}}+{\mathrm{I}}_{2\mathrm{m}}\right)\mathrm{\omega }$

${\mathrm{L}}_{\mathrm{f}}=\left(\frac{8{\mathrm{mL}}^{2}}{12}+\frac{{\mathrm{mL}}^{2}}{9}+\frac{{\mathrm{mL}}^{2}}{18}\right)\mathrm{\omega }=\frac{5}{6}{\mathrm{mL}}^{2}\mathrm{\omega }$

$\mathrm{mvL}=\frac{5}{6}{\mathrm{mL}}^{2}\mathrm{\omega }$

$\mathrm{\omega }=\frac{6\mathrm{v}}{5\mathrm{L}}$

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