\[\begin {array} {1 1} (a)\;\frac{v}{4}R & \quad (b)\;\frac{v}{3}R \\ (c)\;\frac{2v}{3}R & \quad (d)\;\frac{3v}{4R} \end {array}\]

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centre of mass of the loop - bullet system is $R/2$ from centre of loop.

By momentum conservation :

$mv=2m V_{cm}$

$V_{cm}=\large\frac{v}{2}$

Applying angular momentum conservation about the centre of mass.

$MVR/2=I_cw$

$I_c$ for ring $= mr^2+m(r/2)^2$

$I_c$ of bullet $= m (r/2)^2$

moment of inertia of the system $I_c= \large\frac{6mR^2}{4}$

$\qquad= \large\frac{3}{2} m$$R^2$

$mv \large\frac{R}{2}= \frac{3}{2}$$ mR^2w$

$w= \large\frac{v}{3}$$R$

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