A thin uniform rod of length L is bent at its mid point as shown in the figure. The distance of centre of mass from the point 'O' is

 

From the question,

${\mathrm{x}}_{\mathrm{cm}}=\frac{{\displaystyle \frac{\mathrm{m}}{2}\left(\frac{\mathrm{L}}{4}\right)+\frac{\mathrm{M}}{2}\frac{\mathrm{L}}{4}\mathrm{cos\theta }}}{\mathrm{m}}\hat{\mathrm{i}}$

${\mathrm{x}}_{\mathrm{cm}}=\frac{\mathrm{L}}{8}(1-\mathrm{cos\theta })\hat{\mathrm{i}}$

${\mathrm{y}}_{\mathrm{cm}}=\frac{\frac{\mathrm{m}}{2}\left(\frac{\mathrm{L}}{4}\right)\mathrm{sin\theta }\times \mathrm{m}}{\mathrm{m}}\hat{\mathrm{j}}$

${\mathrm{y}}_{\mathrm{cm}}=\frac{\mathrm{L}}{8}\mathrm{sin\theta }\hat{\mathrm{j}}$

Center of mass separation from O = $\sqrt{{{\mathrm{x}}_{\mathrm{com}}}^2+{{\mathrm{y}}_{\mathrm{com}}}^2}$

$=\frac{\mathrm{L}}{8}\sqrt{{(1-\mathrm{cos\theta })}^2+{\sin }^2\mathrm{\theta }}$

$=\frac{\mathrm{L}}{8}\sqrt{1-2\mathrm{cos\theta }+{\sin }^2\mathrm{\theta }+{\cos }^2\mathrm{\theta }}$

$=\frac{\mathrm{L}}{4}\sin \frac{\mathrm{\theta }}{2}$

Hence the correct answer is $\frac{\mathrm{L}}{4}\sin \frac{\mathrm{\theta }}{2}.$