A particle of mass $\mathrm{m}$ is suspended from point $\mathrm{O}$ and undergoes circular motion in horizontal plane as conical pendulum as shown in figure.

 

If $$\begin{gathered} \overrightarrow{r} be the ins\tan \tan eous position vector of the particle about O and \overrightarrow{v} \\ be the ins\tan \tan eous velocity , then \overrightarrow{L_O}=m(\overrightarrow{r}\times \overrightarrow{v}) whose direction continuously changes \\ but magnitude remains cons\tan t . \end{gathered}$$

About the centre angular momentum $L_c=mv(l \sin \theta ) and direction is along CO.$

$\left(\mathrm{C}\right)$ $\mathrm{T} \cos \mathrm{\theta }= \mathrm{mg}..........\left(1\right)$
     $\mathrm{T} \sin \mathrm{\theta }= {\mathrm{m\omega }}^2\mathrm{lsin\theta }..........\left(2\right)$

        ${\mathrm{m\omega }}^2 \mathrm{l} \mathrm{cos\theta }=\mathrm{mg}$          

 ${\mathrm{\omega }}^2 = \frac{\mathrm{g}}{\mathrm{l} \cos \mathrm{\theta }}, \mathrm{\omega }=\sqrt{\frac{\mathrm{g}}{\mathrm{l} \cos \mathrm{\theta }}}$

  $\mathrm{V}=\mathrm{\omega }\left(\mathrm{l} \mathrm{sin\theta }\right)= \sqrt{\frac{\mathrm{gl}}{\mathrm{cos\theta }}}\sin \mathrm{\theta }$

  ${\mathrm{F}}_{\mathrm{av}}\times ∆\mathrm{t}= ∆\mathrm{P}= {\mathrm{P}}_2-{\mathrm{P}}_1$

  $=2 \mathrm{msin\theta }\sqrt{\frac{\mathrm{gl}}{\cos \mathrm{\theta }}}$

  ${\mathrm{F}}_{\mathrm{av}} =\frac{{\mathrm{P}}_2-{\mathrm{P}}_1}{∆\mathrm{t}}= \frac{2\mathrm{m} \mathrm{sin\theta }\sqrt{{\displaystyle \frac{\mathrm{gl}}{\cos \mathrm{\theta }}}}}{{\displaystyle \raisebox{1ex}{$\mathrm{\pi }$}\!\left/ \!\raisebox{-1ex}{$\mathrm{\omega }$}\right.}}$

  ${\mathrm{F}}_{\mathrm{av}} =\frac{2\mathrm{m} \mathrm{sin\theta }}{\mathrm{\pi }}\sqrt{\frac{\mathrm{gl}}{\mathrm{cos\theta }}}\times \sqrt{\frac{\mathrm{g}}{\mathrm{l} \cos \mathrm{\theta }}} \Rightarrow {\mathrm{F}}_{\mathrm{av}}=\frac{2\mathrm{mg} \mathrm{tan\theta }}{\mathrm{\pi }}$

 $\mathrm{L}$ will be constant about line $\mathrm{OC}$ as shown in diagram.