The centre C of a thin uniform rod AB of mass m and length l is rigidily fixed to axle  $OO\text{'}$ , such that the angle between the rod and axle  $OO\text{'}$  is  $\theta ={37}^o$ (see figure). An external torque is applied on the axle  $OO\text{'}$  such that the system rotates with angular  velocity  ${\omega }_0$. Friction is absent, then choose the INCORRECT
 

Let angular momentum of rod about c is  $L_C$
Let us consider an element of length dr at a distance r from C
So angular momentum of this element about C,
 $d{L}_C=\left(\frac{M}{l}dr\right)\left(\omega r\right)\left(r\sin \theta \right)$
So   $L_C={\int }_{-1/2}^{1/2}\frac{M}{L}dr\left(\omega r\right)r\sin \theta$
 $=\frac{m{l}^2}{12}\omega \sin \theta =\frac{m{l}^2\omega }{20}$
Let  $L_{OO\text{'}}$  be the angular momentum of rod about   $OO\text{'}$
So  $L_{OO\text{'}}=k\omega$
 $$\begin{gathered} =\left(\frac{m{l}^2}{12}{\sin }^2\theta \right)\omega \\ {L}_{OO\text{'}}=\frac{3m{l}^2\omega }{100} \end{gathered}$$
Now Angular momentum about c has two components. Let  $L_H$  and  $L_{OO\text{'}}$  are component of angular momentum about C along horizontal and  $OO\text{'}$  (vertical)
 
 $\tan \theta =\frac{L_H}{L_{OO\text{'}}}\Rightarrow L_H=L_{OO\text{'}}\cot \theta$
so  $\frac{d{L}_H}{dt}=L_{OO\text{'}}\cos e{c}^2\theta \frac{d\theta }{dt}$