A bob of mass m attached to an inextensible string of length $$l$$ is suspended from a vertical support. The bob rotates in a horizontal circle with an angular speed $$\omega$$ rad/s about the vertical. About the point of suspension:

We consider a bob of mass $$m$$ tied to an inextensible string of length $$l$$. The upper end of the string is fixed at the support point $$O$$, and the bob is executing uniform circular motion in a horizontal plane with constant angular speed $$\omega$$. Because the string makes a fixed angle $$\theta$$ with the vertical, the horizontal radius of the circle is $$r_h=l\sin\theta$$ and the bob’s linear speed is

$$v=\omega r_h=\omega l\sin\theta.$$

About the point of suspension $$O$$ the position vector of the bob is

$$\vec r=\;l\sin\theta\;\hat i\cos\phi+l\sin\theta\;\hat j\sin\phi-l\cos\theta\;\hat k,$$

where the azimuth $$\phi$$ increases uniformly with time as $$\phi=\omega t$$. The momentum of the bob is $$\vec p=m\vec v$$ with

$$\vec v=\frac{d\vec r}{dt}=l\sin\theta\,\omega\bigl(-\hat i\sin\phi+\hat j\cos\phi\bigr).$$

We now apply the basic definition of angular momentum:

$$\vec L=\vec r\times\vec p.$$

Carrying out the cross-product component by component,

$$\begin{aligned} \vec L &= m\bigl(\vec r\times\vec v\bigr) \\ &=m l^2\omega\Bigl[\sin\theta\cos\theta\cos\phi\$$, $$\hat i +\sin\theta\cos\theta\sin\phi\$$, $$\hat j +\sin^2\theta\$$, $$\hat k\Bigr]. \end{aligned}$$

The magnitude is therefore

$$ \begin{aligned} |\vec L| &=m l^2\omega\sqrt{\sin^2\theta\cos^2\theta(\cos^2\phi+\sin^2\phi)+\sin^4\theta} \\ &=m l^2\omega\sqrt{\sin^2\theta\cos^2\theta+\sin^4\theta} \\ &=m l^2\omega\sin\theta\sqrt{\cos^2\theta+\sin^2\theta} \\ &=m l^2\omega\sin\theta, \end{aligned} $$

which is constant because $$m,\,l,\,\omega,$$ and $$\theta$$ are fixed. Next we examine the direction. The $$\hat k$$ component of $$\vec L$$ is fixed (equal to $$m l^2\omega\sin^2\theta$$), but the horizontal components are proportional to $$\cos\phi$$ and $$\sin\phi$$. As $$\phi$$ increases, the horizontal part of $$\vec L$$ revolves in the horizontal plane, so the overall direction of $$\vec L$$ keeps changing even though its length stays the same.

To confirm this with torque, we recall the formula $$\vec\tau=\vec r\times\vec F$$. About $$O$$ the tension is radial, so $$\vec r\times\vec T=0$$. The only non-zero torque comes from the weight $$\vec F_g=m\vec g$$. Because $$\vec g$$ is vertical, $$\vec\tau=\vec r\times m\vec g$$ lies entirely in the horizontal plane and is always perpendicular to $$\vec L$$. Since $$\frac{d\vec L}{dt}=\vec\tau\perp\vec L$$, the torque can only turn $$\vec L$$ without changing its magnitude.

Thus the angular momentum about the suspension point maintains a constant magnitude while its direction keeps revolving.

Hence, the correct answer is Option C.