91Ó°ÊÓ

The first step is to understand the basic concept of angular momentum. Angular momentum (\(L\)) can be defined as the rotational equivalent of linear momentum and can be calculated as the product of the moment of inertia (\(I\)) and the angular velocity (\(\omega\)), i.e., \(L = I \cdot \omega\). However, in this case, the problem does not provide information about \(I\) or \(\omega\), so we rely on the external force (\(F\)) acting on it. The external force generating here would be due to gravitational pull which is \(F = m \cdot g\).

The change in angular momentum is caused by an external torque (\(\tau\)). For a particle moving in a circular path, the torque is the cross product of the position vector (\(r\)) of the particle and the force (\(F\)) acting on it, i.e., \(\tau = r \times F\). In the current problem, the force is due to gravity, acting vertically downward and points towards the center of the circle, therefore making an angle of 90 degrees. Also, the sinus of 90 degrees is 1. Thus, \(\tau = r \times F = r \times (m \cdot g) = r \times (m \cdot g) \cdot \sin 90 = m \cdot g \cdot l\), where \(l\) is the length of the lever arm (in this case equivalent to \(r\)). The rate of change of angular momentum (\(\frac{dL}{dt}\)) is, consequently, equivalent to the torque, i.e., \(\frac{dL}{dt} = \tau = m \cdot g \cdot l\)

Finally, the answer should be compared with the options given in the problem. The expression obtained for the rate of change of angular momentum is \(m \cdot g \cdot l\). Among the available options, the correct answer is therefore (C) \(m \cdot g \cdot l\).