The question focuses on the concept of angular momentum in the context of rotational motion. Angular momentum is a key physical quantity in rotational dynamics, and it has a specific direction dictated by the right-hand rule.
**Explanation:**
When a mass rotates about a fixed point, it possesses angular momentum. Angular momentum (\mathbf{L}) is given by the cross product of the position vector (\mathbf{r}) and the linear momentum (\mathbf{p}) of the object:
\mathbf{L} = \mathbf{r} \times \mathbf{p}
According to the right-hand rule, if you point the fingers of your right hand in the direction of the position vector (\mathbf{r}) and curl them in the direction of the linear momentum (\mathbf{p}), your thumb points in the direction of the angular momentum (\mathbf{L}).
In this scenario, since the mass is rotating in a plane, the direction of the angular momentum is perpendicular to the plane of rotation. This is because the right-hand rule dictates that the cross product of vectors perpendicular to each other results in a vector that points perpendicularly away from their plane.
Therefore, the angular momentum of a rotating mass in a plane is directed along a line perpendicular to the plane of rotation.
**Conclusion:** The correct answer is a line perpendicular to the plane of rotation. This aligns with the standard physical principle governing angular momentum.
The center of mass of a thin rectangular plate (fig - x) with sides of length \( a \) and \( b \), whose mass per unit area (\( \sigma \)) varies as \( \sigma = \sigma_0 \frac{x}{ab} \) (where \( \sigma_0 \) is a constant), would be 