Step 1: The angular momentum \( \mathbf{L} \) of a particle about the origin is given by the cross product:
\[ \mathbf{L} = \mathbf{r} \times m \mathbf{u} \]
where \( \mathbf{r} \) is the position vector and \( \mathbf{u} \) is the velocity vector.
Step 2: The magnitude of the angular momentum is given by:
\[ L = |\mathbf{r}| \cdot m |\mathbf{u}| \cdot \sin \theta \]
where \( \theta \) is the angle between \( \mathbf{r} \) and \( \mathbf{u} \).
Step 3: Since \( b \) is constant and the particle moves in a straight line, the angular momentum varies with \( \theta \), and the correct expression is:
\[ L = |\mathbf{r}| \cdot |\mathbf{u}| \cdot \sin \theta. \]
A uniform circular disc of radius \( R \) and mass \( M \) is rotating about an axis perpendicular to its plane and passing through its center. A small circular part of radius \( R/2 \) is removed from the original disc as shown in the figure. Find the moment of inertia of the remaining part of the original disc about the axis as given above.
\( \text{M} \xrightarrow{\text{CH}_3\text{MgBr}} \text{N} + \text{CH}_4 \uparrow \xrightarrow{\text{H}^+} \text{CH}_3\text{COCH}_2\text{COCH}_3 \)
Identify the ion having 4f\(^6\) electronic configuration.