Solution:
Angular momentum \(=\mathrm{mvr}\)
$$
\begin{aligned}
&=k i l o g r a m \times \frac{\text { metre }}{\text { second }} \times \text { metre } \\
&=k i l o g r a m \times \frac{(\text { metre })^{2} \times \text { second }}{\text { second } \times \text { second }} \\
&=\left(k g \times \frac{(\text { metre })^{2}}{\text { second }^{2}}\right) \times \text { sec } \\
&=\text { joule } \times \text { second }
\end{aligned}
$$
