Obtain the relation between torque of a system of particles and angular momentum.

The total angular momentum of a system of particles is the vector sum of the angular momenta of individual particles. For a system of $n$ particles,
$\overrightarrow{L} = \overrightarrow{l_{1}} + \overrightarrow{l_{2}} + \overrightarrow{l_{3}} + \ldots + \overrightarrow{l_{n}} = \sum_{i=1}^{n} \overrightarrow{l_{i}}$
where $\overrightarrow{l_{i}} = \overrightarrow{r_{i}} \times \overrightarrow{p_{i}}$ is the angular momentum of the $i^{\text{th}}$ particle,$\overrightarrow{r_{i}}$ is its position vector,and $\overrightarrow{p_{i}}$ is its linear momentum.
Differentiating the total angular momentum with respect to time $t$:
$\frac{d\overrightarrow{L}}{dt} = \sum_{i=1}^{n} \frac{d\overrightarrow{l_{i}}}{dt} = \sum_{i=1}^{n} \left( \frac{d\overrightarrow{r_{i}}}{dt} \times \overrightarrow{p_{i}} + \overrightarrow{r_{i}} \times \frac{d\overrightarrow{p_{i}}}{dt} \right)$
Since $\frac{d\overrightarrow{r_{i}}}{dt} = \overrightarrow{v_{i}}$ and $\overrightarrow{v_{i}} \times \overrightarrow{p_{i}} = \overrightarrow{v_{i}} \times (m\overrightarrow{v_{i}}) = 0$,the expression simplifies to:
$\frac{d\overrightarrow{L}}{dt} = \sum_{i=1}^{n} (\overrightarrow{r_{i}} \times \overrightarrow{F_{i}}) = \sum_{i=1}^{n} \overrightarrow{\tau_{i}} = \overrightarrow{\tau}_{ext} + \overrightarrow{\tau}_{int}$
Since internal forces occur in equal and opposite pairs along the same line of action (Newton's third law),their net torque $\overrightarrow{\tau}_{int} = 0$. Thus,the relation is:
$\frac{d\overrightarrow{L}}{dt} = \overrightarrow{\tau}_{ext}$