Obtain the relation between angular momentum of a particle and torque acting on it.

By definition of angular momentum of a particle,$\vec{l} = \vec{r} \times \vec{p}$.
Differentiating this equation with respect to time $t$,we get $\frac{d\vec{l}}{dt} = \vec{r} \times \frac{d\vec{p}}{dt} + \frac{d\vec{r}}{dt} \times \vec{p}$.
Since $\frac{d\vec{p}}{dt}$ is the rate of change of linear momentum,which equals the force $\vec{F}$,and $\frac{d\vec{r}}{dt} = \vec{v}$,we have $\frac{d\vec{l}}{dt} = \vec{r} \times \vec{F} + \vec{v} \times \vec{p}$.
Because $\vec{v}$ and $\vec{p}$ are in the same direction,their cross product $\vec{v} \times \vec{p} = 0$.
Therefore,$\frac{d\vec{l}}{dt} = \vec{r} \times \vec{F} = \vec{\tau}$.
Thus,the time rate of change of angular momentum is equal to the torque acting on the particle.