Answer:
As shown in Fig. suppose a particle has velocity \[\vec{\upsilon }\] at point P at some instant t. Its angular momentum about any arbitrary point O is \[\vec{l}=\vec{r}\times m\vec{\upsilon }\] The magnitude of angular momentum is \[l=m\upsilon r\sin \theta \] where \[\theta \] is the angle between \[\vec{r}\] and \[\vec{\upsilon }\] . Now the particle has a constant velocity. Although the position of the particle changes with time, the line of direction of velocity \[\vec{\upsilon }\] remains same and hence \[OM=r\sin \theta \]is a constant. Consequently, the magnitude \[l\] remains constant. Further, the direction of \[\vec{l}\] is perpendicular to the plane of \[\vec{r}\] and \[\vec{\upsilon }\] . It is into the plane of the paper. This direction does not change with time. Thus, \[\vec{l}\] remains the same in magnitude and direction and is therefore conserved. As the velocity of the particle remains constant, so no external force and hence no external- torque is acting on the particle.
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