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(D) The angular momentum $\vec{L}$ of a particle about a point is given by the cross product $\vec{L} = \vec{r} 	imes \vec{p} = \vec{r} 	imes (m\vec

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Remains zero

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(D) The angular momentum $\vec{L}$ of a particle about a point is given by the cross product $\vec{L} = \vec{r} 	imes \vec{p} = \vec{r} 	imes (m\vec{v})$.Since the particle is moving along a straight line and the fixed point lies on the same line,the position vector $\vec{r}$ and the velocity vector $\vec{v}$ are always collinear (parallel or anti-parallel).The angle $	heta$ between $\vec{r}$ and $\vec{v}$ is either $0^\circ$ or $180^\circ$.Therefore,the magnitude of angular momentum is $L = |\vec{r}| |m\vec{v}| \sin(	heta) = 0$,because $\sin(0^\circ) = 0$ and $\sin(180^\circ) = 0$.Thus,the angular momentum remains zero throughout the motion.

$A$ particle is moving along a straight line with increasing speed. Its angular momentum about a fixed point on this line

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