Q 11-06-212JEE MainJEE Main 2019 (10 Apr, Shift 2)Easy
The time dependence of the position of a particle of mass $m = 2$ is given by $\vec r(t) = 2t\,\hat i - 3t^2\,\hat j$. Its angular momentum, with respect to the origin, at time $t = 2$ is:
Answer: (C) $-48\,\hat k$
$\vec v = 2\hat i - 6t\,\hat j$. At $t = 2$: $\vec r = 4\hat i - 12\hat j$ and $\vec v = 2\hat i - 12\hat j$.
$$\vec L = m\,\vec r\times\vec v = 2\left[(4)(-12) - (-12)(2)\right]\hat k = 2(-48 + 24)\hat k = -48\,\hat k$$
Solution by Sreeraj P, M.Sc Physics