Q 11-06-090JEE MainJEE Main 2024 (29 Jan, Shift 2)Medium
A body of mass $5\ \text{kg}$ moving with a uniform speed $3\sqrt2\ \text{m s}^{-1}$ in the $X$–$Y$ plane along the line $y = x + 4$. The angular momentum of the particle about the origin will be ______ $\text{kg m}^2\,\text{s}^{-1}$.
Numerical value type. Enter your answer.
Answer: 60
Perpendicular distance of the line $x - y + 4 = 0$ from the origin:
$$d = \frac{|4|}{\sqrt{1^2 + 1^2}} = 2\sqrt2\ \text{m}$$
$$L = mvd = 5\times3\sqrt2\times2\sqrt2 = 60\ \text{kg m}^2\,\text{s}^{-1}$$
Solution by Sreeraj P, M.Sc Physics