JEE Main 2025 — 22 January, Morning Shift — Question 64

The position vectors of two 1 kg particles, (A) and (B), are given by

$\vec{r}_{A}=\left(\alpha_{1} t^{2} \hat{i}+\alpha_{2} t \hat{j}+\alpha_{3} t \hat{k}\right) m$ and $\vec{r}_{B}=\left(\beta_{1} \hat{t}+\beta_{2} \mathrm{t}^{2} \hat{j}+\beta_{3} t \hat{k}\right) \mathrm{m}$, respectively ;

$\left(\alpha_{1}=1 \mathrm{~m} / \mathrm{s}^{2}, \alpha_{2}=3 \mathrm{~nm} / \mathrm{s}, \alpha_{3}=2 \mathrm{~m} / \mathrm{s}, \beta_{1}=2 \mathrm{~m} / \mathrm{s}\right.$,

$\beta_{2}=-1 \mathrm{~m} / \mathrm{s}^{2}, \beta_{3}=4 \mathrm{p} \mathrm{m} / \mathrm{s}$ ),

where t is time, n and p are constants, At $\mathrm{t}=1 \mathrm{~s},\left|\overrightarrow{\mathrm{~V}}_{\mathrm{A}}\right|=\left|\overrightarrow{\mathrm{V}}_{\mathrm{B}}\right|$ and velocities $\vec{V}_{A}$ and $\vec{V}_{B}$

of the particles are orthogonal to each other. At $t=1 \mathrm{~s}$, the magnitude of angular

momentum of particle (A) with respect to the position of particle (B) is $\sqrt{\mathrm{L}} \mathrm{kgm}^{2} \mathrm{~s}^{-1}$. The value of $L$ is $\qquad$ .