JEE Main 2025 — Rotational Motion Question with Solution

The position vectors of two 1 kg particles, (A) and (B), are given by ${\overrightarrow{\mathrm{r}}}_{\mathrm{A}}=\left({\alpha }_1{\mathrm{t}}^2\hat{i}+{\alpha }_2\mathrm{t}\hat{j}+{\alpha }_3\mathrm{t}\hat{k}\right)\mathrm{m}$ and ${\overrightarrow{\mathrm{r}}}_{\mathrm{B}}=\left({\beta }_1\mathrm{t}\hat{i}+{\beta }_2{\mathrm{t}}^2\hat{j}+{\beta }_3\mathrm{t}\hat{k}\right)\mathrm{m}$, respectively; $\left({\alpha }_1=1\mathrm{m}/{\mathrm{s}}^2,{\alpha }_2=3\mathrm{nm}/\mathrm{s},{\alpha }_3=2\mathrm{m}/\mathrm{s},{\beta }_1=2\mathrm{m}/\mathrm{s},{\beta }_2=-1\mathrm{m}/{\mathrm{s}}^2,{\beta }_3=4\mathrm{pm}/\mathrm{s}\right)$, where t is time, n and p are constants. At $t=1\mathrm{s},\left|\overrightarrow{V_A}\right|=\left|{\vec{V}}_B\right|$ 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 _______ .