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2025 26 questions

1. Q492025 · 22 Jan 2025 · Shift 1· Numerical answer

   The position vectors of two $1\text{ kg}$ particles, (A) and (B), are given by ${\vec{r}}_{\mathrm{A}}=\left({\alpha }_1{t}^2\hat{i}+{\alpha }_{2}t\hat{j}+{\alpha }_{3}t\hat{k}\right)\text{m}$ and ${\vec{r}}_{\mathrm{B}}=\left({\beta }_{1}t\hat{i}+{\beta }_2{t}^2\hat{j}+{\beta }_{3}t\hat{k}\right)\text{m}$, respectively; $\left({\alpha }_1=1{\text{ m/s}}^2,{\alpha }_2=3n\text{ m/s},{\alpha }_3=2\text{ m/s},{\beta }_1=2\text{ m/s},{\beta }_2=-1{\text{ m/s}}^2,{\beta }_3=4p\text{ m/s}\right)$, where $t$ is time, $n$ and $p$ are constants. At $t=1\text{ s}$, $\left|{\vec{V}}_{\mathrm{A}}\right|=\left|{\vec{V}}_{\mathrm{B}}\right|$ and velocities ${\vec{V}}_A$ and ${\vec{V}}_B$ of the particles are orthogonal to each other. At $t=1\text{ s}$, the magnitude of angular momentum of particle (A) with respect to the position of particle (B) is $\sqrt{\mathrm{L}}\mathrm{kg}{\mathrm{m}}^2{\mathrm{s}}^{-1}$. The value of $\mathrm{L}$ is _________.

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   Answer: 90