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NEET UG PHYSICS MCQ: SYSTEM OF PARTICLES

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Analyzing +4 | -1
The position vectors of two 1 kg particles, (A) and (B), are given by r→A=(α1t2i^+α2tj^+α3tk^)m\overrightarrow{\mathrm{r}}_{\mathrm{A}}=\left(\alpha_1 \mathrm{t}^2 \hat{\mathrm{i}}+\alpha_2 \text{t}\hat{\mathrm{j}}+\alpha_3 \text{t} \hat{\mathrm{k}}\right) \mathrm{m} and r→B=(β1ti^+β2t2j^+β3tk^)m,\overrightarrow{\mathrm{r}}_{\mathrm{B}}=\left(\beta_1 \mathrm{t} \hat{\mathrm{i}}+\beta_2 \mathrm{t}^2 \hat{\mathrm{j}}+\beta_3 \mathrm{t} \hat{\mathrm{k}}\right) \mathrm{m}, respectively; (α1=1 m/s2,α2=3 nm/s,α3=2 m/s,β1=2 m/s,β2=−1 m/s2,β3=4pm/s),\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{p} \mathrm{m} / \mathrm{s}\right), where t is time, n and p are constants, At t=1 s,\text{t}=1 \mathrm{~s}, ∣V→A∣=∣V→B∣\left|\overrightarrow{\mathrm{V}}_{\mathrm{A}}\right|=\left|\overrightarrow{\mathrm{V}}_{\mathrm{B}}\right| and velocities V⃗A\mathrm{\vec{V}_A} and V⃗B\mathrm{\vec{V}_B} of the particles are orthogonal to each other. At t=1 s,\text{t}=1 \mathrm{~s}, the magnitude of angular momentum of particle (A) with respect to the position of particle (B) is L kgm2 s−1.\sqrt{\mathrm{L}} \mathrm{~kgm}^2 \mathrm{~s}^{-1}. The value of 'L' is
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Correct option: A
90
Option A (90) is correct because after calculating the velocities and ensuring orthogonality, the angular momentum magnitude squared (L) is found to be 90 kgm²/s².

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