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Rotational Motion question

2025 · 22 Jan · Shift 1 · Q73
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  5. /2025 · 22 Jan · Shift 1 · Q73

Rotational Motion question

2025 · 22 Jan · Shift 1 · Q73

JEE MainPhysicsRotational MotionNumerical+4 / −1
The position vectors of two 1 kg particles, (A) and (B), are given by r→A=(α1t2i^+α2tj^+α3tk^)m and r→B=(β1t^i^+β2t2j^+β3tk^)m, respectively; \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} \text { and } \overrightarrow{\mathrm{r}}_{\mathrm{B}}=\left(\beta_1 \hat{\mathrm{t}} \hat{i}+\beta_2 \mathrm{t}^2 \hat{j}+\beta_3 \mathrm{t} \hat{k}\right) \mathrm{m} \text {, respectively; }rA​=(α1​t2i^+α2​tj^​+α3​tk^)m and rB​=(β1​t^i^+β2​t2j^​+β3​tk^)m, respectively;  (α1=1 m/s2,α2=3nm/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{n} \mathrm{m} / \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)(α1​=1 m/s2,α2​=3nm/s,α3​=2 m/s,β1​=2 m/s,β2​=−1 m/s2,β3​=4pm/s), where t is time, n and ppp are constants. At t=1 s,∣VA→∣=∣VB→∣t=1 \mathrm{~s},\left|\overrightarrow{V_A}\right|=\left|\overrightarrow{V_B}\right|t=1 s,​VA​​​=​VB​​​ and velocities V⃗A\vec{V}_AVA​ and V⃗B\vec{V}_BVB​ of the particles are orthogonal to each other. At t=1 st=1 \mathrm{~s}t=1 s, the magnitude of angular momentum of particle (A) with respect to the position of particle (B) is Lkgm2 s−1\sqrt{\mathrm{L}} \mathrm{kgm}^2 \mathrm{~s}^{-1}L​kgm2 s−1. The value of L is ‾\underline{\hspace{2cm}}​.
Numerical answer
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