The position vectors of two 1 kg particles, (A) and (B), are given by \overrightarrow{r}_A= (\alpha_1 t^2 \stackrel{\^}{i}+\alpha_2 t \stackrel{\^}{j}+\alpha_3 t \stackrel{\^}{k}) {\m} and \overrightarrow{r}_B= (\beta_1 t \stackrel{\^}{i}+\beta_2 t^2 \stackrel{\^}{j}+\beta_3 t \stackrel{\^}{k}) {\m}, respectively; (\alpha_1=1 {\m} \/ { s}^2, \alpha_2=3 {\m} \/ { s}^2, \alpha_3=2 {\m} \/ { s}, \beta_1=2 {\m} \/ { s}., \beta_2=-1 {\m} \/ { s}^2, \beta_3=4 p {\m} \/ { s} ), where t is time, n and p are constants. At t=1 { s},\overrightarrow{| {V}_A |}=\overrightarrow{| {V}_B |} and velocities \overrightarrow{V}_A and \overrightarrow{V}_B of the particles are orthogonal to each other. At t=1 { s}, the magnitude of angular momentum of particle (A) with respect to the position of particle (B) is \sqrt{L} {\kgm}^2 { s}^{-1}. The value of L is ______ . [22 Jan 2025 Shift 1]

JEE Main 22 Jan 2025 Shift 1 Paper

SECTION - B
Numerical Value Type Questions: This section contains 5 Numerical based questions. The answer to each question should be rounded-off to the nearest integer.

Section: Physics

Question : 46 of 75

Marks: +1, -0

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

→

A=(α1t2

+α2t

+α3t

)m and ‌‌

→

B=(β1t

+β2t2

+β3t

)m, respectively; (α1=1m∕ s2,α2=3m∕ s2,α3=2m∕ s,β1=2m∕ s., β2=−1m∕ s2,β3=4pm∕ s ), where t is time, n and p are constants. At t=1 s,|

→

A|=|

→

B| and velocities

→

A and

→

B of the particles are orthogonal to each other. At t=1 s, the magnitude of angular momentum of particle (A) with respect to the position of particle (B) is √Lkgm2 s−1. The value of L is ______ .

[22 Jan 2025 Shift 1]

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