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JEE MAIN - Physics (2025 - 22nd January Morning Shift - No. 24)

The position vectors of two 1 kg particles, (A) and (B), are given by $$ \overrightarrow{\mathrm{r}}_{\mathrm{A}}=\left(\alpha_1 \mathrm{t}^2 \hat{i}+\alpha_2 \mathrm{t} \hat{j}+\alpha_3 \mathrm{t} \hat{k}\right) \mathrm{m} \text { and } \overrightarrow{\mathrm{r}}_{\mathrm{B}}=\left(\beta_1 \hat{\mathrm{t}} \hat{i}+\beta_2 \mathrm{t}^2 \hat{j}+\beta_3 \mathrm{t} \hat{k}\right) \mathrm{m} \text {, respectively; } $$ $\left(\alpha_1=1 \mathrm{~m} / \mathrm{s}^2, \alpha_2=3 \mathrm{n} \mathrm{m} / \mathrm{s}, \alpha_3=2 \mathrm{~m} / \mathrm{s}, \beta_1=2 \mathrm{~m} / \mathrm{s}, \beta_2=-1 \mathrm{~m} / \mathrm{s}^2, \beta_3=4 \mathrm{pm} / \mathrm{s}\right)$, where t is time, n and $p$ are constants. At $t=1 \mathrm{~s},\left|\overrightarrow{V_A}\right|=\left|\overrightarrow{V_B}\right|$ 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 _________.
Answer
90

Explanation

Given, $${m_A} = 1\,kg = {m_B}$$

$$\overrightarrow {{r_A}} = ({\alpha _1}{t^2}\widehat i + {\alpha _2}t\widehat j + {\alpha _3}t\widehat k)\,m$$

$$\overrightarrow {{r_B}} = ({\beta _1}t\widehat i + {\beta _2}{t^2}\widehat j + {\beta _3}t\widehat k)\,m$$

$$({\alpha _1} = 1\,m/{s^2},{\alpha _2} = 3n\,m/s,{\alpha _3} = 2\,m/s$$

$${\beta _1} = 2\,m/s,{\beta _2} = - 1\,m/{s^2},{\beta _3} = 4p\,m/s)$$

$$\overrightarrow {{V_A}} = {{d\overrightarrow {{r_A}} } \over {dt}} = 2t\widehat i + 3n\widehat j + 2\widehat k$$

$$\overrightarrow {{V_B}} = {{d\overrightarrow {{r_B}} } \over {dt}} = 2\widehat i - 2t\widehat j + 4p\widehat k$$

$$\overrightarrow {{V_B}} \,.\,\overrightarrow {{V_B}} = 0$$ (As $$\overrightarrow {{V_A}} \bot \overrightarrow {{V_B}} $$ given)

$$ \Rightarrow 4t - 6nt + 8p = 0$$

At $$t = 1$$, $$4 - 6n + 8p = 0$$

$$ \Rightarrow 2 - 3n + 4p = 0$$

$$ \Rightarrow 3n = 2 + 4p$$ ..... (1)

given, $$\left| {\overrightarrow {{V_A}} } \right| = \left| {\overrightarrow {{V_B}} } \right|$$

$$ \Rightarrow 4 + 9{n^2} + 4 = 4 + 4 + 16{p^2}$$

$$ \Rightarrow {(2 + 4p)^2} = 16{p^2}$$ (from (1)

$$ \Rightarrow 4 + 16{p^2} + 16p = 16{p^2}$$

$$ \Rightarrow p = - {1 \over 4}$$

$$ \Rightarrow 3n = 2 + 4\left( { - {1 \over 4}} \right) = 1 \Rightarrow n = {1 \over 3}$$

Now, $$\overrightarrow L = {m_A}\left( {{{\overrightarrow r }_{A/B}} \times {{\overrightarrow v }_A}} \right)$$

at $$t = 1\,\sec ,$$

$$\overrightarrow {{r_{{A \over B}}}} = \left( {{\alpha _1} - {\beta _1}} \right)\widehat i + \left( {{\alpha _2} - {\beta _2}} \right)\widehat j + \left( {{\alpha _3} - {\beta _3}} \right)\widehat k$$

$$ \Rightarrow \overrightarrow {{r_{{A \over B}}}} = (1 - 2)\widehat i + (3n + 1)\widehat j + (2 - 4p)\widehat k$$

$$ = - \widehat i + 2\widehat j + 3\widehat k$$

$$\overrightarrow L = \left| {\matrix{ {\widehat i} & {\widehat j} & {\widehat k} \cr { - 1} & 2 & 3 \cr 2 & 1 & 2 \cr } } \right|$$

$$ = \widehat i(4 - 3) - \widehat j( - 2 - 6) + \widehat k( - 1 - 4)$$

$$\overrightarrow L = \widehat i + 8\widehat j - 5\widehat k$$

$$ \Rightarrow \left| {\overrightarrow L } \right| = \sqrt {{1^2} + {8^2} + {{( - 5)}^2}} = \sqrt {1 + 64 + 25} $$

$$ \Rightarrow \left| {\overrightarrow L } \right| = \sqrt {90} $$ kg m$^2$ s$^{-1}$ = $\sqrt L$ (given)

So, L = 90

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