Verified Solution • 2025–26 Session moderate

QJEEM25Phb8bd: JEE Main 2025 Physics - Class 11 JEEM25Phb8bd Solution: The position vectors of two 1 kg particles, (A) and (B), are…

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; } (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) , where t is time, n and pp are constants. At t=1 s,∣VA→∣=∣VB→∣t=1 \mathrm{~s},\left|\overrightarrow{V_A}\right|=\left|\overrightarrow{V_B}\right| and velocities V⃗A\vec{V}_A and V⃗B\vec{V}_B of the particles are orthogonal to each other. At t=1 st=1 \mathrm{~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} . The value of L is _________.

Step-by-Step Verified Solution

amp;VA=(2ti^+3nj^+2k^)amp;V→B=(2i^−2t^+4pk^)amp;V→A⋅ V→B=0amp;4−6n+8p=0amp;2−3n+4p=0amp;3n=2+4pamp;∣ V→A∣=∣V→B∣amp;4+9n2+4=4+4+16p2amp;p=−14⇒n=13amp; L→2=mA(r→A/B×V→A)amp;r→A/B=(α1−β1)i^2+(α2−β2)j^+(α3−β3)amp;=(1−2)i^+(1+1)j^+3k^\begin{aligned}& \mathrm{V}_{\mathrm{A}}=(2 t \hat{\mathrm{i}}+3 \mathrm{n} \hat{j}+2 \hat{\mathrm{k}}) \\& \overrightarrow{\mathrm{V}}_{\mathrm{B}}=(2 \hat{\mathrm{i}}-2 \hat{\mathrm{t}}+4 \mathrm{p} \hat{\mathrm{k}}) \\& \overrightarrow{\mathrm{V}}_{\mathrm{A}} \cdot \overrightarrow{\mathrm{~V}}_{\mathrm{B}}=0 \\& 4-6 \mathrm{n}+8 \mathrm{p}=0 \\& 2-3 \mathrm{n}+4 \mathrm{p}=0 \\& 3 \mathrm{n}=2+4 \mathrm{p} \\& \left|\overrightarrow{\mathrm{~V}}_{\mathrm{A}}\right|=\left|\overrightarrow{\mathrm{V}}_{\mathrm{B}}\right| \\& 4+9 \mathrm{n}^2+4=4+4+16 \mathrm{p}^2 \\& \mathrm{p}=\frac{-1}{4} \quad \Rightarrow \mathrm{n}=\frac{1}{3} \\& \overrightarrow{\mathrm{~L}}^2=\mathrm{m}_{\mathrm{A}}\left(\overrightarrow{\mathrm{r}}_{\mathrm{A} / \mathrm{B}} \times \overrightarrow{\mathrm{V}}_{\mathrm{A}}\right) \\& \overrightarrow{\mathrm{r}}_{\mathrm{A} / \mathrm{B}}=\left(\alpha_1-\beta_1\right) \hat{\mathrm{i}}^2+\left(\alpha_2-\beta_2\right) \hat{\mathrm{j}}+\left(\alpha_3-\beta_3\right) \\& =(1-2) \hat{\mathrm{i}}+(1+1) \hat{\mathrm{j}}+3 \hat{\mathrm{k}}\end{aligned}

amp;=∣i^amp;j^amp;k^−1amp;2amp;32amp;1amp;2∣=i^+8j^−5k^amp;=1+64+25=90\begin{aligned}& =\left|\begin{array}{ccc}\hat{\mathrm{i}} & \hat{\mathrm{j}} & \hat{\mathrm{k}} \\-1 & 2 & 3 \\2 & 1 & 2\end{array}\right|=\hat{\mathrm{i}}+8 \hat{\mathrm{j}}-5 \hat{\mathrm{k}} \\& =\sqrt{1+64+25}=\sqrt{90}\end{aligned}

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