JEE Main 2025 — Rotational Motion Question with Solution

Given, $$m_A=1kg=m_B$$

$$\overrightarrow{r_A}=({\alpha }_1{t}^2\hat{i}+{\alpha }_{2}t\hat{j}+{\alpha }_{3}t\hat{k})m$$

$$\overrightarrow{r_B}=({\beta }_{1}t\hat{i}+{\beta }_2{t}^2\hat{j}+{\beta }_{3}t\hat{k})m$$

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

at $$t=1\sec ,$$

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

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

$$=-\hat{i}+2\hat{j}+3\hat{k}$$

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

$$=\hat{i}(4-3)-\hat{j}(-2-6)+\hat{k}(-1-4)$$

$$\overrightarrow{L}=\hat{i}+8\hat{j}-5\hat{k}$$

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

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

So, L = 90