List I describes four systems, each with two | vedclass

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$(I)$ $v_{ BA }^2=v_A^2+v_{ B }^2-2 v_{A B} \cos 	heta$

As $\omega_{\hat{A}}=\omega_{ B }, 	heta=90^{\circ}$ remains constant.

Also, $v _{ A }

Vedclass · Accepted Answer

$I \rightarrow S , II \rightarrow T , III \rightarrow P , IV \rightarrow R$

Vedclass · Answer

$(I)$ $v_{ BA }^2=v_A^2+v_{ B }^2-2 v_{A B} \cos 	heta$

As $\omega_{\hat{A}}=\omega_{ B }, 	heta=90^{\circ}$ remains constant.

Also, $v _{ A }= v _{ B }=1 m / s$

So, $v_{ BA }=\sqrt{2} m / s$

$(II)$ $\overrightarrow{ u }_{ A }=\frac{5 \pi}{2} \hat{ i }+\frac{5 \pi}{2} \hat{ j }$

$=\overrightarrow{ v }_A=\frac{5 \pi}{2} \hat{ i }+\left(\frac{5 \pi}{2}-10 \cdot \frac{\pi}{3}ight) \hat{ j }$

$=\frac{5 \pi}{2} \hat{ i }-\frac{5 \pi}{6} \hat{ j }$

$\overrightarrow{ u }_B=-\frac{5 \pi}{2} \hat{ i }+\frac{5 \pi}{2} \hat{ j }$

$\overrightarrow{ u }_{ B }=-\frac{5 \pi}{2} \hat{ i }-\left(\frac{5 \pi}{6}+1ight) \hat{ j }$

$\overrightarrow{ v }_{B . A}=-5 \pi \hat{ i }-\hat{ j }$

$v _{B A}=\sqrt{25 \pi^2+1}$

$(III)$ $x _{ A }=\sin t$

$v _{ A }=\cos t =\frac{1}{2} m / s$

$x _{ B }=\cos t$

$v _{ B }=-\sin t =-\frac{\sqrt{3}}{2} m / s$

$v _{ BA }=-\frac{\sqrt{3}}{2}-\frac{1}{2}$

$(IV)$ $\overrightarrow{ v }_{ A } \& \overrightarrow{ v }_{ B }$ are always perpendicular

So, $\left|\overrightarrow{ v }_{ BA }ight|=\sqrt{ v _{ A }^2+ v _B^2}=\sqrt{10} m / s$

$List I$ describes four systems, each with two particles $A$ and $B$ in relative motion as shown in figure. $List II$ gives possible magnitudes of then relative velocities (in $ms ^{-1}$ ) at time $t=\frac{\pi}{3} s$.

Which one of the following options is correct?

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