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