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Q.
A particle moves through angular displacement $ \theta $ on a circular path of radius $r$. The linear displacement will be :
BHUBHU 2005
Solution:
$\Delta \vec{ r }=\vec{ r }_{2}-\vec{ r }_{1}$,
where $r_{2}=r_{1}=r$
Hence, $\Delta r=\sqrt{r_{2}^{2}+r_{1}^{2}-2 r_{2} r_{1} \cos \theta}$
$=2 r \sin \frac{\theta}{2}$