A particle is moving uniformly in a circular path of radius r. When it moves through an angular displacement `theta`, then the magnitude of the corresponding linear displacement will be

Options

A. 2 r cos `((theta)/(2))`
B. 2 r cot `((theta)/(2))`
C. 2 r tan `((theta)/(2))`
D. 2 r sin `((theta)/(2))`

Answer

D

Explanation

The figure is shown as below

`"ln"DeltaAOB," "sin(theta)/(2)=(AB)/(AO)" "(becauseAO=r)`
`AB=AO sin.(theta)/(2)rArrAB=r sin.(theta)/(2)`
`AC=AB+BC" "(becauseAB=BC)`
`r=(rsintheta)/(2)+rsin.(theta)/(2), Ac=rsin.(theta)/(2)+rsin.(theta)/(2)`
`AC=2rsin.(theta)/(2)`
So, the magnitude of the corresponding linear displacement will be `2rsin.(theta)/(2)`.