question_answer

A wheel of radius 1 meter rolls forward half a revolution on a horizontal ground. The magnitude of the displacement of the point of the wheel initially in contact with the ground is             [BCECE 2005]            

A)                         \[2\pi \]

B)                                  \[\sqrt{2}\pi \]

C)             \[\sqrt{{{\pi }^{2}}+4}\]    

D)             \[\pi \]

Correct Answer: C

Solution :

                Horizontal distance covered by the wheel in half revolution = \[\pi R.\]             So the displacement of the point which was initially in contact with ground = AA' = \[\sqrt{{{(\pi R)}^{2}}+{{(2R)}^{2}}}\]                 =\[R\sqrt{{{\pi }^{2}}+4}\]\[=\sqrt{{{\pi }^{2}}+4}\]             \[(As\,R=1m)\]