question_answer

A wheel of radius 1m 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 :

A)  \[2\pi \]                              

B)         \[\sqrt{2\pi }\]

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

D)         \[\pi \]

Correct Answer: C

Solution :

When wheel rolls half a revolution, the point (P) of the wheel which is in contact with the ground initially, moves at the top of the wheel as shown. Horizontal displacement of wheel, \[x=\pi r,\]where r being the radius. Vertical displacement of point P, \[y=2r\] Net displacement\[=\sqrt{{{x}^{2}}+{{y}^{2}}}\]                                 \[=\sqrt{{{(\pi r)}^{2}}+{{(2r)}^{2}}}\] \[=r\sqrt{{{\pi }^{2}}+4}\] \[=\sqrt{{{\pi }^{2}}+4}\]              \[(\because \,r=1\,m)\]