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 P                 \[y=2r\] Net displacement \[=\sqrt{{{x}^{2}}+{{y}^{2}}}\]                 \[=\sqrt{{{(\pi r)}^{2}}+{{(2\,r)}^{2}}}\]                 \[=r\sqrt{{{\pi }^{2}}+4}\] \[=\sqrt{{{\pi }^{2}}+4}\] \[(\because \,r=1\,m)\]