question_answer

A wheel of radius 1 m rolls forward half are revolution on a horizontal ground. The magnitude of the displacement of the point of the wheel initially in contact with the ground is  

A) \[\pi \]

B) \[2\pi \]

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

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

Correct Answer: D

Solution :

[d]   Linear distance moved by wheel in half revolution \[=\pi r\]. Point \[{{P}_{1}}\]after half revolution reaches at \[{{P}_{2}}\] vertically 2m above the ground. \[\therefore \,\,Displacement\,{{P}_{1}}{{P}_{2}}\] \[\sqrt{{{\pi }^{2}}{{r}^{2}}+{{2}^{2}}}=\sqrt{{{\pi }^{2}}+4}\]          \[[\because \,\,\,r=1m]\]