question_answer

A wheel of radius 1 m 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) \[\pi \]                           

B) \[2\pi \]

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

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

Correct Answer: D

Solution :

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