Question

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. $ 2\pi $
• B. $ \sqrt{2\pi } $
• C. $ \sqrt{{{\pi }^{2}}+4} $
• D. $ \pi $

Answer 1

Answer: (C)

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) $