Q. A wheel of radius $1\ m$ rolls forward half a revolution on a horizontal ground. The magnitude of the displacement of the point of wheel initially in contact with the ground is
A) $2\pi$
B) $\sqrt{2}\pi$
C) $\sqrt{\pi^{2} + 4}$
D) $\pi$
Explanation
(c) Horizontal distance covered by the wheel in half revolution $= \pi R$
So the displacement of the point which was initially in contact with ground
$= AA^{'} = \sqrt{(\pi R)^{2} + (2R)^{2}}$
$= R\sqrt{\pi^{2} + 4} = \sqrt{\pi^{2} + 4}\ \ \ \lbrack$As$\ R = 1m\rbrack$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.