Question

A wheel of radius 1 meter 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. π

Answer 1

Answer: (C)

$\sqrt{\pi^{2} + 4}$

Answer 2

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' = $\sqrt{(\pi R)^{2} + (2R)^{2}}$

=$R\sqrt{\pi^{2} + 4} = \sqrt{\pi^{2} + 4}$ $(AsR = 1m)$