System of Particles and Rotational Motion - Result Question 14

14. 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

[2002]

(a) $\pi$

(b) $2 \pi$

(c) $\sqrt{2} \pi$

(d) $\sqrt{\pi^{2}+4}$

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Answer:

Correct Answer: 14. (d)

Solution:

  1. (d)

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 $P_1 P_2$

$=\sqrt{\pi^{2} r^{2}+2^{2}}=\sqrt{\pi^{2}+4} \quad[\because r=1 m]$