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

2 π

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\sqrt{2} π

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\sqrt{π^{2} + 4}

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π

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Solution

The correct option is C $\sqrt{π^{2} + 4}$
Horizontal distance covered by the wheel in half revolution = $π$ R

So the displacement of the point which was initially in contact with ground = $A A^{'} = \sqrt{(π R)^{2} + (2 R)^{2}}$
$= R \sqrt{π^{2} + 4} = \sqrt{π^{2} + 4} (A s R = 1 m)$

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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

The correct option is C √π2+4Horizontal distance covered by the wheel in half revolution = π R So the displacement of the point which was initially in contact with ground = AA′=√(πR)2+(2R)2 =R√π2+4=√π2+4 (As R=1 m)

The correct option is C $\sqrt{π^{2} + 4}$
Horizontal distance covered by the wheel in half revolution = $π$ R

So the displacement of the point which was initially in contact with ground = $A A^{'} = \sqrt{(π R)^{2} + (2 R)^{2}}$
$= R \sqrt{π^{2} + 4} = \sqrt{π^{2} + 4} (A s R = 1 m)$