The displacement of the point of the wheel initially in contact with the ground, when the wheel rolls forward half a revolution will be (radius of the wheel is $R$ )

\frac{R}{\sqrt{π^{2} + 1}}

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R \sqrt{π^{2} + 4}

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2 π R

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π R

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Solution

The correct option is B $R \sqrt{π^{2} + 4}$

Horizontal distance covered by the wheel in half revolution $= π R$
So, displacement of the point which was initially in contact with the ground $= \sqrt{(π R)^{2} + (2 R)^{2}} = R \sqrt{π^{2} + 4}$

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The displacement of the point of the wheel initially in contact with the ground, when the wheel rolls forward half a revolution will be (radius of the wheel is R)

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

The displacement of the point of the wheel initially in contact with the ground, when the wheel rolls forward half a revolution will be (radius of the wheel is $R$ )

The correct option is B $R \sqrt{π^{2} + 4}$

Horizontal distance covered by the wheel in half revolution $= π R$
So, displacement of the point which was initially in contact with the ground $= \sqrt{(π R)^{2} + (2 R)^{2}} = R \sqrt{π^{2} + 4}$