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A wheel of radius 1 m rolls forward half...

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 on contact with the ground is.

A

`2 pi`

B

`sqrt(2)pi`

C

`sqrt(pi^(2)+4)`

D

`pi`

Text Solution

Verified by Experts

The correct Answer is:
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 `=A A' sqrt((pi R)^(2)+(2R)^(2))`
`=R sqrt(pi^(2)+4)=sqrt(pi^(2)+4))` (as R = 1 m)
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