**The answer is B: 2R.**

Consider a ring rolling without sliding.

Any particle of the ring is at the highest point of its path when it is diametrically opposite the point where the ring touches the surface. The distance between this particle and the point is 2R.

Since the ring rolls without sliding, the linear velocity of the point of contact between ring and surface is zero. This means this point serves as an instantaneous center of rotation, or the center of the curvature of the path. In other words, in any given moment, all particles of the ring are rotating around the point of contact between the ring and the surface.

Therefore, the radius of the curvature of the path of the point "on top" of ring is the distance between this point and the center of rotation, or 2R.

**Q. A ring of radius `R` rolls without sliding with a constant velocity .The radius of curvature o f the path followed by any particle of the ring at the highest point of its path will be:- A) `R` B) `2R` C) `4R` D) None.**

When a ring moves by rotation without slipping on a surface the highest points P of the ring will have a speed `2v` and the lowest points will have the speed 0. Thus the highest points of the ring will rotate about the lowest points of the ring with an angular frequency `omega` and a radius of gyration `2R` .

`2v =omega*2R`

It is like when the motion of the upper points is composed of a linear translation with speed `v` of the center of ring and a rotation about the center of the ring with the same angular speed `omega` (see the figure below)

`v_("tot") = v +omega*R = v+v =2v`

Therefore the radius of the curvature of a particle at the highest point of the ring is `2R` . **The correct answer is B**.

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