Question

a marbal of mass m and radius Rlooped in ..what is minimum
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Answer 1

Your expression for the velocity looks right; but we have to get a few other things taken care of.

First - the center of the marble doesn't move from 0 to 2R, it moves from r to 2R-r - so the potential energy due to this is smaller than mg(2R) which is what you had in your expression.

On the other hand, you need to take account of the energy of the sphere rolling (which is stated explicitly). The moment of inertia for a solid sphere (the usual case for a "marble") is

I=25mr2

This leads to rolling energy

E=12Iω2=1225mr2(vr)2=15mv2

Thus your energy equation has to be corrected to

12mv2+mg(2R−2r)+15mv2=mgh

Also - note that the marble is moving in a path with radius R-r not radius R; you need to take that into account when you compute the limiting velocity ("fast enough to stick to the track") - you have to put R-r where you have R in your velocity equation:

mv2R−r=mgv=g(R−r)−−−−−−−√

Combining these:

710mg(R−r)+2mg(R−r)=mghh=2710(R−r)

Note that it is sometimes said that "you can ignore the rotational energy of the marble if it is very small", but that is emphatically not true - the rotational energy for a solid sphere is always 2/5 of the (linear) kinetic energy, regardless of the size of the marble. It can therefore only be ignored for the case of a (frictionless) sliding object.

Finally - since no physics problem is complete without a diagram:

enter image description here

This shows clearly where the R-r term is coming from.

Thus if you ignore the rolling, the diagram explains the "correct" answer (which in a way was your question). If you include rolling, then you need to modify the solution as shown.