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PLEASE GIVE PROPER EXPLANATION.
This question will be faulty until you give some relation between radius of curvatures of pt. A,B and C.
Now,if radius of curvatures at all 3 pts are assumed to be equal,then:
For A,
N+(mv^2/r)=mg; because centrifugal for will act along the radial vector joining A and centre of curvature and in opposite N=mg-(mv^2/r) direction,so will get added to normal reaction force(N) which is along the same direction.
For B,
N=mg+(mv^2/r); because N will act in up direction while centrifugal force and weight will act in forward direction.
For C,
Same as A.
Therefore,N(A)=N(C)<N(B).
NOTE:This question has been solved assuming the plane of motion to be vertical.
If the plane of motion is horizontal,N(A)=N(B)=N(C)=mg.
IS THE ANSWER N2>N1>N3?
in this question if you consider your reference frame on the car and then apply the concept of centrifugal force.
now,at point A,
mg=n+centrifugal
at B,
mg+centrifugal=n
so the normal at b is definately more than at a.for c we have to note that the centrifugal depends on radius and hence more the radius less the centrifugal.
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