Correct Answer is: (a) maximum at A
Let rA, rB and rC be the radii of curvature of the road surface at A, B and C respectively. Let NA, NB and NC be the normal reactions of the road on the body at A, B and C respectively. The centres of curvature of the road surface lie above the road at A and C, and below the road at B.
∴ NA - mg = mv 2/rA, mg - NB = mv 2/rB, NC - mg = mv 2/rC.
Clearly, NA and NC are > NB. Also, as rA < rC, from the figure,
NA = mg + mv2 / rA > NC = mg + mv2 / rC .