IIT JEE Advanced 2006 Paper 1

Using conservation of energy, we can write the following:
EA = mghA+KA (1)
EB = KB (2)
EC = mghC+KC (3)

Since EA = EB = EC, we have the following:
KB > KA
KB > KC
From Eqs. (1) and (3), we get
EA−EC = 0
That is,
mg(hA−hC)+[KA−KC] = 0
hA−hC =

KC−KA

mg

From kinetic energy KB, only translational part gets converted into potential energy at point C, we get hC < hA. Therefore, KC > KA