A bob of mass m is suspended by a light string of length L. It is imparted a horizontal velocity v0 at the lowest point A such that it completes a semi-circular trajectory in the vertical plane with the B string becoming slack only on reaching Q the topmost point, C.Now match the Column-I with Column-II and mark the correct choice from the codes given below.Codes

There are two external forces on the bob: gravity (mg) ard tension (T) in the string. The tension does no work as displacement is always perpendicular to the string. Total mechanical energy (E) of the system is conserved.If we take potential energy of the system to be zero at the lowest point A, thenAt A, E = 12mv02 ---------(i)From Newton's second lawTA-mg = mv02L-----------(ii)Where TA is the tension in the string at A.At the highest point C, the string slackens as the tension in the string (TC) becomes zeroC, E = 12mvC2+mg(2L)-----(iii)From Newton's second lawmg = mvC2L-------(iv)Where vC is the velocity at C,From (iv), vC = gL(iii)-qFrom (iii), E = 12m(gL)+2mgL = 52mgL-----(v)(i)-rAt B, B , E = 12mvB2+mg(L)Where, vB is the velocity at B.or 12,vB2 = E-mg(L) = 52mgL-mgL                                                            (Using (v))12mvB2 = 32mgL∴   vB = 3gL(ii)-sThe ratio of kinetic energies at B and C is∴  KBKC = 12mvB212mvC2 = 3gLgL = 31(iv)-p