The correct option is A A−r,B−s,C−q,D−p
There are two external forces on the bob: gravity (mg( and 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, then
At A,E=12mv20....(i)
From Newton's second law
TA−mg=mv20L....(ii)
where TA is the tension in the string at A.
At the highest point C, to complete the circle tension in string will be minimum (zero).
At C,E=12mv2C+mg(2L).....(iii)
From Newton's second law
mg=mv2CL....(iv)
From (iv), vC=√gL;C−q
From (iii), E=12m(gL)+2mgL=52mgL.....(v)
Using (i), 12mv50=52mgL,
v0=√5gL;∴A−r....(vi)
At B,E=12mv2B+mg(L)
or 12mv2B=E−mg(L)=52mg−mgL (Using (v))
12mvB2=32mgL
∴vB=√3gL;∴B−s
The ratio of kinetic energies at B and C is
∴KBKC=12mv2B12mv2C=3gLgL=31;∴D−p.