Given that velocities at different points are critical
so velocity at top point (T)
⇒ VT=√gR...(1)
From the diagram
in right triangle
△OMA⇒MA=OP=OPcosθ
At point A, θ=60°
⇒HA=OG−OP⇒HA=R−Rcos60°=R2...(2)
in right triangle
△OMB⇒MB=QP=QPcos(180−θ)
At point B, θ=120°
⇒HB=OG+OQ⇒HB=R+Rcos60°=3R2...(4)
Using energy conservation equation at point T and A
⇒12mVT2+mg(2R)=12mVA2+mgH1...(2)
putting the value of eqn (1), eqn (2) in eqn (4)
⇒12m(√gR)2+mg(2R)=12mVA2+mgR2
⇒VA=2√gR
Using energy conservation equation at point T and B
⇒12mVT2+mg(2R)=12mVB2+mgH2...(2)
putting the value of eqn (1), eqn (3) in eqn (4)
⇒12m(√gR)2+mg(2R)=12mVB2+mg3R2
⇒VB=√2gR
Ratio of VA and VB=√2:1
So the correct option is (D)