Tension in the string connected to the 4 kg block =T2 Tension in each side of the movable pulley (connected to 3 kg blocks) =T1 Since, the pulley is ideal and frictionless, the net upward force on the pulley is ‌T2=2T1 ‌4g−T2=4a(‌ downward acceleration ‌a).....(i) ‌T2=4g−4a ‌T1=3g−3a(‌ acceleration upwards ‌a)....(ii) ‌T2=2T1........(iii) From Eqs. (i) and (ii) ‌4g−4a=2(3g−3a) ‌4g−4a=6g−6a ‌2a=2g ⇒a=g Substitute a=g in Eq. (ii) T1=3g−3g=0(‌ invalid ‌) Assuming equilibrium a=0 Then, T1=3g (From Eq. (ii)) ‌T2=2T1 ‌‌