If the system of blocks shown in the figure is released from rest, the ratio of the tensions T1 and T2 is (Neglect the mass of the string shown in the figure)
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 ‌‌