The pulley arrangements shown in figure are identical, the mass of the rope being negligible. In case I, the mass m is lifted by attaching a mass 2m to the other end of rope with a constant downward force F = 2mg, where g is acceleration due to gravity. The acceleration of mass m in case I is

The pulley arrangements shown in figure are identical, the mass of the

1.	The pulley arrangements shown in figure are identical, the mass of the rope being negligible. In case I, the mass m is lifted by attaching a mass 2m to the other end of rope with a constant downward force F = 2mg, where g is acceleration due to gravity. The acceleration of mass m in case I is
Answer» zero more than that in CASE less than that in case II equal to that in case II Solution :`F=ma, a=g((m_(1)-m_(2))/(m_(1)+m_(2))), F-T=0 and` `T=2mg` also `T-mg=ma^(1)` Finally ` ALT a^(1)`