Option 1 -> When accelerating upward, the normal force must overcome gravity AND provide the upward acceleration, so N = m(g+a), making apparent weight greater than actual weight.
Option 2 -> This occurs when the lift accelerates downward, where N = m(g-a), reducing apparent weight.
Option 3 -> This happens when the lift moves at constant velocity or is at rest, where acceleration is zero.
Option 4 -> This occurs only in free fall when acceleration equals -g, making the normal force zero.
Hence, Option 1: The apparent weight of the lift shall be more than its actual weight -> When a lift accelerates upward, the floor must exert an additional upward force to provide the acceleration on top of supporting the weight against gravity, resulting in apparent weight = m(g+a) which exceeds the actual weight (mg) -> correct