A man standing on a lift that is descending does not feel any weight because

Analyze the question: A man in a descending lift feels weightless. We need the reason.

Concept: What you "feel" as weight is the normal reaction (R) from the surface you stand on. For a lift accelerating downward with acceleration a:

mg − R = ma R = m(g − a)

Solution:

• If the lift accelerates downward at a < g, R is smaller than mg, so the man feels lighter.
• If the lift accelerates downward at a = g (for example, if the cable breaks and it is in free fall), then R = m(g − g) = 0.
• With R = 0, the floor pushes on the man with no force, so he feels weightless. This is apparent weightlessness.

Checking the options:

• A. Air-tightness has nothing to do with weight. Wrong.
• B. Being in a vacuum does not remove gravity. Objects still have weight there. Wrong.
• C. The lift's acceleration equals g, so R = 0. Correct.
• D. Gravity still acts on the man. If it did not, he would not accelerate downward at all. Wrong.

Answer: C. The lift's acceleration is the same as the acceleration due to gravity.

Tip: In lift problems, remember R = m(g − a) for downward acceleration and R = m(g + a) for upward acceleration. Weightlessness means R = 0, which happens when a = g downward. Gravity never disappears; only the feeling of weight does.