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A lift moves downward with an acceleration a less than g. The apparent weight of a person of mass m inside it is
AMore than the actual weight mg
BExactly equal to the actual weight mg
CLess than the actual weight mg
DAlways zero during the motion
Answer & Solution
Correct answer: C. Less than the actual weight mg
1. For downward acceleration, the net downward force is mg − R = ma.
2. So R = mg − ma = m(g − a).
3. Since a is positive, R is less than mg, so apparent weight is less than the actual weight.
4. Rule out zero: that occurs only in free fall when a equals g.
_Source: Samacheer Kalvi (TN SCERT) Class 10 Science, Unit 1 "Laws of Motion", p.18_
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