Laws Of Motion Question 242

Question: Mass of a person sitting in a lift is 50 kg. If lift is coming down with a constant acceleration of 10 $ m/sec^{2}. $ Then the reading of spring balance will be $ (g=10m/sec^{2}) $ [RPET 2003; Kerala PMT 2005]

Options:

A) 0

B) 1000N

C) 100 N

D) 10 N

Show Answer

Answer:

Correct Answer: A

Solution:

$ R=m(g-a)=0 $



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