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 $
BETA
