The ratio of the weight of the man in a stationary lift and when it is moving downward with uniform acceleration 'a' 3:2. The value of a 'a' is : (g = acceleration due to gravity)

a) 3/2 g b) g c) 2/3 g d) g/3

Dear Student, 

The weight in stationary lift = mg 
The weight in the lift moving downwards with the acceleration 'a' = (mg-ma)
Given, mgmg-ma=32gg-a=32a=g3
hence, the correct option is d.
Regards 
 

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