# 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, $\frac{mg}{mg-ma}=\frac{3}{2}\phantom{\rule{0ex}{0ex}}⇒\frac{g}{g-a}=\frac{3}{2}\phantom{\rule{0ex}{0ex}}⇒a=\frac{g}{3}$
hence, the correct option is d.
Regards

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