The ratio of the weight of a man in a stationary lift & when it is moving downward with uniform acceleration `a` is $3 : 2$ The value of `A` is

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Solution

The weight of the man in lift = mg

Weight in it when moving downwards with uniform acceleration is mg-ma

We have given that

a weight of the man in lift / Weight in it when moving downwards with a uniform acceleration $= 3 : 2$

$\frac{m g}{m g - m a} = \frac{3}{2}$

$\frac{G}{g - a} = \frac{3}{2}$

$2 g = 3 g - 3 a$

$g = 3 a$

$a = \frac{g}{3}$

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