The ratio of the weight of a man in a stationary lift and when it is moving downward with uniform acceleration ‘a’ is 3:2. The value of ‘a’ is (g – Acceleration due to gravity on the earth)

$\frac{3}{2} g$

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$\frac{g}{3}$

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$\frac{2}{3} g$

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$g$

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Solution

The correct option is B
$\frac{g}{3}$

$\frac{w e i g h t o f a m a n i n s t a t i o n a r y l i f t}{w e i g h t o f a m a n i n d o w n w a r d m o v i n g l i f t} = \frac{m g}{m (g - a)} = \frac{3}{2} \frac{g}{g - a} = \frac{3}{2} \Rightarrow 2 g = 3 g - 3 a o r a = \frac{g}{3}$

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The ratio of the weight of a man in a stationary lift and when it is moving downward with uniform acceleration ‘a’ is 3:2. The value of ‘a’ is (g – Acceleration due to gravity on the earth)

The correct option is B g3 weight of a man in stationary liftweight of a man in downwardmoving lift=mgm(g−a)=32gg−a=32⇒2g=3g−3a or a=g3

The correct option is B
$\frac{g}{3}$

$\frac{w e i g h t o f a m a n i n s t a t i o n a r y l i f t}{w e i g h t o f a m a n i n d o w n w a r d m o v i n g l i f t} = \frac{m g}{m (g - a)} = \frac{3}{2} \frac{g}{g - a} = \frac{3}{2} \Rightarrow 2 g = 3 g - 3 a o r a = \frac{g}{3}$