The apparent weight of a person in a lift moving downwards is half his apparent weight in the same lift moving upwards with the same magnitude of acceleration. Acceleration of the lift is:

g

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g / 4

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g / 2

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g / 3

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Solution

The correct option is C $g / 3$
Apparent weight of a person in lift will be $m (g + a)$ while moving upwards whereas it will be $m (g - a)$ while moving downwards.

Hence, as per the given information,

$m (g + a) = 2 m (g - a)$

$g + a = 2 g - 2 a$

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

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The apparent weight of a person in a lift moving downwards is half his apparent weight in the same lift moving upwards with the same magnitude of acceleration. Acceleration of the lift is:

The correct option is C g/3Apparent weight of a person in lift will be m(g+a) while moving upwards whereas it will be m(g−a) while moving downwards.Hence, as per the given information,m(g+a)=2m(g−a)g+a=2g−2aHence, a=g3

The correct option is C $g / 3$
Apparent weight of a person in lift will be $m (g + a)$ while moving upwards whereas it will be $m (g - a)$ while moving downwards.

Hence, as per the given information,

$m (g + a) = 2 m (g - a)$

$g + a = 2 g - 2 a$

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