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Q.

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. So the acceleration of the lift is:

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a

g

b

\frac{g}{4}

c

\frac{g}{2}

d

\frac{g}{3}

answer is D.

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Detailed Solution

Concept: we know that when lift moves upward or downward, it accelerates or de-accelerates. Which denotes that the lift behaves as a non-inertial frame. So if we want to observe the phenomena inside the lift, we should convert this non-inertial frame to an inertial frame. Because we can apply Newton’s laws of motion only for inertial frames of reference.
To convert a non-inertial frame to an inertial frame, we use pseudo force on the person.
Apparent weight is the virtual weight of man that makes the sense of heaviness or weightlessness during the motion of lift. Apparent weight is given by-
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

So, a

= \frac{g}{3}

Hence, option 4 is correct.

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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. So the acceleration of the lift is: