When a lift is going up with uniform acceleration, the apparent weight of a person is $W_{1}$
When the lift is stationary, his apparent weight is $W_{2}$
When the lift falls freely his apparent weight is $W_{3}$
When the lift is going down with uniform acceleration which is less than the acceleration due to gravity, his apparent weight is $W_{4}$
The increasing order of these four weights is

W_{1}, W_{3}, W_{2}, W_{4}

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W_{3}, W_{4}, W_{2}, W_{1}

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W_{3}, W_{2}, W_{4}, W_{1}

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W_{2}, W_{3}, W_{4}, W_{1}

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Solution

The correct option is C $W_{3}, W_{4}, W_{2}, W_{1}$
When lift accelerates up, pseudo force acts downwards, hence it increases apparent weight. $W_{1} > m g$
When lift is stationary, $W_{2} = m g$
When lift falls freely, it accelerates with g downwards, causing an upwards pseudo force in the frame of the lift equal to mg. Hence total force is 0. So weight is 0. $W_{3} = 0$
When lift accelerates down, pseudo force acts upwards, hence it decreases apparent weight $W_{4} < m g$ , but also the acceleration is less than g, therefore $W_{4} = m (g - a) > 0$
Hence, $W_{3} < W_{4} < W_{2} < W_{1}$

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