The apparent weight of a man in a lift is $W_{1}$ when lift moves upwards with some acceleration and is $W_{2}$ when it is accelerating down with same acceleration. The true weight of the man and acceleration of lift is

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True weight of the man $\frac{W_{1} - W_{2}}{2}$

True weight of the man $\frac{W_{1} + W_{2}}{2}$

Acceleration $(\vec{a}) = 0$

$\vec{a} = g (\frac{W_{1} - W_{2}}{W_{1} + W_{2}})$

answer is A, C.

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Detailed Solution

$1) W_{1} = m (g + a), W_{2} = m (g - a) W_{1} + W_{2} = 2 m g \Rightarrow W_{1} + W_{2} = 2 W (∵ W = m g) \Rightarrow \frac{W_{1} + W_{2}}{2} = W$

$2) \frac{W_{1}}{W_{2}} = \frac{m (g + a)}{m (g - a)} = \frac{g + a}{g - a} \frac{g}{a} = \frac{W_{1} + W_{2}}{W_{1} - W_{2}} \Rightarrow a = g (\frac{W_{1} - W_{2}}{W_{1} + W_{2}})$

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