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The apparent weight of a person inside a lift is $W_{1}$ when lift moves up with a certain acceleration and is $W_{2}$ when lift moves down with same acceleration. The weight of the person when lift moves up with constant speed is

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$\frac{W_{1} - W_{2}}{2}$

$2 W_{2}$

$\frac{W_{1} + W_{2}}{2}$

$2 W_{1}$

answer is A.

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Detailed Solution

Lift is going up with acceleration ‘a’ :

$W_{1} = m (g + a)$ ________ (1)

Lift is going down with acceleration ‘a’ :

$W_{2} = m (g - a)$ ________ (2)

(1) + (2)

$\Rightarrow m g = \frac{W_{1} + W_{2}}{2}$

mg is the weight of the person (mg) $= \frac{W_{1} + W_{2}}{2}$

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