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

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

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

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{2 W}_{1}

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{2 W}_{2}

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Solution

The correct option is D $\frac{W_{1} + W_{2}}{2}$
$W_{1} = m (g + a)$ (psuedo force is downwards)
$W_{2} = m (g - a)$ (psuedo force is upwards)
$∴ m g = \frac{W_{1} + W_{2}}{2}$

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The apparent weight of a person inside a lift is W1 when lift moves up with certain acceleration and is W2 when lift moves down with same acceleration. The weight of person when lift moves up with constant speed is

The correct option is D W1+W22W1=m(g+a) (psuedo force is downwards)W2=m(g−a) (psuedo force is upwards)∴mg=W1+W22

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

The correct option is D $\frac{W_{1} + W_{2}}{2}$
$W_{1} = m (g + a)$ (psuedo force is downwards)
$W_{2} = m (g - a)$ (psuedo force is upwards)
$∴ m g = \frac{W_{1} + W_{2}}{2}$