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Standard VIII

Physics

Density

The apparent ...

Question

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 A $\frac{W_{1} + W_{2}}{2}$
When lift moves up,

$W_{1} - m g = m a$

when lift moves down,

$m g - W_{2} = m a$

Hence,

$W_{1} + W_{2} = 2 m g$

$⟹ m g = \frac{W_{1} + W_{2}}{2}$

When lift moves with constant velocity,

$W = m g$

so,

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

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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 A W1+W22When lift moves up,W1−mg=mawhen lift moves down,mg−W2=maHence,W1+W2=2mg⟹mg=W1+W22When lift moves with constant velocity,W=mgso,W=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 A $\frac{W_{1} + W_{2}}{2}$
When lift moves up,

$W_{1} - m g = m a$

when lift moves down,

$m g - W_{2} = m a$

Hence,

$W_{1} + W_{2} = 2 m g$

$⟹ m g = \frac{W_{1} + W_{2}}{2}$

When lift moves with constant velocity,

$W = m g$

so,

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