If a coin is dropped in a lift it takes $t_{1}$ time to reach the floor and takes $t_{2}$ time when lift is moving up with constant acceleration, when which one of the following relation is correct?

t_{1} = t_{2}

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t_{1} > t_{2}

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t_{2} > t_{1}

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t_{1} >> t_{2}

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Solution

The correct option is B $t_{1} > t_{2}$
Time $t_{1}$ for stationary lift $= \sqrt{\frac{2 h}{g}}$
When lift is moving up with constant acceleration, then
$t_{2} = \sqrt{\frac{2 h}{g + a}}$
$∴ t_{1} > t_{2}$ .

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If a coin is dropped in a lift it takes t1 time to reach the floor and takes t2 time when lift is moving up with constant acceleration, when which one of the following relation is correct?

The correct option is B t1>t2Time t1 for stationary lift =√2hgWhen lift is moving up with constant acceleration, thent2=√2hg+a∴t1>t2.

If a coin is dropped in a lift it takes $t_{1}$ time to reach the floor and takes $t_{2}$ time when lift is moving up with constant acceleration, when which one of the following relation is correct?

The correct option is B $t_{1} > t_{2}$
Time $t_{1}$ for stationary lift $= \sqrt{\frac{2 h}{g}}$
When lift is moving up with constant acceleration, then
$t_{2} = \sqrt{\frac{2 h}{g + a}}$
$∴ t_{1} > t_{2}$ .