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Question

A person in lift which ascends up with acceleration  $$10\ m /s^{2}$$ drops a stone from a height $$10\ m$$. The time of descent is $$(g = 10  m/s^{2})$$


A
1s
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B
2s
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C
1.5s
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D
0.5s
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Solution

The correct option is A $$1  s$$
Since, the lift is ascending up against the force of gravity with a = $$10  m/s^2$$, we have
Apparent weight of the man $$ M(g+a)$$ or the acceleration is $$20  m/s^2$$(also called pseudo acceleration).
Now, the stone is dropped from height $$ h=10  m$$ and hence time of descent $$t=\sqrt{\dfrac{2h}{g+a}}$$
$$t=\sqrt{\dfrac{2 \times 10}{20}}=1  sec$$

Physics

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