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A pendulum clock is mounted in an elevator car which starts going up with a constant acceleration `w`, with `w lt g`. At a height `h` the acceleration of the car reverses, its magnitude remaining constant. How soon after the start of the motion will the clock show the right time again ? |
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Answer» Correct Answer - `sqrt((2h)/(a)) sqrt((1+n)-sqrt(1-n))/(1-sqrt(1-n)),"where" n=a//g` |
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