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In the figure shown, blocks `P` and `Q` are in contact but do not stick to each other. The lower face of `P` behaves as a plane mirror. The springs are in their natural lengths. The system is released from rest. Then the distance between `P` and `Q` when `Q` is at the lowest point first time will be A. `(2mg)/(K)`B. `(mg)/(K)`C. `(4mg)/(K)`D. 0 |
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Answer» Correct Answer - A Both blocks loose contact immediately after the release `T_(P) = 2pi sqrt((m)/(4K)) .T_(Q) = 2 pi sqrt((m)/(K))` `:. T_(Q) = 2T_(P)` Q comes at lowest position at time `(T_(Q))/(2)` travelling a distance `(2mg)/(K)` downwards. In time `(T_(Q))/(2)`, i.e. time period of `P(T_(P))` the block `P` come back to original position `:.` Time distance P and Q is `(2 mg)/(K)` |
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