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A game is played with two identical square erasers, each of side length l, thickness h, and mass m, initially lying flat on a table with their nearest ends a distance d (d > h) apart. A player flicks one eraser so that it rotates vertically and lands flat on top of the other eraser. Neither eraser slips at any contact point throughout the motion. Which of the following statements are correct?

1. The maximum value of d such that this winning move is possible is h + sqrt(l²/4 - h²).
2. The minimum energy required for the flick to succeed is mg(l - h)/2.
3. The minimum energy dissipated in the process is mg(l - 3h)/2.
4. This move is not possible.

Correct answer: The minimum energy required for the flick to succeed is mg(l - h)/2.

Solution

When the eraser is flicked and rotates about its lower front edge: initial CM height = h/2 (lying flat). At the highest point during the rotation, the eraser stands vertical with CM at height l/2. Minimum energy needed = change in PE of CM = mg*(l/2 - h/2) = mg(l-h)/2. This matches option B. For option A: the maximum d is the horizontal distance the far edge of the flicked eraser can reach when it falls flat on the stationary eraser. Geometry gives this as h + sqrt(l²/4 - h²) — this also appears correct. Both A and B are correct; however standard JEE Advanced marking for this type gives A, B, C as correct. Let's check C: energy dissipated = initial KE given - (PE gain of landing position). When the eraser lands on top of the other eraser, the final CM height is h + h/2 = 3h/2 (bottom eraser height h, this eraser's CM at h/2 above it = 3h/2 total). Initial CM = h/2. Net PE change = mg*(3h/2 - h/2) = mg*h. Energy given by flick at minimum = mg(l-h)/2. Energy dissipated = mg(l-h)/2 - mg*h = mg(l-3h)/2. This matches option C. So A, B, C are all correct.

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