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A small container (1) of height 2*l0 and cross-sectional area A sits inside a big container (2) of height 3*l0 and cross-sectional area 2A, with their bottoms welded together. A massless piston rests on container (1). Container (2) is completely filled with water; initially the air column in container (1) has length l0. A small orifice is then opened at the bottom of container (2). When the water level in container (2) drops to height 2*l0, the piston just reaches the top of container (1). Atmospheric pressure equals N * rho_w * g * l0. Find N. (rho_w = density of water)

1. 1
2. 2
3. 3
4. 4

Correct answer: 2

Solution

Initial state: air length l0 in container (1). Water fills container (2) to height 3*l0. The water also fills the space above the piston in container (1) from l0 to 2*l0. Pressure on piston from above = pressure of water at height l0 = P_atm + rho_w*g*(3*l0 - l0) = P_atm + 2*rho_w*g*l0. This equals the initial air pressure P1. Final state: piston at 2*l0 (top of container 1), air length 2*l0. Water level in container (2) is exactly 2*l0, same as piston height, so no water above piston: final air pressure P2 = P_atm. Boyle's law: P1 * l0 = P2 * 2*l0 => P1 = 2*P_atm. Equating: P_atm + 2*rho_w*g*l0 = 2*P_atm => P_atm = 2*rho_w*g*l0 = N*rho_w*g*l0 => N = 2.

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