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(A) According to Archimedes' principle,the weight of the floating ice cube is equal to the weight of the water displaced by it.Let $M$ be the mass of

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It will fall

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(A) According to Archimedes' principle,the weight of the floating ice cube is equal to the weight of the water displaced by it.Let $M$ be the mass of the ice and $V_{air}$ be the volume of the air bubble.The total volume of the ice cube is $V_{ice} + V_{air}$.The weight of the ice cube is $W = Mg$.Since it is floating,the weight of the displaced water is equal to the weight of the ice cube: $W_{displaced} = Mg$.The volume of the displaced water is $V_{disp} = \frac{Mg}{\rho_w g} = \frac{M}{\rho_w}$,where $\rho_w$ is the density of water.When the ice melts,it turns into water of mass $M$,occupying a volume $V_{melted} = \frac{M}{\rho_w}$.However,the air bubble was previously displacing water equal to its own volume $V_{air}$ to keep the cube afloat.When the ice melts,the air bubble is released,and the water level drops because the volume previously occupied by the air bubble $(V_{air})$ is no longer displacing water.Therefore,the water level will fall.

An ice cube contains a large air bubble. The cube is floating on the horizontal surface of water contained in a trough. What will happen to the water level when the cube melts?

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