A wooden block, with a coin placed on its top | vedclass

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(D) Initially, the block and the coin float together. The total weight is $W = (M_{block} + m_{coin})g$. According to Archimedes' principle, the buoya

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both $l$ and $h$ decrease

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(D) Initially, the block and the coin float together. The total weight is $W = (M_{block} + m_{coin})g$. According to Archimedes' principle, the buoyant force equals the total weight, so the volume of water displaced is $V_{displaced} = (M_{block} + m_{coin}) / \rho_{water}$.When the coin falls into the water, it sinks to the bottom. The block now only supports its own weight $M_{block}g$. The new volume of water displaced by the block is $V'_{block} = M_{block} / \rho_{water}$.The coin, now at the bottom, displaces water equal to its own volume $V_{coin} = m_{coin} / \rho_{coin}$.Since the density of the coin $\rho_{coin}$ is much greater than the density of water $\rho_{water}$, the volume of water displaced by the coin when submerged $(m_{coin} / \rho_{coin})$ is less than the volume of water it displaced when floating on the block $(m_{coin} / \rho_{water})$.Therefore, the total volume of water displaced decreases, causing the water level $h$ to decrease.Since the block now supports less weight, it rises, meaning the submerged depth $l$ of the block decreases.

$A$ wooden block, with a coin placed on its top, floats in water as shown in the figure. The distances $l$ and $h$ are shown. After some time, the coin falls into the water. Then:

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