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

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(A) Initially,the block and the coin float together. The total weight of the block and the coin is balanced by the buoyant force,which is equal to the

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$l$ decreases and $h$ decreases

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(A) Initially,the block and the coin float together. The total weight of the block and the coin is balanced by the buoyant force,which is equal to the weight of the displaced water. Let $M$ be the mass of the block and $m$ be the mass of the coin. The total weight is $(M+m)g$. The volume of water displaced is $V_{disp} = (M+m)/\rho_w$,where $\rho_w$ is the density of water.When the coin falls into the water,the block alone floats. The weight of the block is $Mg$. The volume of water displaced by the block is $V'_{disp} = M/\rho_w$. The coin sinks to the bottom and displaces a volume of water equal to its own volume,$V_{coin} = m/\rho_c$,where $\rho_c$ is the density of the coin.The total volume of water displaced is now $V_{total} = V'_{disp} + V_{coin} = M/\rho_w + m/\rho_c$. Since the density of the coin $\rho_c$ is much greater than the density of water $\rho_w$,$m/\rho_c The submerged depth of the block $l$ is proportional to the weight of the block alone $(Mg)$,which is less than the weight of the block plus the coin $(Mg+mg)$. Thus,the submerged depth $l$ also decreases.

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

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