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

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(D) Initially,the block floats with the coin. According to the law of floatation,the total weight of the block and the coin is balanced by the buoyant

Vedclass · Accepted Answer

Both $l$ and $h$ decrease

Vedclass · Answer

(D) Initially,the block floats with the coin. According to the law of floatation,the total weight of the block and the coin is balanced by the buoyant force,which is equal to the weight of the water displaced by the submerged part of the block. 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,it sinks (assuming its density is greater than water). The block now only needs to support its own weight $Mg$. The new volume of water displaced by the block is $V'_{disp} = M/\rho_w$. Since $V'_{disp} Regarding $h$,the total volume of water displaced initially was $V_{disp} = (M+m)/\rho_w$. After the coin falls,the block displaces $M/\rho_w$ and the coin displaces its own volume $V_c = m/\rho_c$. The new total volume of water displaced is $V'_{total} = M/\rho_w + m/\rho_c$. Since the density of the coin $\rho_c > \rho_w$,it follows that $m/\rho_c < m/\rho_w$. Therefore,$V'_{total} < V_{disp}$. As the total volume of water displaced decreases,the water level $h$ also decreases. Thus,both $l$ and $h$ decrease.

$A$ 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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