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(C) Let the volume of the block be $V$. When it is placed in the liquid,four-fifths of its volume is submerged.According to the principle of floatatio

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$4m$

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(C) Let the volume of the block be $V$. When it is placed in the liquid,four-fifths of its volume is submerged.According to the principle of floatation,the weight of the block equals the weight of the liquid displaced:$Mg = \frac{4}{5} V \rho g$  --- $(I)$When a coin of mass $m$ is placed gently on the top surface of the block,the block is just submerged. This means the total weight of the block and the coin is equal to the weight of the liquid displaced by the entire volume $V$ of the block:$(M + m)g = V \rho g$  --- $(II)$Dividing equation $(I)$ by equation $(II)$:$\frac{Mg}{(M + m)g} = \frac{\frac{4}{5} V \rho g}{V \rho g}$$\frac{M}{M + m} = \frac{4}{5}$$5M = 4(M + m)$$5M = 4M + 4m$$M = 4m$

$A$ container of large surface area is filled with a liquid of density $\rho$. $A$ cubical block of side edge $a$ and mass $M$ is floating in it with four-fifths of its volume submerged. If a coin of mass $m$ is placed gently on the top surface of the block,it is just submerged. $M$ is:

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