A container of large surface area is filled with liquid of density \[\rho \]. A cubical block of side edge a and mass M is floating in it with four-fifth of its volume submerged. If a coin of mass m is placed gently on the top surface of the block is just submerged. M is
A. \[4m/5\]
B. \[m/5\]
C. \[4m\]
D. \[5m\]

Hint: Archimedes’s principle states that the weight of the body that is partially or completely submerged in the liquid is equal to the weight of the liquid displaced by the volume of the body. Using Archimedes’s principle, express the weight of the displaced fluid by four-fifth of the volume of the block and again express the weight of the displaced fluid by the total volume of the block. Solve the two equations to get the mass of the block.

Formula used:
Buoyant force or weight of the liquid displaced, \[{F_B} = V{\rho _l}g\],
where, V is the volume, \[{\rho _l}\] is the density of the liquid and g is the acceleration due to gravity.

Complete step by step answer:
We know that from Archimedes’s principle, the weight of the body that is partially or completely submerged in the liquid is equal to the weight of the liquid displaced by the volume of the body. Let the volume of the block is V. We have given that the four-fifth of the block is immersed in the liquid.
Using Archimedes’s principle, we can write,
\[Mg = \dfrac{4}{5}V{\rho _l}g\] …… (1)
Here, M is the mass of the block, g is the acceleration due to gravity and \[{\rho _l}\] is the density of the liquid.

Now, when the coin of mass m is placed gently on the top of the block, the whole block is just submerged in the liquid. Again, using Archimedes principle for this case, we get,
\[\left( {M + m} \right)g = V{\rho _l}g\] …… (2)
Dividing equation (1) by equation (2), we get,
\[\dfrac{{Mg}}{{\left( {M + m} \right)g}} = \dfrac{{\dfrac{4}{5}V{\rho _l}g}}{{V{\rho _l}g}}\]
\[ \Rightarrow \dfrac{M}{{M + m}} = \dfrac{4}{5}\]
\[ \Rightarrow \dfrac{4}{5}\left( {M + m} \right) = M\]
\[ \Rightarrow \dfrac{4}{5}m = \dfrac{1}{5}M\]
\[ \therefore M = 4m\]
Thus, the mass of the block is four times the mass of the coin.

So, the correct answer is option C.

Note:Archimedes’s principle also states that the weight of the liquid displaced by the body is equal to the buoyant force provided by the liquid. The weight of the liquid displaced consists of the volume of the body and density of the liquid. Students often take the density of the body in the buoyant force term which is the wrong expression.