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A cubical block of side \[0.5m\] floats on water with \[30\% \] of its volume under water. What is the maximum weight that can be put on the block without fully submerging it underwater? (take density of water \[ = {10^3}kg/{m^3}\])
(A) \[65.4kg\]
(B) \[87.5kg\]
(C) \[30.1kg\]
(D) \[46.3kg\]

Last updated date: 20th Jun 2024
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Hint: The extra mass needed to be added must at maximum be equal to the mass of the water displaced. The volume submerged is equal to the volume rise (or displaced) of the water.
Formula used: In this solution we will be using the following formulae;
\[m = \rho V\] where \[\rho \] is density and \[V\] is volume

Complete Step-by-Step Solution:
A cubical block is said to float on water with a particular percentage of the volume submerged. Generally, if we keep adding mass on this cubical block, more volume of the block will be submerged. at a certain mass, the block will just almost be completely submerged such that any extra mass added will have the block sinking. We are to determine this maximum mass above which the block becomes completely submerged.
To do so, we must note that an object placed in a fluid experiences a buoyancy effect which is equal to the weight of the liquid displaced by the object. For the extra weight which would be submerged, we have that
\[{W_o} = {W_w}\]
But \[W = mg\]
since \[m = \rho V\] where \[\rho \] is density and \[V\] is volume. Hence,
\[W = {m_o}g = {m_w}g = {\rho _w}{V_w}g\] where the subscript O and w signifies object and water respectively
\[{m_o} = {\rho _w}{V_o}\] where \[{V_o}\] is the maximum volume of an object to be submerged.
Since, 30 percent is submerged, then 70 percent more is the maximum to be submerged. hence,
\[{m_o} = {\rho _w}\left( {0.7{L^3}} \right)\] since the volume of a cube is \[{L^3}\].
Hence, by inserting all known values, we get
\[{m_o} = 1000\left( {0.7{{\left( {0.5} \right)}^3}} \right)\]
\[{m_o} = 87.5kg\]

Hence, the correct option is B

Note: For clarity, from
\[W = {m_o}g = {m_w}g = {\rho _w}{V_w}g\]
\[ \Rightarrow m = {\rho _w}{V_w}\] , the equation became
\[m = {\rho _w}{V_o}\].
This can occur because a special dimension must be conserved. When the object is dipped, it must occupy a space, and hence, the water must displace by the exact same dimension of space. So we say that the volume submerged is equal to the liquid displaced.