Question

A cube of wood supporting \[200\,{\text{gm}}\] mass just in water (\[\rho = 1\,{\text{g/cc}}\]). When the mass is removed, the cube rises by \[2\,{\text{cm}}\]. The volume of cube is
A. \[1000\,{\text{cc}}\]
B. \[800\,{\text{cc}}\]
C. \[500\,{\text{cc}}\]
D. None of these

Answer 1

Hint:Use the Archimedes’ principle for the floating or immersed objects. This principle will give the relation between the density of the water, mass of the cube and the extra mass on the cube and the edge of the cube. Also use the formula for volume of a cube to determine the volume of cube in the water.

Formulae used:
The density \[\rho \] of an object is given by
\[\rho = \dfrac{M}{V}\] …… (1)
Here, \[M\] is the mass of the object and \[V\] is the volume of the object.
The volume \[V\] of a cube is given by
\[V = {a^3}\] …… (2)
Here, \[a\] is the length of the edge of the cube.

Complete step by step answer:
The cube of wood is supporting a mass of \[200\,{\text{g}}\] in the water. According to the Archimedes’ principle, the weight of the liquid displaced by an object floating or immersed in the liquid is equal to the weight of the object immersed in the liquid. Rewrite equation (1) for the density of the wood in the water when the wood is supporting the mass.
\[{\rho _{wood}} = \dfrac{{{M_{wood}}}}{{{V_{wood}}}}\]
Rearrange the above equation for \[{M_{wood}}\].
\[{M_{wood}} = {\rho _{wood}}{V_{wood}}\] …… (3)
The volume \[{V_{wood}}\] of the wood block is
\[{V_{wood}} = {a^3}\] …… (4)
Here, \[a\] is the length of the edge of the wood block.
Substitute \[{a^3}\] for \[{V_{wood}}\] in equation (3).
\[{M_{wood}} = {\rho _{wood}}{a^3}\]

The weight of the water displaced due to the block in the water is equal to the weight of the block and the weight of the mass.
\[{M_{mass}}g + {M_{wood}}g = {M_{water}}g\]
The mass of the water displaced is also equal to the mass of the wood.
Substitute \[200\,{\text{g}}\] for \[{M_{mass}}\], \[{\rho _{wood}}{a^3}\] for \[{M_{wood}}\] and \[{\rho _{water}}{a^3}g\] for \[{M_{water}}\] in the above equation.
\[\left( {200\,{\text{g}}} \right)g + {\rho _{wood}}{a^3}g = {\rho _{water}}{a^3}g\]
\[ \Rightarrow \left( {200\,{\text{g}}} \right) + {\rho _{wood}}{a^3} = {\rho _{water}}{a^3}\]
Rearrange the above equation for \[{\rho _{wood}}{a^3}\].
\[{\rho _{wood}}{a^3} = {\rho _{water}}{a^3} - 200\] …… (5)

The edge of the wood rises by 2 cm above the water when the mass is removed.Then the volume of the cube in the water also changes which is \[{a^2}\left( {a - 2} \right)\].Hence, the weight of the wood immersed in the liquid is equal to the weight of the water displaced.
\[{M_{wood}}g = M_{water}^1g\]
Substitute \[{\rho _{wood}}{a^3}\] for \[{M_{wood}}\] and \[{\rho _{water}}{a^2}\left( {a - 2} \right)\] for \[M_{water}^1\] in the above equation.
\[{\rho _{wood}}{a^3}g = {\rho _{water}}{a^2}\left( {a - 2} \right)g\]
\[ \Rightarrow {\rho _{wood}}{a^3} = {\rho _{water}}{a^2}\left( {a - 2} \right)\]
Substitute \[{\rho _{water}}{a^2}\left( {a - 2} \right)\] for \[{\rho _{wood}}{a^3}\] in equation (5).
\[{\rho _{water}}{a^2}\left( {a - 2} \right) = {\rho _{water}}{a^3} - 200\]
\[ \Rightarrow {\rho _{water}}{a^3} - {\rho _{water}}{a^2}\left( {a - 2} \right) = 200\]

Substitute \[1\,{\text{g/cc}}\] for \[{\rho _{water}}\] in the above equation.
\[\left( {1\,{\text{g/cc}}} \right){a^3} - \left( {1\,{\text{g/cc}}} \right){a^2}\left( {a - 2} \right) = 200\]
\[ \Rightarrow {a^3} - {a^2}\left( {a - 2} \right) = 200\]
\[ \Rightarrow {a^3} - {a^3} + 2{a^2} = 200\]
\[ \Rightarrow {a^2} = 100\]
\[ \Rightarrow a = 10\,{\text{cm}}\]
Hence, the length of the edge of the wood is \[10\,{\text{cm}}\].
Calculate the volume of the cube.Substitute \[10\,{\text{cm}}\] for \[a\] in equation (4).
\[{V_{wood}} = {\left( {10\,{\text{cm}}} \right)^3}\]
\[ \therefore {V_{wood}} = 1000\,{\text{cc}}\]
Therefore, the volume of the cube is \[1000\,{\text{cc}}\].

Hence, the correct option is A.

Note:Since the cube rises by 2 cm when the mass is removed, the volume of the cube in the water changes to \[{a^2}\left( {a - 2} \right)\] as only dimension of only one edge is changed and the other two are the same. As all the units are in the CGS system of units, there is no need to convert the units of physical quantities from CGS to SI systems of units.