The cubes of equal mass one made of iron and the other of aluminium are immersed in water and weighed under such case
A) The weight of the aluminium cube will be less than that of the iron cube.
B) The two weights will be equal
C) The weight of the iron cube will be less than that of the aluminium cube
D) The data provided is insufficient

Hint:When a body of the same mass is immersed or partially immersed in the water then a buoyant force starts acting on the body. If the body is floating then this buoyant force is equal to the weight of the body.
In the given problem, we have two different cubes of different metals. One is iron and the other one is aluminium. The masses of both cubes are equal. So, if these cubes are immersed in the water a force starts acting on both cubes in the opposite direction of the weight of the cubes. Hence, the net force will depend only on the densities of the cubes.

Formula used:We applying Archimedes principle so the formula of net weight will be-
Net weight$ = mg - $buoyant force
Net weight $ = mg - m'g$
But \[m' = {\rho _w} \bullet v\]
Where \[{\rho _w}\]is the density of the water.

Complete step by step solution: -
When a body is immersed in water its weight is greater than the buoyant force experienced by the body.
In this question, we have an aluminium cube and an iron cube of the same mass. So the weights of the both cubes will be equal.
According to the Archimedes when a body immersed or partially immersed in a liquid it displaced the volume of liquid equal to the volume of the immersed portion of the body.
In the above question, aluminium and iron cubes are immersed in the water. So a buoyant force experienced by both cubes.
If ${\left( {{F_B}} \right)_I}$ and ${\left( {{F_B}} \right)_A}$ be the buoyant forces experienced by iron and aluminium cubes respectively and \[{\rho _w}\] be the density of water. \[{\rho _I}\] and \[{\rho _A}\] be the densities of iron cube and aluminium cube respectively. Let ${V_I}$ and ${V_A}$ be the volume of water displaced by iron and aluminium cubes respectively.
So, net force on iron cube-
$ \Rightarrow {F_I} = mg - {\left( {{F_B}} \right)_I}$
Here ${\left( {{F_B}} \right)_I} = m'g$ where $m'$ is the mass of the cube on water
$ \Rightarrow {F_I} = mg - m'g$
We know that $m' = {\rho _w}{V_I}$
$
   \Rightarrow {F_I} = mg - {\rho _w}{V_I}g \\
   \Rightarrow {F_I} = g\left( {m - {\rho _w}{V_I}} \right) \\
$
${V_I}$Is the volume of the cube. So ${V_I} = \dfrac{m}{{{\rho _I}}}$
\[ \Rightarrow {F_I} = g\left[ {m - {\rho _w}\dfrac{m}{{{\rho _I}}}} \right]\]
\[ \Rightarrow {F_I} = mg\left[ {1 - \dfrac{{{\rho _w}}}{{{\rho _I}}}} \right]\] ……………………..(i)
Similarly, we can find the net force for r aluminium cube which will be-
\[ \Rightarrow {F_A} = mg\left[ {1 - \dfrac{{{\rho _w}}}{{{\rho _A}}}} \right]\] ………..(ii)
Now comparing equation (i) and (ii) by remembering that iron is denser than aluminium.
\[{\rho _I} > {\rho _A}\]
So, ${F_I} > {F_A}$
Hence, the weight of the iron cube will be greater than the weight of the aluminium cube.

Therefore, option A is correct.

Note: -The density of iron cube is greater than the density of the aluminium cube. So the weight of the iron cube is greater than the aluminium cube whenever the mass is the same or not. If we have critical velocity then we can also find the viscosity of both cubes.