A tank conatins iodine and water as shown in the adjoining figure. As aluminium cube of side 5cm is the equlibrium as shown. Calculate the fraction of volume of cube inside the iodine? Take
Relative density of iodine =4.927
Relative density of Aluminium =2.7

Explanation

Let the cube be x m inside the iodine. Then, depth of cube in water =(0.05 -x)m
Using Archimedes. principle, buoyant force on cube due to iodine is
`F_(1)=(0.05)^(2) xx x 4.927 xx 10^(3) xx 9.8N`
=120.71x
Buoyant force on cube due to water is
`F_(2)=(0.05)^(2) xx (0.05 -x) xx 10^(3) xx 9.8N`
=1.225 -24.5 x
When the cube is in equilibrium, `F_(1)+F_(2)="Weight of aluminium cube "`
`120.71 x+1.225 -24.5x=(0.05)^(3) xx 2.7 xx 10^(3) xx 9.8`
`rArr =96.21x=2.085`
`rArr x=0.022 m =2.2cm`