A uniform cube of mass `M` is floating on the surface of a liquid with three fourth of its volume immersed in the liquid (density of liquid is `rho`). Now a cubical cavity is made inside the cube and the cavity is filled by a material having twice the density of the original cube. The minimum dimension (sides) of this cavity such that this cube is now completely immersed in the liquid is `[(4M)/(nrho)]^(1/3)`. Find the value of `n`

Answer

9

Explanation

Side of cube `=a`
`Mg=3/4a^(3)rhog`
If side of cavity is `b`.
New mass of cube
`=(a^(3)-b^(3))3/4rho+b^(3)6/4rho=(a^(3)+b^(2))3/4rho`
`a^(3)rhog=(a^(3)+b^(3))3/4rhog`
`b^(3)=(a^(3))/3,b=[1/3]^(1/3)a`
`b=[4/9M/(rho)]^(1/3)`