An aluminium cube of side `20cm` floats in mercury. How much farther will the block sink when tempereture rises from `27^(@)C` to `77^(@)C` ? Density of aluminium and mercury at `27^(@)C` are `2.7` and `13.6g//cc` while the coeficient of volume expansion of mercury and linear expansion of aluminium are `1.8xx10^(-4)//^(@)C` and `23xx10^(-6)//^(@)C` respectively.

Explanation

KEY IDEA
As the temperature rises, the block expands and the density of the liquid decreases, so it like we are putting a bigger block of same weight in a lighter liquid . So , to apply the same buoyant force , a larger fraction of the volume should sink into the liquid .

Calculations : If side of cube, density of liquid (mercury) and length of the floating cube inside the liquid are L.
`sigma , h`, and L., `sigma. , h. ` at 300 K and 350 K , respectively , then by principle of floatation
`m =V_("in") sigma =V_("in")^(.) sigma^(.) ` (with `m = V_(rho) = rhoL^3`)
This is , as `V_("in") = hA = hL ^(2)`,
`h L^2sigma = h^(.) (L.) sigma^(.)`
or `[(h.)/h]=[L/L^(.)]^2xx[sigma/sigma^(.)]`
Now as L. `= L (1+ alpha_2DeltaT) and sigma^(.) = sigma (1+ gamma_L DeltaT)` , therefore ,
`[(h.)/h]=((1+gamma_LDeltaT))/((1+alpha_sDeltaT))=((1+gammaDeltaT))/((1+2alpha_sDeltaT))=1+(gamma_L-2alpha_s)DeltaT`
So, `Deltah, =(h.-h)=h(gamma_L-2alpha_s)xxDeltaT`
Also, from the equation `m=V_("in")sigma , ` we have
`rhoL^(3) = hL^(2) sigma `
That is , `h=[rho/sigma]L=[(2.7)/(13.6)]xx20=3.970cm`
So, `Deltah = 3.970 (1.8xx10^(-4)-2xx23xx10^(-6))(350-300)`
`= 2.66 xx10^(-2)cm`
As expected the block sinks deeper.