# Thermal expansion and fluid statics

An aluminum cube floats in a bowl of liquid mercury at 0oC. If the temperature is raised to 27.5oC, by what percent will the fraction of volume submerged change?

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An aluminum cube floats in a bowl of liquid mercury at 0oC. If the temperature is raised to 27.5oC, by what percent will the fraction of volume submerged change?

Answer:

With increase in temperature the density of both aluminum and mercury changes.

The density of a substance at t0C is given by the formula

where is the coefficient of volume expansion of the substance and 0 is the density at 0oC.

Let at 00C the volume of the aluminum cube is V0 and the volume of the mercury displaced is V1, then according to law of flotation, the weight of the displaced liquid is equal to the weight of the body and hence

V1 0m g = V0 0 g

Where 0m is the density of mercury at 00C and 0 is that of aluminum. Gives the fraction of the volume submerged at 00C

f1 = V1/V0 = 0/0m

Let at t0C the new volume of the aluminum cube is V. And the density of mercury will be

where is the coefficient of volume expansion of mercury.

Hence the volume of cube submerged v2 is now given by

V2 m g = V g where is the density of aluminum at t0C, given by

where is the coefficient of volume expansion of aluminum.

The new fraction of the volume submerged is

f2 = V2/V = /m =

hence the percent change in the fraction of the volume submerged is given by

100*(f2 - f1)/f1 = 100[(f2/f1) - 1] = 100

= 100

= 0.3076 %

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