91Ó°ÊÓ

The fraction submerged is equal to the ratio of the average density of the object to the density of the fluid in which the object is submerged. This can be mathematically expressed as;

$\mathbf{fractionsubmerged}\mathbf{=}\frac{{\mathbf{ÒÏ}}_{\mathbf{obj}}}{{\mathbf{ÒÏ}}_{\mathbf{fl}}}$…â¶Ä¦â¶Ä¦â¶Ä¦(1)

Here,${\mathit{ÒÏ}}_{\mathbf{o}\mathbf{b}\mathbf{j}}$is the density of the object and ${\mathit{ÒÏ}}_{\mathbf{f}\mathbf{l}}$ is the density of the fluid. The density of the fluid varies with the temperature. So, the effect of temperature on Archimedes’ principle can be understood by comparing the fraction of submerged copper in 0°C of water with 0°C of water.

The density of water at 0°C is 1000 Kg/m³. We need to calculate the density at 95°C. Density is the ratio of mass to volume. The mass of water will be the same as at 0°C, that is, 1000 Kg. But the volume changes due to thermal expansion. This change in volume can be calculated as;

$\mathrm{Δ}V=\mathrm{β}V\mathrm{Δ}T$…â¶Ä¦â¶Ä¦â¶Ä¦â¶Ä¦(2)

Here, is the volume change, is the coefficient of volume expansion, is the initial volume, andis the temperature change.

In water,

$$\begin{gathered} \mathrm{β}=210×{10}^{--6} °C^{-1} \\ V=1 m^3 \\ \mathrm{Δ}T=95°C-0°C \\ =95°C \end{gathered}$$

Substitute these values in equation (2).

$$\begin{gathered} \mathrm{Δ}V=\left(210×{{10}^{-}-}^6 °C^{-1}\right)×1 m^3×95°C \\ =0.01995 m^3 \end{gathered}$$

Therefore,

role="math" localid="1655701794625" $$\begin{gathered} \text{density of water at 95}°C=\frac{mass}{v+\mathrm{Δ}v} \\ =\frac{1000Kg}{1.01995m^3} \\ =980.440Kg/m^3 \end{gathered}$$

Fraction submerged can be calculated in both situations by substituting the corresponding values in equation (1),

$$\begin{gathered} \text{fraction submerged at}0°C=\frac{{\mathrm{ÒÏ}}_{copper}}{1000Kg/m^3} \\ \text{fraction submerged at95}°C=\frac{{\displaystyle {\mathrm{ÒÏ}}_{copper}}}{{\displaystyle 980.440Kg/m^3}} \end{gathered}$$

Divide these two equations.

$$\begin{gathered} \frac{\text{fraction submerged at}0°C}{\text{fraction submerged at95}°C}=\frac{980.440Kg/m^3}{1000Kg/m^3} \\ =0.980 \end{gathered}$$

The fraction of copper submerged at 0°C of water is 0.980 times greater than the copper submerged at 95°C.

Therefore, there is a small variation in the fraction submerged with a temperature rise.