Problem Statement
A sphere of aluminium of radius 1 cm is placed in a container of mercury (both at 20°C). Mercury and aluminium have the same density at 20°C. When the temperature is raised to 100°C, will the sphere sink or float? $\gamma_{Al}=7.2\times10^{-5}$ /°C, $\gamma_{Hg}=1.82\times10^{-4}$ /°C.
Given Information
- $r=1\,\text{cm}$, $T_0=20\,\text{°C}$
- $\Delta T=80\,\text{°C}$
- $\gamma_{Al}=7.2\times10^{-5}\,\text{/°C}$
- $\gamma_{Hg}=1.82\times10^{-4}\,\text{/°C}$
- Initial: $\rho_{Al}=\rho_{Hg}$
Physical Concepts & Formulas
If the sphere expands less than mercury over the same temperature rise, the mercury becomes less dense relative to the sphere and the sphere sinks. Compare $\gamma_{Al}$ with $\gamma_{Hg}$.
- $\rho(T)=\rho_0/(1+\gamma\Delta T)$
- Sphere sinks if $\rho_{Al}(100°)>\rho_{Hg}(100°)$, i.e. if $\gamma_{Al}<\gamma_{Hg}$
Step-by-Step Solution
Step 1 — New density of aluminium: $\rho_{Al}$ decreases.
$\rho_{Al}(100)=\rho_0/(1+7.2\times10^{-5}\times80)=\rho_0/(1+5.76\times10^{-3})$
Step 2 — New density of mercury: $\rho_{Hg}$ also decreases.
$\rho_{Hg}(100)=\rho_0/(1+1.82\times10^{-4}\times80)=\rho_0/(1+1.456\times10^{-2})$
Step 3 — Compare densities: Mercury expands more, so $\rho_{Hg}(100)<\rho_{Al}(100)$.
$\rho_{Al}(100)>\rho_{Hg}(100) \Rightarrow \text{sphere sinks}$
Worked Calculation
$$\gamma_{Al}=7.2\times10^{-5}<\gamma_{Hg}=1.82\times10^{-4}\Rightarrow\text{Al expands less, Al denser than Hg at 100°C}$$
Answer
$$\boxed{\text{The aluminium sphere sinks}}$$
The aluminium sphere sinks because mercury expands more than aluminium, making the mercury less dense than the sphere at 100°C.
Physical Interpretation
At 20°C the densities are equal. At 100°C, mercury has expanded by 1.456% and aluminium by only 0.576%, so mercury is now less dense and no longer supports the sphere. This illustrates buoyancy as density ratio, not absolute density.
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