An aluminium sphere is immersed in water at 10^ ^ ^ C . If the...

The buoyant force on the completely immersed sphere is given by $F_B = V \rho_w g$, where $V$ is the volume of the aluminium sphere and $\rho_w$ is the density of water.

When the temperature is increased by $\Delta T$, the new volume of the sphere becomes $V' = V(1 + \gamma_{Al} \Delta T)$ and the new density of water becomes $\rho_w' = \dfrac{\rho_w}{1 + \gamma_w \Delta T}$, where $\gamma_{Al}$ and $\gamma_w$ are the coefficients of volume expansion of aluminium and water respectively.

The new buoyant force is F_B' = V' \rho_w' g = V \rho_w g \left(\dfrac{1 + \gamma_{Al} \Delta T}{1 + \gamma_w \Delta T} \right).

Since the coefficient of volume expansion of liquids is generally much greater than that of solids, $\gamma_w > \gamma_{Al}$.

Therefore, \dfrac{1 + \gamma_{Al} \Delta T}{1 + \gamma_w \Delta T}$< 1$, which implies $F_B' < F_B$.

Thus, the force of buoyancy will decrease.