Question

A steel ball floats in a vessel with mercury.How will the volume of the part of the ball submerged in mercury change if a layer of water
completely converging the ball is poured above the mercury?

Answer 1

Let the volume of the steel ball be $V$, and let the volume o fits part immersed in mercury be $v_0$ before water is poured and $V_1$ after water covers the ball completely. The value of $V_0$ can be found from the condition
${\rho }_{st}V={\rho }_{mer}{V}_0$,
where ${\rho }_{st}$ and ${\rho }_{mer}$ are the densities of steel and mercury. Since the pressure of water is transmitted through mercury to the lower part of the baal, the buoyant force exerted on it by water is ${\rho }_w(V-V_1)g$, where ${\rho }_w$ is the density of water, while the buoyancy of mercury is ${\rho }_{mer}{V}_{1g}$. The condition of floating for the ball now becomes
${\rho }_{st}V={\rho }_{mer}{V}_1+{\rho }_w(V-V_1)$
whence
$V_1=\frac{{\rho }_{st}-{\rho }_w}{{\rho }_{mer}-{\rho }_w}V$
Thus, the ratio of the volumes of the parts of the ball submerged in mercury in the former and latter cases is
$\frac{V_0}{V_1}=\frac{{\rho }_{st}}{{\rho }_{mer}}\frac{{\rho }_{mer}-{\rho }_w}{{\rho }_{st}-{\rho }_w}=\frac{1-{\rho }_w/{\rho }_{mer}}{1-{\rho }_w/{\rho }_{st}}$
Since ${\rho }_{mer}>{\rho }_{st},V_0>V_1$ i.e. the volume of the parts of the ball immersed in mercury will becomes smaller when water is poured.