Floating in Mercury

consider 3rd case, replace water with more mercury and imagine the result, the answer will become pretty obvious.

Without the water, the weight of the mercury displaced by the ball and the "air" (or none) displaced by the ball equals the weight of the ball.

$W1_{Mercury}+W_{Air}=W_{Ball}$

With the water, the weight of the mercury displaced by the ball and the weight of the water displaced by the ball equals the weight of the ball.

$W2_{Mercury}+W_{Water}=W_{Ball}$

or

$W1_{Mercury}+W_{Air}=W2_{Mercury}+W_{Water}$

Neglecting $W_{Air}$ , we have

$W2_{Mercury}=W1_{Mercury}-W_{Water}$

so that the fraction of the ball in the mercury is reduced.

This assumes that 1) the ball is in hydrostatic equilibrium, and 2) is less dense than mercury but denser than both water and air.

Let's figure this out for a steel ball of unit radius

Let the relative densities of mercury, steel, water, and air be approximately $13.5, 8, 1, 0$ . Then, using the formula for the volume of a Spherical Cap , we have two equations, the first one with just mercury and air, and the second with mercury and water

$13.5\dfrac{\pi}{3}h^2(3r-h)+0\dfrac{\pi}{3}(2-h)^2(3r-(2-h))=8\dfrac{4\pi}{3}r^3$

$13.5\dfrac{\pi}{3}h^2(3r-h)+1\dfrac{\pi}{3}(2-h)^2(3r-(2-h))=8\dfrac{4\pi}{3}r^3$

with $h$ being $1.12409$ and $1.08017$ with the mercury and air and with mercury and water. Thus, relative to the level of the mecury, the steel ball is slightly higher when submerged in water.

In the equilibrium "before", the buoyant force is equal to the weight of displaced fluid (Archimedes' Principle).

If the ball would remain in place, the weight of displaced fluid would be greater "after", since water is more dense than air. Thus the buoyant force increases, forcing the ball upward.

Mercury is denser than water. If the ball floats in mercury, it will definitely float in a liquid less denser than mercury, in this case, water.

In the 'after' case, should the ball stay in the same position, the downwards pressure on the mercury would be higher than that on the ball because the water is shallower above the ball. Thus the ball would slightly rise.

The buoyant force before and after is the same, because in both cases the ball is supposed to be floating. Then, we have: $V_{i}\cdot \rho_{Hg}=V_{f}\cdot \rho_{Hg}+V_{water}\cdot \rho_{water}$ $\Rightarrow V_{f}=V_{i}-\frac{V_{water}\cdot \rho_{water}}{\rho_{Hg}}$ Where $V_{i}$ and $V_{f}$ are the initial and final volume of the ball in mercury, respectively. And we can, clearly see that the final volume is less than the initial volume, so we can deduce the ball has risen its position.

Part of the ball is now in water. It exoeriences upward force equal to volume of water displaced. Therefore goes up.

Buoyant force is always equal to the weight of the displaced fluid. In both cases, forces are balanced and thus the buoyant force equals the weight of the ball. The weight of the ball is constant, which means that the buoyant force must also be constant (as the ball has no net force in either case). In the second case, there is more total volume of fluid that the ball dispaces, and so it must displace LESS of the denser fluid which would have a greater weight per volume of displacement. The ball must rise, so that buoyant force can reimain constant.

SURFACE TENSION , MERCURY HAVE HIGH COHESIVE FORCE (FORCE BETWEEN MERCURY MOLECULES) THAN ADHESIVE FORCE (FORCE BETWEEN MERCURY AND GLASS) , THUS MERCURY DONT WET GLASS , WHEREAS WATER WET GLASS , THAT IS , WATER HAVE MORE ATTRACTION TOWARDS ANY MATERIAL (SUCH AS GLASS OR ANY BALL) THAN MERCURY THUS WHEN WATER IS POURED , WATER WILL ATTRACT THE BALL , THUS BALL WILL ARISE.

Suppose the position of the ball is constant after,then the part in the air turns into the part in the water whose density is greater.And because the buoyancy equals the gravity of the ball,it is impssible that the ball stay still which means that the total buoyancy after exceeds the gravity of the ball.As a result,the ball has to rise to reduce the buoyancy so that it equals the gravity of the ball.