Question

for two liquids having different density the buoyant force is​

Answer 1

Explanation:

There's a key factor you're missing when you're considering the added weight on the ball.

The fluid B wont just be pushing down on the top of the object floating in liquid A, it also pushes down on liquid A itself. Due to the hydrostatic principle, this therefore increases the pressure of liquid A. This added pressure increases the force that liquid A is exerting on the object in the upwards direction. This is already enough to cancel out the effects of the added weight on top of the object.

As far as the additional height goes, this is because fluid B is more dense than air. Using Archimedes' principle, we can see that this should necessarily increase the buoyant force on the object because

FBuoyant=ρfluidVg

(where V is volume displaced by the submerged object, ρfluid is the density of the fluid and g is acceleration due to gravity.

We can see that when the density of the surrounding fluid increases (and it should be safe to assume fluid B is more dense than the air formerly above fluid A), the buoyant force increases, which will result in more lift on the submerged object. As it rises, the effective density of the fluid will decrease, as more and more of the object gets surrounded by the less dense fluid B, instead of fluid A. (Buoyancy then becomes Fbuoyant=(ρAVA+ρBVB)g where VA and VB are the volumes that each fluid acts on respectively) If fluid B is more dense than the object, it will continue to rise until it floats on fluid B; if fluid B is less dense than the object, it will rise until Fbuoyant=Fgravity and it reaches an equilibrium position between the two fluids.