Question

Write the formula for buoyant force in a liquid and explain the terms in it. A, B, and C are three science students. They have $3$ objects of relative densities $0.9,1,$ and $1.5$ but with the same volume. Student A took the object with RD $=0.9$, student B took the object with RD $=1$, and student C took the last object. All three of them completely immersed their objects in water and removed their hands. What can be the observations of the three students about the force of buoyancy acting on the three objects:

1. When they immersed the objects completely in water.
2. When they removed their hands after immersing the objects.

Answer 1

${\text{F}}_{\text{B}}={\text{V}}_{\text{liq disp}}\times {\text{d}}_{\text{liq}}$
${\text{F}}_{\text{B}}=$ Force of buoyancy on the object
${\text{V}}_{\text{liq disp}}=$ Volume of liquid displaced
${\text{d}}_{\text{liq}}$ = Density of liquid in which the object is immersed

According to Archimedes' principle, the force of buoyancy acting on an object is equal to the weight of the fluid displaced by the object, whether it is partially or fully immersed in the fluid.

It is given that all the objects have same volume.
In the first case, all the objects are completely immersed in the water. So, the volume of water displaced will be the same. Also, density of water is $1\text{g / cc}$. Therefore, the weight of water displaced is also same for all the three objects.

When the hands are removed, due to the difference in the RD of the objects, student A's object will float on water, student B's object will get submerged completely, and student C's object will completely sink to the bottom.

In this case, the buoyancy force acting on the objects of students B and C will be the same because they have same volume immersed in water. But student A's object will experience lesser buoyant force than the other two because its RD is lesser than 1 and it floats on the surface.