Complete Question:
Two objects, T and B, have identical size and shape and have uniform density. They are carefully placed in a container filled with a liquid. Both objects float in equilibrium. Less of object T is submerged than of object B, which floats, fully submerged, closer to the bottom of the container. Which of the following statements is true?
- Object T has a greater density than object B.
- Object B has a greater density than object T.
- Both objects have the same density.
Answer:
Object B has a greater density than object T
Explanation:
Any object partially or completely submerged in a liquid, experiments an upward force, equal to the weight of the volume displaced by the liquid. This force is called the buoyant force, and can be expressed as follows:
Fb = ρl * Vs*g
where ρl is the density of the liquid, and Vs is the submerged volume.
This force must be compared with the weight of the object, which is always downward, and can be expressed as follows:
Fg = ρb* Vb * g
where ρb, is the density of the object, and Vb is the total volume of the object, regardless which portion is submerged.
For object B, as it floats fully submerged, this means that both forces are equal in magnitude:
Fg = Fb⇒ ρb* Vb * g = ρl * Vs*g
As Vb = Vs (the object is fully submerged) this means that ρb =ρl.
For object T, as it floats partially submerged, this means that Fg < Fb:
Fg= ρt* Vt * g < Fb = ρl * Vs*g.
Now, we know that ρb =ρl, so we can replace in the equation above:
ρT* Vt * g < ρb*Vs*g
Simplifying common terms, and replacing Vs by KVt (where K is the fraction of the total volume which is submerged, i.e. K<1), we have:
ρt*Vt < ρb*K*Vt ⇒ ρt / ρb < K < 1 ⇒ ρt < ρb ⇒ ρb > ρt
