<u>Answer:</u>
According to the <u>Archimedes’ Principle</u>, <em>a body totally or partially immersed in a fluid at rest, experiences a vertical upward thrust equal to the mass weight of the body volume that is displaced.</em>
In this case we have an object partially immersed in water.
If we observe in <u>the image attached</u>, this object does not completely fall to the bottom because the net force acting on it is zero, this means <u>it is in equilibrium</u>.
This is due to <u>Newton’s first law of motion:</u> if a body is in equilibrium the sum of all the forces acting on it is equal to zero
(1)
Here, <u>the object experiences two kind of forces</u>:
A force downwards because of its weight:
<h2>
(2)
</h2>
Where
is the mass of the object and
the acceleration of gravity, which in this case (on earth) is 
And a force applied upwards because of the buoyancy:
<h2>

(3)
</h2>
Where;
is the <u>density of Water</u> which is
And
is <u>the volume of the immersed part of the object</u>
Well, if the total net force exerted to the object is zero:
Note
has a negative sign because this force is applied downwards
This means the buoyancy is equal to the weight:
<h2>

(4)
</h2>
(5)
Dividing by
in both sides:
(6)
Now, <u>the density</u> of an object or fluid is the relation between the mass and the volume, in the case of the object of this problem is:
<h2>

(7)
</h2>
Where
is the <u>Volume of the object</u>.
From (7) we can find the mass of the object:
<h2>

(8)
</h2>
and substitute this value in (6), in order to get a relation between the density of the object and the density of the water, as follows:
(9)
<h2>

</h2><h2 />
Then;
<h2>

</h2><h2 />
This result is the same as
which represents the 
Therefore the answer is:
<h2>

of the mass of the object is below the water line or <u>immersed in water
</u></h2>