A cube with dimension w= 10cm and specific density 2.0 is dropped into a beaker. The beaker has 3 fluids (water, oil, and glop) of specific density 1.0, 1.8, and 2.5. The fluids do not mix. The thickness of each fluid is greater than w. Determine where the cube will be in static equilibrium.

Explanation:

The cube will be partially dipped in the density of 1.8 and 2.5 because the density of cube is in between 1.8 and 2.5 .

Let the cube is completely dipped with thickness d in density of 1.8  and thickness 10-d in liquid with specific  density of 2.5

Let cross sectional area be A

volume of cube = A x 10

weight = 10A x 2  x g

upthrust by density 1.8

= A X d x g x 1.8

upthrust in density 2.5

= A X (10- d) x g x 2.5

For equilibrium

Total upthrust = total weight.

A X d x g x 1.8 +A X (10- d) x g x 2.5 = 10A 2  x g

1.8d + 25 - 2.5 d = 20

.7d = 5

d = 7.14 cm

so cube will float with 7.14  cm in specific density of 1.8 and 2.86 cm in the specific density of 2.5 .  

atomic = no of protons.

mass number-atomic number = neutron number