Physics buoyancy problem

A block of wood is placed into some water and floats when the sides are submerged by 8cm. The same block of wood is placed into an unknown liquid and floats when the sides are submerged by 6.2cm. What is the density of the unknown liquid?

Hello Jane,

Just in case your teacher wants you to SHOW that the depth is inversely proportional to the density..... You might want to know haw to do the problem correctly.

The Buoyant (B) force is equal to the weight of the displaced liquid. And because of equilibrium, we know the buoyant force is equal to the weight of the block of wood (mg)

So B = mg for both situations.

Lets call Dw the density of water and Du the density of the unknown liquid. Lets call Hw the depth the block sinks to in the water and Hu the depth that it sinks to in the unknown liquid. And the cross sectional area of the block is A, of course.

So in the water, the volume of liquid displaced by the block is Hw * A and the mass of the displaced water is Dw * Hw * A, so the weight of the displaced water is mass times gravity so....

B = Dw * Hw * A * g

And for the unknown liquid, the weight of the displaced liquid is

B = Du * Hu * A * g

Now we can set them equal

Dw * Hw * A * g = Du * Hu * A * g

Cancel out the A's and the g's

Dw * Hw = Du * Hu

Du = (Hw/Hu) * Dw

Du = (8 cm / 6.2 cm) * 1000 kg/m^3

Du = 1290 kg/m^3

Good Luck

The amount submerged is inversely proportional to the density
Density of water/density of unknown fluid = 6.2/8
or density of unknown fluid = (8/6.2)*density of water