According to Archimedes'principle, the mass of a floating object equals the mass of thefluid displaced by the object. Use the principle to solve thefollowing problems:
a. A wodden cylinder 30cm high floats verticallyin a tub of water (density = 1 g/cm3). The top ofthe cylinder is 14.1cm above the surface of the liquid. What is thedensity of the wood?
b. The same cylinder floats vertically in a liquid of unknowndensity. The top of the cylinder is 20.7cm above the surface of theliquid. What is the liquid density?

According to equilibrium of forces, the upthrust in the floating object is also equal to the force acting vertically downward on the object (weight). Mathematically,

U = W

Upthrust is given as

U = ρvg

Where ρ = density of the cylinder

v = volume of the cylinder

g = acceleration due to gravity

Weight is given as

W = mg

Where m = mass of the immersed cylinder.

But mass given in terms of density and volume is

m = PV

Where P = density of water

V = volume of immersed cylinder.

Hence,

W = PVg

Since 14.1cm of the 30cm cylinder is immersed, we can find the volume of the immersed part of the cylinder in terms of the volume of the cylinder.

Volume of immersed cylinder = [1 - (14 1/30)] × volume of cylinder

V = (1 - 0.47)v

V = 0.53v

Equating the Upthrust and weight (equilibrium of forces), we have

U = W

ρvg = 0.53Pvg

ρ × v × g = 0.53 × 1 × v × g

ρ = 0.53g/cm^3

B. Given that the density of the cylinder is now known to be 0.53g/cm^3.

The top of the cylinder is now 20.7cm above the unknown liquid, which means that the volume of the portion of the cylinder above the unknown liquid is now

20.7/30 × total volume = 0.69v

Hence, the volume of the immersed cylinder, V, is 1 - 0.69v = 0.31v

Using the equilibrium of forces formula,

U = W

ρvg = PVg

Note: P is now the volume of the unknown liquid.

=> 0.53vg = 0.31Pvg

=> P = 0.53/0.31

P = 1.71g/cm^3