Question

According to Archimedes’ principle, the mass of a floating object equals the mass of the fluid displaced by the object. Use this principle to solve the following problems. (a) A wooden cylinder 30.0 cm high floats vertically in a tub of water (density 1:00 g/cm3 ). The top of the cylinder is 13.5 cm above the surface of the liquid. What is the density of the wood? (b) The same cylinder floats vertically in a liquid of unknown density. The top of the cylinder is 18.9 cm above the surface of the liquid. What is the liquid density? (c) Explain why knowing the length and width of the wooden objects is unnecessary in solving Parts (a) and (b).

Answer 1

Part a)

`\rho = 0.55 g/cm^3`

Part b)

`\rho_L = 1.49 g/cm^3`

Part c)

Since we know that the base area will remain same always

so here the length and width of the object is not necessary to obtain the above data in such type of questions

Step-by-step explanation:

Part a)

As we know that when cylinder float in the water then weight of the cylinder is counter balanced by the buoyancy force

So here we know

buoyancy force is given as

`F_b = \rho_w V_(sub) g`

`F_b = (1 g/cm^3) (30 - 13.5) Ag`

`F_b = 16.5 Ag`

Now we know that the weight of the cylinder is given as

`W = \rho (30 cm)A g`

now we have

`\rho (30 cm) A g = 16.5 A g`

`\rho = 0.55 g/cm^3`

Part b)

When the same cylinder is floating in other liquid then we will have

`F_b = \rho_L (30 - 18.9 )A g`

so we have

`\rho_L (11.1) Ag = 0.55(30) Ag`

`\rho_L = 1.49 g/cm^3`

Part c)

Since we know that the base area will remain same always

so here the length and width of the object is not necessary to obtain the above data in such type of questions