Density and floating in equilibrium – problems and solutions

1. A block placed into two liquids with different types. In liquids A, 0.6 part of an object is in the liquid. In liquids B, 0.5 part of an object is in the liquid. Determine the ratio of the density of liquid A to liquid B.

Known :

Density of block = xDensity and floating in equilibrium – problems and solutions 1

Part of the object in liquid A = 0.6

The density of liquid A = y

Part of object in liquid B = 0.5

The density of liquid B = z

Wanted: The ratio of the density of liquid A to liquid B (y: z)

Solution :

Density and floating in equilibrium – problems and solutions 2

Substitute x in equation 2 with x in equation 1 :

Density and floating in equilibrium – problems and solutions 3

2. An object placed in liquid P, 0.5 part of the object is in liquid. When the object placed in liquid Q, the object is in liquid, as shown in the figure below. What is the ratio of the density of liquid P to liquid Q?

Known :

Density of object = xDensity and floating in equilibrium – problems and solutions 4

Part of object in liquid P = 1/2 = 0.5

The density of liquid P = y

Part of the object in liquid Q = 1

The density of liquid Q = z

Wanted: The ratio of the density of liquid P to liquid Q (y: z)

Solution :

Density and floating in equilibrium – problems and solutions 5

Substitute x in equation 2 with x in equation 1 :

Density and floating in equilibrium – problems and solutions 6

3. A wood floating in oil (density of oil is 800 kg/m3). 4/5 parts of the wood are in the oil. What is the density of the wood?

Known :

Density of oil (ρ) = 800 kg/m-3

Part of wood in oil = 4/5

Wanted: Density of wood

Solution :

Density and floating in equilibrium – problems and solutions 7

The density of wood is 640 kg/m3.

4. The weight of a block is 50 N. In water, the weight of the block is 30 N. If the density of water is 103 kg/m-3 what is the density of block.

Known :

Weight of block in air (w) = 50 N

Weight of block in water (wwater) = 30 N

Density of water (ρwater) = 103 kg/m3 = 1000 kg/m3

Acceleration due to gravity (g) = 10 m/s2

Mass of block (m) = w/g = 50/10 = 5 kg

Wanted : Density of block (ρblock)

Solution :

Buoyant force :

FA = w – wwater = 50 – 30 = 20 N

Volume of block in water :

FA = ρ g V

20 = (1000)(10) V

20 = (10,000) V

V = 20/10,000

V = 0.002 m3

Density of block :

ρ = m/V = 5 / 0.002 = 2500 kg/m3

5. A wood block dipped in water. 25% of its part is above the surface of water and 75% is in water. Density of water is 1 g/cm3 and acceleration due to gravity is 10 m/s2. Determine the density of the blocks.

A. 0.025 kg/m3

B. 0.075 g/cm3

C. 0.250 kg/m3

D. 0.750 g/cm3

Known :

Part of block above the surface of water = 25% = 25/100 = 0.25

The part of block in water = 75% = 75/100 = 0.75

The density of water (ρwater) = 1 g/cm3 = 1000 kg/m3

Acceleration due to gravity = 10 m/s2

Wanted : Density of block (ρblock)

Solution :

Volume of block in water = the ratio of the density of block to the density of water

Density and floating in equilibrium – problems and solutions 1

The correct answer is D.

Comments