A hydrometer floats in water with 6.0 of it's graduated stem unimmersed, and in oil of relative density 0.8 with 4.0cm of the stem unimmersed. Calculate the length of the stem unimmersed when the hydrometer is placed in a liquid of refractive density 0.9.
Well, this hydrometer seems like quite the adventurer, exploring different liquids and all! Let's put our thinking cap on and figure out the answer.
To solve this, we can use the concept of relative densities. The formula for relative density is:
Relative Density = Density of Substance / Density of Water
In this case, we have the relative density of oil as 0.8 and the relative density of the unknown liquid as 0.9. Since water has a relative density of 1, we know that the density of the oil is 0.8 times the density of water, and the density of the unknown liquid is 0.9 times the density of water.
Now, let's calculate the lengths of the stems unimmersed in each liquid. For water, we have 6.0 cm unimmersed, and for oil, we have 4.0 cm unimmersed.
Using the concept of relative densities, we can set up a proportion:
6 cm (unimmersed in water) / 4 cm (unimmersed in oil) = 1 (density of water) / 0.8 (density of oil)
Now, we can solve for the length of the stem unimmersed in the unknown liquid:
x (unknown length) / 4 cm (unimmersed in oil) = 1 (density of water) / 0.9 (density of unknown liquid)
Cross-multiplying and solving for x, we get:
x = (4 cm) * (1 / 0.9)
x ≈ 4.44 cm
Therefore, the length of the stem unimmersed when the hydrometer is placed in the liquid with a relative density of 0.9 is approximately 4.44 cm.
I hope that helps! Just remember, this hydrometer is quite the daredevil. Maybe it's time for it to settle down and find a nice liquid to call home.
To calculate the length of the stem unimmersed when the hydrometer is placed in a liquid of refractive density 0.9, we can use Archimedes' principle.
1. First, let's start by calculating the volume of the hydrometer that is submerged in each liquid:
- In water: The length of the stem unimmersed in water is 6.0 cm, so the length immersed in water is 10.0 cm (assuming the total length of the stem is 16.0 cm). Since we know that the density of water is 1.0 g/cm³, the volume of water displaced or submerged by the hydrometer in water is 10.0 cm³.
- In oil: The length of the stem unimmersed in oil is 4.0 cm, so the length immersed in oil is also 10.0 cm. Since we know that the relative density of oil (ρᵣ) is 0.8, the density of oil (ρ) can be calculated as follows: ρ = ρᵣ × ρᵢ, where ρᵢ is the density of the liquid in which the hydrometer is being placed. Thus, the density of oil is 0.8 × 1.0 g/cm³ = 0.8 g/cm³. Therefore, using Archimedes' principle, the volume of oil displaced or submerged by the hydrometer in oil is also 10.0 cm³.
2. Now, let's calculate the relative density of the liquid in which the hydrometer is being placed. Using Archimedes' principle again, we know that the volume of the hydrometer immersed in a liquid is directly proportional to the relative density of that liquid. Mathematically, we can express this as:
(volume immersed in liquid 1) / (volume immersed in liquid 2) = (relative density of liquid 1) / (relative density of liquid 2)
Let's denote the length of the stem unimmersed in the new liquid as "x" cm. The total length of the hydrometer stem is 16.0 cm.
Applying the formula, we get:
(10.0 cm)/(10.0 cm + x cm) = 0.8 / 0.9
3. Solving the equation above for x, we can cross-multiply and simplify:
0.9 × 10.0 cm = 0.8 × (10.0 cm + x cm)
9.0 cm = 8.0 cm + 0.8x cm
0.8x cm = 1.0 cm
x cm = 1.0 cm / 0.8
x cm = 1.25 cm
Therefore, the length of the stem unimmersed when the hydrometer is placed in a liquid of relative density 0.9 is 1.25 cm.
To solve this problem, we need to use the concept of relative density and the principle of buoyancy.
The relative density is defined as the ratio of the density of a substance to the density of a reference substance. In this case, the reference substance is water.
Let's assume that the length of the stem unimmersed when the hydrometer is placed in a liquid of refractive density 0.9 is "x" cm.
According to the problem,
For water: the length of the stem unimmersed = 6.0 cm
For oil: the length of the stem unimmersed = 4.0 cm
We can use the formula for relative density to find the density of the liquid in question in terms of the density of water:
Relative Density = Density of the liquid / Density of water
Since the relative density of oil is given as 0.8, we can find the density of oil as follows:
0.8 = Density of oil / Density of water
Density of oil = 0.8 * Density of water
Now, using the principle of buoyancy, we can say that the weight of the liquid displaced by the hydrometer is equal to the weight of the hydrometer itself.
When placed in water:
Weight of the hydrometer = Buoyant force = Weight of water displaced
When placed in oil:
Weight of the hydrometer = Buoyant force = Weight of oil displaced
And similarly, when placed in the liquid of refractive density 0.9, we can say:
Weight of the hydrometer = Buoyant force = Weight of the liquid displaced
Since the hydrometer has the same weight in all three situations, we can equate the buoyant forces:
Weight of water displaced = Weight of oil displaced = Weight of liquid displaced
Now, let's use this information to solve for the length of the stem unimmersed in the liquid of refractive density 0.9:
1) Weight of water displaced:
The weight of water displaced is equal to the weight of the immersed part of the hydrometer in water:
Weight of water displaced = (Length of stem immersed in water / Total length of the stem) * Weight of the hydrometer
Since it's given that 6.0 cm of the stem is unimmersed in water:
Weight of water displaced = (6.0 / Total length of the stem) * Weight of the hydrometer
2) Weight of oil displaced:
The weight of oil displaced is equal to the weight of the immersed part of the hydrometer in oil:
Weight of oil displaced = (Length of stem immersed in oil / Total length of the stem) * Weight of the hydrometer
Since it's given that 4.0 cm of the stem is unimmersed in oil:
Weight of oil displaced = (4.0 / Total length of the stem) * Weight of the hydrometer
3) Weight of liquid displaced:
The weight of the liquid displaced is equal to the weight of the immersed part of the hydrometer in the liquid of refractive density 0.9:
Weight of liquid displaced = (Length of stem immersed in the liquid / Total length of the stem) * Weight of the hydrometer
Now, since the hydrometer has the same weight in all three situations, we can equate the three weight expressions:
(6.0 / Total length of the stem) * Weight of the hydrometer = (4.0 / Total length of the stem) * Weight of the hydrometer = (x / Total length of the stem) * Weight of the hydrometer
Simplifying the equation:
6.0 = 4.0 = x
x = 6.0 cm
Therefore, the length of the stem unimmersed when the hydrometer is placed in a liquid of refractive density 0.9 is 6.0 cm.