Mathematically, relative density is expressed as: PANT. Relative density is dimensionless, since it is a ratio between two quantities of same unit. If the ratio is greater than 1, the object will be heavier than the same volume of the reference. If it is less than 1, it will be lighter than the reference. If it is exactly 1, an equal volume of the object and the reference will have the same weight.
It is important to specify the reference material when reporting a relative density, but when the reference material is not specified it is usually understood to be water at 3. Determining the specific gravity same as relative density of an object relative to water entails little effort, as the density of the object only needs to be divided by 1 or , depending on the unit, e.
While it may seem this would result in two different specific gravities based upon which unit is used, this is not the case. According to the above formula, the specific gravity of this object would be. The density of this object is:. Since the specific gravity is, in fact unitless , as long as the same reference is used and consistent units are used the specific gravity will be the exact same number.
Hence specific gravity will also work with gravitational density instead of mass-based density. Specific gravity is a special case of relative density. While specific gravity has a reference density of water, relative density can have any reference density that is used. It is best to specify the reference material when using relative density, using subscripts:. While relative density will not change as long as consistent units are maintained, the relative density is relative to its reference.
The relative density of an object relative to mercury, is different than the relative density with respect to water specific gravity. Taking the relative density with respect to alcohol , the relative densities of ethanol, water and iron are 1. Taking the relative density relative to water, the numbers would be 0. With respect to iron, the numbers are 0. These are all correct relative densities, however it is with different reference points. Note that again, units are not needed and hence these numbers are correct no matter what unit system is used.
Maintaining a consistent set of units is important. The units do not need to even be in terms of "amount of substance" per "volume" directly. For example, suppose someone wanted to find out the specific gravity of a rock.
They could see that the rock deflected a certain spring by 3 inches. Then they put on a reference substance, which deflected the spring by 5 inches. The first rock made the water in a certain graduated cylinder rise by 20 mm, and the reference made it rise by 34 mm.
The relative density between these two objects can easily be determined without having to figure out several constants which would be needed to determine the density directly like the spring constant or the cross sectional area of the cylinder.
Now if they knew the actual density of the reference substance, then they can find the actual density of the test object. Changes in temperature affect the densities of materials, and the relative densities between two materials need to be corrected at certain temperatures. It is best to record the temperatures of the two materials, expressed here:.
One important use of relative density is hydrometers. It is relatively easy to show that the displacement of a hydrometer is approximately directly proportional to the change of relative density for small changes with respect to any reference that the hydrometer is placed into.
First the hydrometer is placed into the reference fluid shown in light blue. The hydrometer is then marked at the height at which the water's level stands at blue line.
Assumptions :. At this line, the specific gravity is 1. When the hydrometer is placed in the new fluid, shown as green, a new mark is placed.
Note that the hydrometer has dropped slightly, hence the specific gravity is lighter in this fluid then in the first fluid. If Varsity Tutors takes action in response to an Infringement Notice, it will make a good faith attempt to contact the party that made such content available by means of the most recent email address, if any, provided by such party to Varsity Tutors.
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Possible Answers:. Correct answer:. Explanation : Write the density formula. The density of air is: Find the volume of the freezer in meters. Solve for the mass of the air. Report an Error. Explanation : To answer this question, it's best to start by drawing a free body diagram so that we can identify the forces acting on the cube.
Next, plug this value into the above equation for the mass of the cube. Explanation : Use the information given to find the density of the raft. Therefore the magnitude of the gravity force is equal to the magnitude of the buoyant force: The mass of the raft is equal to its density times its volume.
We start using subscripts to keep track of which densities and volumes we mean: The 's cancel, and the is of the raft's total volume: Setting the magnitudes equal again at the Great Salt Lake:. Correct answer: None; it is unitless. Explanation : Specific gravity is a ratio of a material's density to that of the density of water. Possible Answers: No. It may float or sink depending on the temperature.
Correct answer: Yes. Explanation : Using density equation: For a sphere: If diameter is , then radius is. Converting to and plugging in values: This is less than the density of water, so it will float. Explanation : In this question, we're presented with a scenario in which an object of a given density is floating in water. Next, we can rearrange the equation to obtain the following: Next, it's important for us to recognize that we can rewrite the expression for the mass of the object.
Since density is equal to mass divided by volume, we can rearrange this in terms of density and volume as follows: We can then make one final rearrangement Now let's think about this for a moment.
If we take as representing the entire volume of the object, we can solve for the fraction of the object that is above the water like so: Thus, of the object's volume is above the water.
Explanation : We can start with Newton's 2nd law for this problem: Upon release, we only have two forces, gravity and buoyancy: Since the ball is accelerating upward, let's designate an upward force as positive. Then, the first expression becomes: Now we need to determine the mass of water displaced. Therefore we can say: Substituting this into our last expression, we get: Rearranging for density: We have values for each of these except for the mass of water displaced.
Then we can say: Now plugging in values to the density expression, we get:. Explanation : We know that the weight of the object in the air is.
The ratio of the weight of the object to the buoyant force in water will give us the specific gravity of the cube in the following equation: Therefore the correct answer is. Explanation : Since the block is submerged by percent we know that it has a density that is percent that of water.
Explanation : Equation of density is so multiplying the volume by three would decrease the density by. Copyright Notice. View Tutors. Keith Certified Tutor. Stephen Certified Tutor. Moe Certified Tutor.
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