A small solid has an irregular shape and floats in water. Describe how you could measure its density.
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Off the top of my head, the water displacement method might be the best way to measure the density of this object. In this method, the ability of the object to float or sink is of critical importance. When you place the object in water, there should be a rise in the water level because the object impacts the water level causing it to increase. The difference between both levels of water is the density of the object. For example, if you had 20 ml of water as a starting point, and then you placed the object in water and the level rose to 26 ml, the difference between both levels (26-20) is 6 ml. This means that the density would be reflective of the change in water levels.
I am adding on to this because I realized that part of my original answer did not come through. The density as indicated above would be the volume present, and setting it up as mass/ volume would allow you to obtain the density. Sorry about this mixup. I accept full responsibility on it.
The first answer is right in that you should use water. However, this answer does not give all the steps needed to find the density of the object.
If you put the object in water, the amount that the water level goes up is the volume of the object. So, in the example given above, the volume of the object is 6 ml.
However, to find the density, you must also know the mass of the object because mass = density*volume.
So, find the volume using the water test. Then find the mass. Then use the formula above.
To measure the density of an object you need to measure two features of the object - its weight and volume. The weight can be measured by weighing. The volumes can be measured in several ways. For solid objects that are heavier than water and do not dissolve or react with water, the easiest way is to put some water in a container with graduations to show the volume of water plus any submerged material put in the container. Initially some water is filed in the container and the volume of water is ascertained from the graduations. Then the object is put in the water and the total volume of water plus the object is ascertained. The difference of the two reading of volumes gives the volume of the object.
When the object is lighter than water this method does not work because the object will float on water rather than sink completely. In this case there are two alternatives. One is to use, instead of water, some other liquid lighter than the object. The other method is to tie down the object to a heavier sinker of known volume so that the object under study and the sinker together will sink in water. In this case, increase in volume due to sinking of object plus sinker must be adjusted for the volume of sinker to get the volume of the object under study.
In this way, when both weight and volume of the object has been measured, its density can be calculated as weight divided by volume.
The principle of Archimedes enlightened the world with the buoyant property the liquid.The property is that a liquid thrusts up an object sumerged in it with a buoyant force that is equal to the weight of the liquid, whose volume is equal to that of the object completely submerged in it(the liquid).
So the relative density of the irregular shaped object (completely submerged in the liquid) with respect to the liquid is the weight of the object /loss of weight of that object when submerged in the liquid. To get the density of it, multiply the relative density by the density of the liquid.
If the irregular shaped object , which does not sink but floats, could be made to submerge, along with another suitable object of known weight. The loss of weight of the two together could be found say L1. Now find the loss of weight L2 of the other object used, to sumerge the irregular shaped floating object. Now, L2 -L1 is the weight of the liquid whose volume is equal to the volume of the irregular shaped floating object. Now find the relative density of the irregular shaped object by the ratio, weight of the irrgular shaped object/ (L1-L2).Multiply this relative density with that of the liquid to get the density of the irregular shaped object. You can use water instead liquid. The iregular or irreglar it is immeterial. See that there is a way to measure the volume of an irregular shaped object , which is very difficult otherwise.Thanks and respects to Archimedes, the great.
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