What is the volume of cavities?

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The problem involves a piece of iron weighing 6000 N in air and 4000 N in water, with a density of 7.87 g/cc. The buoyant force, equal to the weight difference of 2000 N, allows for the calculation of the volume of water displaced. The total volume of the iron piece, including cavities, is determined to be 764.04 cc, while the volume of the solid iron is calculated as 509.36 cc. By subtracting the volume of solid iron from the total volume, the volume of the cavities is found to be approximately 254.68 cc. Thus, the volume of cavities in the iron piece is around 254.68 cc.
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Hi!
Here's one problem.
A piece of iron(with cavities) weighs 6000 N in air AND 4000 N in water.The density of iron is 7.87 g/cc.What is the volume of cavities?
My assumption is that the weight of water displaced is equal to the buoyant force(which is right because Archimedes said so). The problem is what volume of water is displaced? Is it right to say that the volume of the iron piece without the cavities is the volume of water displaced? :rolleyes:
 
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prasanna said:
Hi!
Here's one problem.
A piece of iron(with cavities) weighs 6000 N in air AND 4000 N in water.The density of iron is 7.87 g/cc.What is the volume of cavities?
My assumption is that the weight of water displaced is equal to the buoyant force(which is right because Archimedes said so). The problem is what volume of water is displaced? Is it right to say that the volume of the iron piece without the cavities is the volume of water displaced?
The volume of the iron (ie. just the iron, without the cavities) is the mass/density (density = 7.87e3 kg/m^3). The mass is the weight (mg) divided by g. The buoyant force is equal to the weight of the water displaced. The volume of water displaced is equal to the volume of the iron and cavities (assuming the cavities are watertight).

Therefore the mass of the iron is 6000/9.8 = 612.25 kg
The volume of just the iron, therefore, is: 612.25/7.87e3 = .07780 m^3
2000 N of water = 2000/9.8 = 204 kg = .204 m^3 = volume of water displaced.


So, what is the volume of the cavities?

AM
 


The volume of cavities in this situation can be determined by using the principle of buoyancy. As you mentioned, the weight of water displaced is equal to the buoyant force, which in this case is equal to the weight difference between the iron piece in air and water. This weight difference is 2000 N (6000 N - 4000 N).

To determine the volume of water displaced, we can use the formula V = m/d, where V is the volume, m is the mass, and d is the density. In this case, the mass of the iron piece is 6000 N (since weight is equal to mass in this case) and the density of iron is 7.87 g/cc. This gives us a volume of 764.04 cc.

However, this volume includes the cavities as well as the solid iron. To find the volume of just the cavities, we can subtract the volume of the solid iron from this total volume. The volume of solid iron can be found by using the same formula, but with the mass of the iron piece in water (4000 N). This gives us a volume of 509.36 cc.

Subtracting this volume from the total volume of water displaced gives us the volume of cavities, which is 254.68 cc. Therefore, the volume of cavities in this piece of iron is approximately 254.68 cc.
 
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