Physics homework - Math Problem Example

The figure below shows two boats, of which one is sunk and the other is afloat. If we consider the floating boat as boat and the sunken boat asboat 2, then we can see that boat 1 has air within it and water around it, whereas boat 2 has water around it as well as inside it. The magnitude of the buoyancy force on an object depends upon the volume of the water that it displaces. In boat 1, the air inside the boat also serves as a means of displacing water. Whereas in boat 2, only the boat’s material itself is displacing the water; its inside is filled with water, thus that volume is not getting displaced.

Thus as boat 1 is displacing more volume of water than boat 2, it has a higher buoyancy force subjected upon it. Clearly, this is evident from the fact that boat 1 is afloat whereas boat two is sunken. 2. The volume of the given ice cube of 10cm dimensions is 1000cm3. It is given that when the ice cube is floating in pure water, 1cm of it is sticking up above the surface of the water. This shows that the weight of the water displaced by the 9cm thick portion of water is equivalent to the weight of the ice cube.

The 9cm thick portion has a volume of 9cm*10cm*10cm i.e. 900cm3. The weight of 900cm3 of pure water is equivalent to 1000cm3 of ice. If we suppose the density of liquid water to be 1g/cm3, then we have 900g of liquid water equivalent to 1000cm3 of ice. This gives us a density of ice as 0.9g/cm3, which is approximately equal to the real value of density of ice (~0.91g/cm3). When the excess 1cm thick part is cut off, the remaining portion of ice has a volume of 900cm3. The weight of 900cm3 of ice is 0.

9g/cm3*900cm3 i.e. 810g. Now, for liquid water, the volume of water that needs to be displaced to have the same weight as that of the ice is 810cm3 (810cm3*1g/cm3 = 810g). This means that only 810cm3 of the total 900cm3 of the remaining ice will get immersed in water. The rest 90cm3 volume will remain above the water level. Considering that the same 10cm*10cm base is taken, with the 9cm dimension considered as height, then the height remaining above the water will be 90cm3/(10cm*10cm) = 0.9cm. Thus, 0.9cm thick portion of the ice cube will still remain on top of the water.

One misconception that this question addresses is that even if you cut off the excess portion from the ice, its density does not change, thus it will always remain afloat with some part above the denser liquid water however small or large its mass may be. 3. For a block of material to sink in a fluid, its density should be greater than that of the fluid. The table shows that only platinum and gold are dense enough to sink in mercury. 4. Dry air is less dense than neon at room temperature and atmospheric pressure.

For a balloon filled with neon to float, the fluid (gas in context) around it should be denser than neon. As air is not denser than neon, thus a neon filled balloon will not float. For an argon atmosphere, we have the opposite case as argon is very much dense than neon. Thus a neon filled balloon will float in an argon atmosphere.