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Volume expansion numerical

In this video, we will solve a numerical on volume expansion. Created by Mahesh Shenoy.

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  • blobby green style avatar for user mohammednazeef94
    The internal diameter of a brass is 15.94cm, and the diameter of a wheel is 16.00cm, both measured at the same temperature. Find the smallest rise in temperature through which the brass ring must be heated in order that it may just slip on to the wheel.
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Video transcript

a glass beaker has a volume of 50ml at 30 degrees Celsius find its volume at a hundred and thirty degrees Celsius given linear expansion coefficient for glass is 4 times 10 to the minus 6 per degree Celsius pause the video see if you can try and solve this yourself first and then we'll solve it together all right let's do it that's write down what's given to us we've been given the volume of the glass so we know volume is 50ml and that is at the temperature 30 degree Celsius we need to find its volume at a hundred and thirty degree Celsius what's in what's important for us is not the temperature but the change in temperature right that's what causes expansion so change in temperature is final temperature 130 minus 30 that's given to us that is a hundred degree Celsius so the temperature is increased by 100 degrees also that's given to us what else is given we've been given alpha L alpha a linear expansion coefficient is 4 times 10 to the minus 6 per degree Celsius inverse and we need to find its volume final volume to do that let's calculate the change in volume first how do we figure out the change in volume the change in volume is given as R for me that's the volume expansion coefficient times we times delta T and we've seen this we've seen this expression in a previous video so if you need a refresher maybe if you're not comfortable with this you a great idea to watch that one first and then come back over here so let's substitute and solve whew but you can see one problem we don't know what alpha V is we need the volume expansion coefficient but that's not given only alpha L is given to us hmm how do we figure this out well in a previous video we actually derived that alpha V is just three times three times all four L and if you're interested in seeing this derivation you can again go back and watch that video but anyways now since we know this we can just substitute and calculate let's do that so Delta V is going to be alpha me which is just three times three times our file and all file is give it to us that is four times ten to the minus six per degree in verse four degree Celsius inverse times the original volume that is 50 MLS which is 50 ml 50 ml times what we have times delta T and delta T is hundred degree Celsius so x times hundred degrees Celsius all right what do we get hmm let's see degree Celsius and degree Celsius inverse cancels right and so what we are what we end up with is 3 times 4 that's 12 12 times 50 or 12 times 5 is is 60 you know what let's just write that now so we have 12 times 50 times hundred and what remains is ml makes sense right Delta V change in volume we should have ml okay let's write this down 12 times 5 is 60 so I've it's 60 but then it is 1 0 here this is 600 and there are two more zeros two more zeros and wait I totally forgot this part there is a 10 to the power minus 6 10 to the power minus 6 so let's put that as well 10 to the power minus 6 and so that gives us let's see let's count six decimals towards the left 1 2 3 4 5 6 so we would get point 0 6 ml that's how much the volume has changed but we need to find the final volume right well we can do that since we know what the initial volume was the initial volume was fifty so initial volume is fifty final volume would be just the initial volume fifty plus the change so it was fifty it increased by point zero six so the new volume would be just fifty point zero six ml and notice the volume changes extremely tiny and so we can justify that alpha is equal to three times alpha L you can only use this when the changes are very tiny which is which is the case in this this particular scenario