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Linear expansion numerical

Let's put our linear expansion coefficient to action. In this video we will explore a numerical on linear expansion and see, step by step, how to tackle such problems. Created by Mahesh Shenoy.

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Video transcript

and aluminium rod has a length of 50 meters at 300 Kelvin find the temperature at which its length would be forty nine point nine meters given the Alpha L value the linear expansion coefficient of aluminium is 25 times 10 to the minus 6 Kelvin inverse now we might be familiar with most of the data like the length and the temperature but we might be wondering what is alpha l what is the meaning of linear expansion coefficient well it just means that if you have an aluminium rod of 1 meter and you [ __ ] it up and increase this temperature by exactly 1 Kelvin then the rod would expand by 25 times 10 to the minus 6 meters that's the meaning of this number and similarly if you were to cool it down by one Kelvin decrease its temperature by 1 Kelvin then the rod would shrink in size by exactly 25 times 10 to my minus 6 meters that is the meaning of alpha L and we have discussed a lot about this in a previous video so if you find this a little bit confusing or maybe you require more clarity on this it would be a great idea to go back and watch that video first and then come back over here alright so with this I want you to pause the video and see if you can you can use this information and logically logically try to solve this problem alright let's solve this let's first write down what's given to us we know that the length of the rod the length of the aluminum rod is 50 meters here 50 meters and the temperature right now the temperature right now is 300 Kelvin find the temperature at which its length would be forty nine point nine meters which means they which means the initial length is 50 the final is forty nine point nine so the change in the length that's important for us the change in the length required would be let's see how much is that forty nine point nine minus fifty and that would be minus point one meters the minus sign is just telling that we want the rod to shrink in size that's all nothing else so we want the road to shrink in size by 0.1 meter the question now is to what temperature we should cool it down right because we want to shrink okay the first thing we'll do is try to connect Delta L temperature length and alpha L everything will try to build and connection between them and the way we like to do that is we can say Delta l equals alpha L Delta l equals alpha L when we have 1 meter long wire or 1 meter long rod and the change in temperature is 1 Kelvin but what if we had 2 meter long wire what would be Delta L in that case well if we had 2 meter long wire we could assume it is made up of two one meter long wires and each one would expand by Delta alpha L and therefore the total expansion will be two times our file does that make sense so the total expansion will be twice similarly if you had three meters the total exemption would be thrice if we have L meters let's do it in general first if we have L meters and the total expansion would be alpha L times L oops times L hope that makes sense right but this would be for one Kelvin rise what would happen if we had two Kelvin rise oh this whole thing would be doubled what if we had 3 Kelvin rise tripled what if we have delta T Kelvin rise Oh in that case we just have to multiply this by delta T so now that we have this connection let's see what we know we know Delta L we know alpha L we also know L we just have to now calculate delta T so we have now reduced the problem to mere algebra so whatever follows in those no longer physics is just algebra so how do we extract delta T from here well we just divide on both sides by alpha L times L so let's do that if we divide on both sides by alpha L times L we get Delta L divided by oops Delta L divided by I want pink color I'll file times L and that would be equal to Delta T and now we can just substitute Delta is 0.1 you can substitute minus point when we can do that no problem minus 0.1 meters divided by our file is just 25 25 times 10 to the minus 6 Kelvin inverse times L and L is 50 all right so let's just quickly solve this what do we get by the way this is 50 meters so the meters and meters cancel we can take the 10 to the minus 6 upstairs that will become 10 to the plus 6 10 to the plus 6 times 0.1 there'll be 10 to the power of 5 so we can write that as 10 to the power 5 so 1 and 5 zeroes divided by 25 times 50 so 1 0 cancels 25 goes 4 times so we get 400 divided by 5 that gives us or there's a negative fighter there's a negative sign so that gives us minus 80 Kelvin what is the minus sign say well the minus sign is just saying that we have to decrease the temperature by 80 Kelvin that makes sense because we already discussed that we have to cool it down so we have to decrease the temperature by 80 Kelvin but the question is what is the to final temperature well the current temperature is 300 we need to decrease the temperature by 80 so what is the final temperature let's just write that down some more space here all right so the final temperature would be the current temperature 300 minus 80 and that would be 220 Kelvin tada that is the answer so that is the temperature at which the whole thing would shrink to 49 point nine meters