you have exams going on next week, they said this coming, should I do it now?  Delta L is what? L alpha delta t, so you will get L times 1 plus alpha delta t, right and similarly you can find the new width to be equal to original width 1 plus alpha delta t, okay so the new area is what? Length into width, okay so the new area, new length into new width, this is equal to L into b is 1 plus alpha delta t whole square, L into b is what? Original area, right so the new area is equal to the original area and then expand this, 1 plus alpha square delta t square plus 2 alpha delta t, okay so this you just ignore, alpha square is very very less, so you will get new area is equal to the original area into 1 plus 2 alpha delta t, fine so it looks very similar to L is equal to L naught 1 plus alpha delta t, okay or you can also write here a n minus a naught, which is delta area, change in area is equal to a naught into 2 alpha delta t, fine so this formula looks very similar to delta L is equal to L naught alpha delta t, okay so if alpha is coefficient of linear expansion, 2 alpha should be coefficient of aerial expansion fine, so we can also name this constant as beta and we see that this is equal to a naught beta delta t, okay where beta is equal to 2 alpha and it is coefficient of aerial expansion and out but then now similarly what do you think the formula will be for volumetric expansion, can you draw parallel and write down the formula for volumetric expansion, delta v is equal to 3 alpha delta, so you not only multiply length and width you also multiply height, so there comes 1 plus alpha delta t whole cube, so you ignore cube and square terms and 3 alpha delta t will come, fine, so here delta v is equal to original volume into 3 alpha delta t, this 3 alpha you can write it as gamma, so v naught gamma delta t, this gamma is coefficient of volumetric expansion and it out still now, so this is the quantitative analysis of expansion, we have quantified all the formulas over here, no doubt see now let me tell you something else here suppose you have a plane, this is a plane fine, so n is equal to l naught alpha delta l is equal to l naught alpha delta t, delta v is b naught alpha delta t, okay what about this if I draw a randomly a line like this, this line expand, if I give need to this substance will this line expand, it will expand, it will expand fine, so if original length is x then delta x will be what, x alpha into delta t, okay very very simple, so all the linear dimensions, all the linear dimension in a substance will follow linear expansion formula, okay, so if the unit is meters centimeter or mm, like unit of length is there, dimension is length then that will expand as linear expansion getting it, so it need not be only external dimension of a substance fine, is it clear, now similarly if I draw a square over here whose area is a naught, this will follow up area expansion fine, similarly if it is a block inside a block there is a small volume that will follow volumetric expansion, fine now let me give you another thing, tell me what will happen here, you have a plate and this plate has a hole in it, if you heat it temperature increases delta t what happens to the diameter of the hole, increase or decrease, if I heat this the hole will increase in size or decrease in size, my question is the material try to occupy this empty space or it will simply move out, which one, it will increase why, 2 pi r will increase, because that is the perimeter of the circle that has a linear dimension that will increase and automatically the diameter increases, now diameter will follow what expansion law, this diameter, linear expansion, fine, d is equal to d naught into 1 plus alpha delta t or delta t is d naught into alpha delta t, fine, have you learned about mechanical properties of solids already, you have done Young's modulus and all that, now tell me one thing, I am just talking about scenarios, I am not getting into the numerical of it right now, so if you have a rod like this, length is like fine, it is clamped between the 2 walls, now you are increasing the temperature by delta t, this rod is not allowed to expand, what will happen, this rod is not allowed to expand, it remains the same length, it remains the same length, but actually what this rod wants to do is increase its length a bit by probably this much, how much that would have been, if you are allowing it to increase, this delta l would have been equal to l alpha delta t, but you are not allowing it to increase, so in a way it is in compressed state, getting in a point and what is the strain in the rod, delta l by l, which is what, strain is delta l by l which is equal to alpha delta t, this is called thermal strain, because the natural length of the rod changes at different temperature, at 20 degrees Celsius length should be this much to have 0 strain, but if you do not allow it to expand, there will be a strain from what it should have been, getting it, so that is the reason why you will see that there is a small gap in the railway tracks, those rails, there is a provision for it to expand if its temperature increases, it will expand little bit, but if you do not allow it to expand, it will become with a curve sort of thing and it will anyway change its length, so like for example, this is the two rails, you are not allowing it to expand, you are clamping it, so if temperature increases, it is just randomly bulge out from here and there, this is thermal strain and there are a lot of nice numerical that are made out of just thermal strain, there comes the Young's modulus also, you know how much stress will be developed, we know that sigma divided by epsilon should be equal to Young's modulus, so sigma should be equal to Young's modulus into strain, so y alpha delta t will be stress and if a rath procession is a, force will be equal to sigma into a, so you should be very comfortable with playing around all these things, but in now since we are looking at for the first time, probably right now we can just focus on the NCRT numerical, in this chapter you will see that many times you know exactly how to solve a numerical, but still you will not be getting the answers, probably because you are not very careful with the calculation, make sure that you are good at calculation and do not use calculator, because if you are an habit of using calculator, you know the problem will be that the difficult chapters probably some difficult question come, you will not be able to probably, you know some hard question not able to solve it and these question it will be pretty that you know exactly how to solve it, still you are not able to solve, so be good at calculation, fine so here is the first numerical, you know how the bullet card's wheel is made, you have been to village, you are like cosmopolitan, you have never seen a bullet card, have you seen the wheel, how it is, there is an iron ring on it, if iron ring is not there you just wear out, so that iron ring is placed very tightly on the wooden wheel, otherwise it will come out, have you ever wondered how it is done, exactly you heat that iron the ring, it expands then you put it on the wooden wheel and then let it cool down, so it will contract and hold that wheel, so the question is on that only, write down a blacksmith fixes, you do not have to write the entire version, just write down the data which is required to solve the question, a blacksmith fixes iron ring on the ring of the wooden wheel of a bullet card, the diameter of the ring, the ring diameter is 5.243 meters, diameter of iron ring is 5.231, diameter of iron ring is 5.231 meters, both these dimensions are given at 27 degrees Celsius, you need to find out what temperature should the ring be heated, so that it fits on the wooden ring, the coefficient of linear expansion for iron alpha is given as 1.2 into 10 raise to power minus 5, what do you think the unit should be, look at delta l is equal to l alpha delta t, so alpha is delta l by l into 1 by delta t meter meter cancels per degree Celsius, this is per degree Celsius, get the value of temperature up to which the iron ring should be heated, so that it gets fitted into the wooden ring from 27 to 30, how much should be change in diameter, delta t should be, so I am saying you know exactly how to solve, still not able to get the answer, how much should be this minus that which is 0.012 meters, this should be equal to diameter of the iron ring into alpha to delta t, so delta t will be what 0.012 divided by diameter of iron ring that is 5.231 into alpha, which is 1.2 into 10 raise to power minus 5, this is delta t, all of you getting this as delta t of the calculation around, this goes in the numerator and it becomes 1200 divided by 5.23, I am just simplifying it, this is delta t, how much this comes out to be, let me do the calculation in front of you, 6, 4, 10, 3, 2, 5, so it is 1.2 into 6.27, so this is 1200 divided by 6.27, so I will take it as 6.3, just a simplification, so whatever I get the answer will be slightly more than that, so it would have been 6, it would have been 200, but this is 6.3, so this you can roughly say around 180 and you can check whether it is true, getting it, delta t is this, so final temperature is what, plus this, 180 plus 227 and if you want to get it exactly, just divide this, you can divide 63, this is like that, you get the answer, 200 by 12, with approximation I am getting 207, let us do a numerical on this thermal stress also, write down a rod of length 2 meter is at a temperature of 20 degree Celsius, so a rod of length 2 meter is at a temperature of 20 degree Celsius, you need to find out the free expansion of the rod, free expansion, as in you are not restricting its expansion, find the free expansion of the rod, if the temperature is increased to 50 degree Celsius, so from T 1, the temperature goes to 50 degree Celsius, you need to find the free expansion, this is the part 1, the value of alpha is 15 into 10 is power minus 6, please solve this, then I will tell you the next one. Just find the expansion, how much it expand, delta l, 0.9 mm, 9 into 20 minus 4, that is correct, so delta l is l alpha delta t, now you need to find out the part, if you prevent it to expand, you are not allowing it to expand, you need to find what is the stress that is developed, how you know Young's modulus is not given, you got the strain, stress, Young's modulus is given as 2 into 10 is power 11 Newton per meter square, find out what is the stress, if you are preventing it completely, same as what we have done earlier, so strain should be equal to alpha delta t, a stress should be what, y alpha delta t, 8 out, normal c part, it is permitted to expand only by 0.4 mm, if you are allowing it to expand only 0.4 mm, now what is the stress, so free expansion is what 0.9 mm, you are allowing it to expand 0.4 mm, how much is the restriction, 0.5, so strain corresponding to 0.5 mm will be there, getting it, when you are completely restricting it, strain corresponding to 0.9 mm was there, when you are allowing it to expand 0.4, 0.5 is the restriction, any doubts, so strain should be equal to what, 0.5 mm, it is 0.5 into 10 is power minus 3 divided by 2 meters, this is strain, this is strain, this into young's modulus is the stress, any doubts, so that is how we deal with the expansion property of the substances, so still there are some varieties of numericals that you have to do it as homework and ask me doubts in the next class, what, stress, strain is this, so we will take next property after the break.