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It is known that solids expand on heating. Consider a case of rod of negligible area of cross-section such that it expands linearly upon heating.Now the increment in the length of the rod is proportional to the change in temperature as well as its length... Thus: $$dl=l*\alpha*\Delta T———(a)$$ ...where $\alpha$ = coefficient of linear expansion of the material with which the rod is made with.

Consider a rod (of length $l_o$)with a variable $\alpha$ such that $\alpha = Kx$...where x is the distance of a point on the rod from one of its end.calculate the the final length of rod when the temperature of the rod is increased by $\Delta$T Now using (a) $$dl=dx*kx*\Delta T———(b)$$ next step is to integrate (b).. $$\int dl =\int kxdx*\Delta T$$ enter image description here

My doubt is regarding the limits to be taken for x.Is it from o to $l_o$?If so what about extra length which is generated slowly by the change in temperature...shouldn’t we include the expansion of that part in to consideration?

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  • $\begingroup$ If the total length of the rod is being heated evenly, then all parts of the rod expand. $\endgroup$ – user207455 May 3 '19 at 6:28
  • $\begingroup$ @SolarMike yeah they do....hence the doubt is arises.. $\endgroup$ – Crypton May 3 '19 at 6:56
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If this was a normal rod with a as it's linear expansivity you would simply use change in length= initial length x a x temperature change. In this problem the only difference is the variable coefficient of linear expansivity and as you have shown that is why you have used integration. So when you are integrating first you have considered the expansion of a small lenght "dl" with linear expansivity Kx. Once you integrate it you get the total expansion of all those small lengths in other words the expansion of this rod. So the equation of linear expansivity stands here as well. So on my opinion you do not have to worry about "extra length which is generated slowly by the change in temperature". Simply apply the limits from 0 to l.

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  • $\begingroup$ Nice thinking thanks... ‘Once integrated you get the total expansion of all those small lengths’..I was not noticing this all time...now it’s clear $\endgroup$ – Crypton May 3 '19 at 7:52
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Let x be the location along the rod (after heating has occurred) of the cross section that was at location $x_0$ before heating, and let the coefficient of linear expansion be a linear function of the initial (material) location $a(x_0)=kx_0$. Then for the segment of the rod between $x_0$ and $x_0+dx_0$, we have after heating that $$dx=(1+\alpha\Delta T)dx_0=(1+kx_0\Delta T)dx_0$$If we integrate this equation from $x_0=0$ to $x_0=l$ we obtain: $$x(l)=l+k\frac{l^2}{2}\Delta T$$This is the final length of the rod.

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