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prakharluv · Answer

The e.m.f E of certain thermocouple depends on the temperature difference θ between it two junction in accordance with the relation E=70θ-θ^2/20 where E is in micro volt, θ in degree Celsius and one junction is at zero degree Celsius. If E may be determined to ± 100μV, the possible error in ℃ when measuring a temperature  of 200℃ is????

theeric · Answer

I'm not familiar with this stuff, but I'll give you a source and my thoughts. I wouldn't count on me being correct, but maybe it will jog your memory! http://ibguides.com/physics/notes/measurement-and-uncertainties See sections 1.2.10 and 1.2.11. $E=10\ \theta-\dfrac{\theta^2}{20}$ For example of the errors, Absolute error of $E$ is $\pm100\ [\mu V]$. Fractional error (a.k.a. relative error) of $E$ is $\dfrac{100\ [\mu V]}{E}$. That error of the other side of the equation will be the sum of the absolute errors of each term. I'll use the $\Delta$ symbol for uncertainty. $\Delta E=\Delta(10\ \theta)+\Delta\left(\dfrac{\theta^2}{20}\right)$ The absolute uncertainty of $\Delta(10\ \theta)$, since $10$ is exact, is $\Delta\theta$. If the absolute error of $\theta$ is $x_\theta$, then the error of that term is $x_theta$. The absolute uncertainty of $\Delta\left(\dfrac{\theta^2}{20}\right)$, since $\frac{1}{20}$ is exact, is $\Delta\left(\theta^2\right)$. And $\Delta\left(\theta^2\right)$ is the two times the fractional error of $\theta$. So the error of this term, if absolute error is $x_\theta$, is $2\left(\dfrac{x_\theta}{\theta}\right)$. So, if we say the absolute error of $\theta$ is $x_\theta$ and absolute error of $E$ is $x_E$, then$$x_E=x_\theta+2\left(\dfrac{x_\theta}{\theta}\right)$$ We know $x_E=100\ [\mu V]$, and we know $\theta=(200-0)\ [\ ^\circ C]$. We would need to find $x_\theta$, if I am correct. Solving for $x_theta$ will not be too difficult for someone good at algebra. Again, I am unfamiliar with this material, but I hope I have helped. Best of luck! :)