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JEE Advanced 2018 Paper 1

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If the measurement errors in all the independent quantities are known, then it is possible to determine the error in any dependent quantity. This is done by the use of series expansion and truncating the expansion at the first power of the error. Forexample, consider the relation z=xyz = \frac{x}{y} . If the errors in x,y and z are Δx,Δy\Delta x, \Delta y and Δz\Delta z ,respectively, then
z±Δz=x±Δxy±Δy=xy(1±Δxx)(1±Δyy)1z \pm \Delta z = \frac{x \pm \Delta x}{y \pm \Delta y} = \frac{x}{y} \left(1 \pm \frac{\Delta x}{x}\right) \left(1 \pm \frac{\Delta y}{y}\right)^{-1}
The series expansion for (1±Δyy)1\left(1 \pm \frac{\Delta y}{y}\right)^{-1} ,to first power in Δyy\frac{\Delta y}{y} , is 1Δyy1 \mp \frac{\Delta y}{y} .The relative errors in independent variables are always added. So the error in z will be Δz=(Δxx+Δyy)\Delta z = \left(\frac{\Delta x}{x} + \frac{\Delta y}{y}\right)
The above derivation makes the assumption that Δxx1,Δyy1\frac{\Delta x}{x} \ll 1, \frac{\Delta y}{y} \ll 1 .
Therefore, the higher power of these quantities are neglected.
Section: Physics
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