JEE Advanced 2018 · Paper 1, Physics question 18

PARAGRAPH “A”

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. For example, consider the relation $z = x/y$. If the errors in $x$, $y$ and $z$ are $\Delta x$, $\Delta y$ and $\Delta z$, respectively, then

$$z \pm \Delta z = \frac{x \pm \Delta x}{y \pm \Delta y} = \frac xy\left(1 \pm \frac{\Delta x}x\right)\left(1 \pm \frac{\Delta y}y\right)^{-1}.$$

The series expansion for $\left(1 \pm \frac{\Delta y}y\right)^{-1}$, to first power in $\Delta y/y$, is $1 \mp (\Delta y/y)$. The relative errors in independent variables are always added. So the error in $z$ will be

$$\Delta z = z\left(\frac{\Delta x}x + \frac{\Delta y}y\right).$$

The above derivation makes the assumption that $\Delta x/x \ll 1$, $\Delta y/y \ll 1$. Therefore, the higher powers of these quantities are neglected.