Error in measurements & instruments(F) – JEE Advanced Previous Year Questions with Solutions

JEE Advanced Previous Year Questions of Physics with Solutions are available at eSaral. Practicing JEE Advanced Previous Year Papers Questions of Physics will help the JEE aspirants in realizing the question pattern as well as help in analyzing weak & strong areas.

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Previous Years JEE Advanced Questions

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 $\Delta \mathrm{x}, \Delta \mathrm{y}$ and $\Delta \mathrm{z}$, respectively, then

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

The series expansion for $\left(1 \pm \frac{\Delta y}{y}\right)^{-1}$, to first power in 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 $\frac{\Delta x}{x}<<1, \frac{\Delta y}{y}<<1$. Therefore, the higher powers of these quantities are neglected.

(There are two questions based on Paragraph “A”, the question given below is one of them)

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 $\Delta \mathrm{x}, \Delta \mathrm{y}$ and $\Delta \mathrm{Z}$, respectively, then

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

The series expansion for $\left(1 \pm \frac{\Delta y}{y}\right)^{-1},$ 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

The above derivation makes the assumption that $\frac{\Delta x}{x}<<1, \frac{\Delta y}{y}<<1 .$ Therefore, the higher powers of these quantities are neglected.

(There are two questions based on Paragraph “A”, the question given below is one of them)