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The purpose of measurement is to find the **true value** of a physical quantity. However, the **measured value** usually differs from the true value. The difference between the true value and the measured value is called **measurement error** (or inaccuracy in measurement). These errors can come from a variety of sources, including limitations in the equipment, the skill of the person making the measurement, and environmental conditions.

The **systematic error** tend to be in one direction (either positive or negative) when measurement is repeated. The **random error** is irregular in nature with respect to its sign as well as magnitude. The **least count** of a measuring instrument is the smallest value that can be measured by it.

The **absolute error** of a measurement is the magnitude of the difference between the true value of the quantity and its measured value. Generally, arithmetic mean of several measurements $a_1, a_2, a_3,\ldots, a_n$ is taken as the true value. Thus, absolute error of $i^\mathrm{th}$ measurement is
\begin{align}
\Delta a_i & =\left|a_i-a_\text{mean} \right| \\
&=\left|a_i-\frac{1}{n}\sum_{i=1}^n a_i\right|.
\end{align}
The **mean absolute error** is the arithmetic mean of the absolute errors i.e.,
\begin{align}
\Delta a_\text{mean}=\frac{1}{n}\sum_1^n \Delta a_i.
\end{align}

The **relative error** (RE) is the ratio of the mean absolute error $\Delta a_\text{mean}$ to the mean value $a_\text{mean}$ of the quantity measured i.e.
\begin{align}
\text{RE}&=\frac{\Delta a_\text{mean}}{a_\text{mean}} \\
&=\frac{1}{a_\text{mean}}\left(\frac{1}{n}\sum_{i=1}^n \Delta a_i\right).
\end{align}

The **percentage error** (PE) is the relative error expressed in percent i.e.,
\begin{align}
\text{PE}=\left(\frac{\Delta a_\text{mean}}{a_\text{mean}}\right)\times100\%.
\end{align}

- Error of a sum or a difference: When two quantities are added or subtracted, the absolute error in the final result is the sum of the absolute errors in the individual quantities.
- Error of a product or a quotient: When two quantities are multiplied or divided, the relative error in the result is the sum of the relative errors in the multipliers.
- The relative error in the physical quantity raised the the power $k$ is the $k$ times the relative error in the individual quantity.

**Problem (IIT JEE 2004):** A wire has a mass $m=0.3\pm 0.003\;\mathrm{g}$, radius $r=0.5\pm 0.005\;\mathrm{mm}$ and length $l=6\pm 0.06\;\mathrm{cm}$. The maximum percentage error in the measurement of its density is

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**Solution:**
The density of a wire of mass $m$, length $l$, and radius $r$ is given by
\begin{align}
\rho=\frac{m}{V}=\frac{m}{\pi r^2 l}.
\end{align}
Differentiate above equation and divide it by $\rho$ to get
\begin{align}
\left.\frac{\Delta \rho}{\rho}\right|_\text{max} =\left|\frac{\Delta m}{m}\right|+2\left|\frac{\Delta r}{r}\right|+\left|\frac{\Delta l}{l}\right|.
\end{align}
From the given data, $m=0.3\;\mathrm{g}$, $\Delta m=0.003\;\mathrm{g}$, $r=0.5\;\mathrm{mm}$, $\Delta r=0.005\;\mathrm{mm}$, $l=6\;\mathrm{cm}$, and $\Delta l=0.06\;\mathrm{cm}$. Substitute the values in above equation to get
\begin{align}
\frac{\Delta \rho}{\rho} &=\frac{0.003}{0.3}+\frac{2\times 0.005}{0.5}+\frac{0.06}{6} \\
&=0.04=4\%.\nonumber
\end{align}

**Problem (IIT JEE 2015): ** The energy of a system as a function of time $t$ is given as $E(t)=A^2\exp(-\alpha t)$, where $\alpha=0.2\;\mathrm{s^{-1}}$. The measurement of A has an error of $1.25\%$. If the error in the measurement of time is $1.50\%$, the percentage error in the value of $E(t)$ at $t=5\;\mathrm{s}$ is $\ldots$.

**Solution:** Differentiate the expression $E(t)=A^2e^{-\alpha t}$ to get
\begin{align}
\mathrm{d} E=2 A e^{-\alpha t}\,\mathrm{d} A-A^2\alpha e^{-\alpha t}\,\mathrm{d} t.
\end{align}
Divide above equation by $E(t)$ and simplify to get
\begin{align}
\frac{\mathrm{d} E}{E}=2\;\frac{\mathrm{d} A}{A}-\alpha \mathrm{d} t=2\;\frac{\mathrm{d} A}{A}-\alpha t\; \frac{\mathrm{d} t}{t}.
\end{align}
The error in measurement of a parameter $x$ is generally defined by
\begin{align}
x_\text{actual}=x_\text{measured}\pm \Delta x,\nonumber
\end{align}
where $\Delta x$ is a small positive number representing measurement error. Let $\Delta A$ and $\Delta t$ be the measurement errors (both positives) in $A$ and $t$. From above equation, $\mathrm{d}E$ is maximum when $\mathrm{d}A=\Delta A$ and $\mathrm{d}t=-\Delta t$. Thus, the percentage relative error in $E(t)$ is given by
\begin{align}
\frac{\Delta E}{E} &=2\;\frac{\Delta A}{A}+\alpha t \;\frac{\Delta t}{t} \\
&=2(1.25\%)+(0.2)(5)(1.5\%) \\
&=4\%.\nonumber
\end{align}