IIT JEE Physics (1978-2016: 39 Years) Topic-wise Complete Solutions

# Units and Measurements

## Fundamental Units

The length, mass, time, electric current, temperature, amount of substance, and luminous intensity are chosen as seven fundamental quantities. The fundamental units of these quantities are metre, kilogram, second, ampere, kelvin, mole, and candela, respectively.

## Errors in Measurement

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., $\Delta a_\text{mean}=\frac{1}{n}\sum_1^n \Delta a_i$. The relative error 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{Relative error}=\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 is the relative error is expressed in percent i.e., $\text{Percentage error}=\left(\Delta a_\text{mean}/a_\text{mean}\right)\times100\%$.

1. 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.
2. 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.
3. 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

1. 1
2. 2
3. 3
4. 4

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}

## Significant Figures

Every measurement involves errors. The measurement result should include all digits that are known reliably plus the first digit that is uncertain. The reliable digits plus the first uncertain digit are known as significant digits or significant figures. The significant figures indicate the precision of measurement which depend on the least count of the measuring instrument. A choice of change of different units does not change the number of significant digits in a measurement. The scientific notation (in the power of 10) is the best way to remove ambiguities in determining the number of significant figure.

In multiplication or division, the final result should retain as many significant figures as are there in original number with the least significant figure.

In a addition or subtraction, the final result should retain as many decimal places as are there in the number with the least decimal places.

Convention for rounding off the uncertain digits. The preceding digit is raised by 1 if the insignificant digit to be dropped is more than 5, and is left unchanged if the latter is less than 5. If insignificant digit to be dropped is 5: if the preceding digit is even, the insignificant digit is simply dropped and, if its is odd, the preceding digit is raised by 1.

Problem (IIT JEE 2003) The edge of a cube is measured to be $1.2\times{10}^{-2}$ m. Its volume should be recorded as

1. $1.7\times{10}^{-6} \mathrm{m}^3$
2. $1.73\times{10}^{-6} \mathrm{m}^3$
3. $1.70\times{10}^{-6} \mathrm{m}^3$
4. $1.728\times{10}^{-6} \mathrm{m}^3$

Solution: The volume $V=(1.2\times{10}^{-2})^3=1.728\times{10}^{-6}\;\mathrm{m^3}$ shall be recorded as $1.7\times{10}^{-6}\;\mathrm{m^3}$ upto two significant figures.

## Dimensional Analysis

By the principle of homogeneity of dimensions, if the dimensions of all the terms are not same, the equation is wrong. The dimensional analysis is also used to deduce relation among the physical quantities.

Problem (IIT JEE 2014): To find the distance $d$ over which a signal can be seen clearly in foggy conditions, a railways engineer uses dimensional analysis and assumes that the distance depends on the mass density $\rho$ of the fog, intensity (power/area) $S$ of the light from the signal and its frequency $f$. The engineer finds that $d$ is proportional to $S^{1/n}$. The value of $n$ is $\ldots$.

Solution: From the given information, \begin{align} d=k\,\rho^a S^b f^c, \end{align} where $k$ is some dimensionless proportionality constant and $a$, $b$, $c$ are unknown parameters. Substitute dimensions of physical quantities in above equation to get \begin{align} [\mathrm{L}^1]=[\mathrm{ML^{-3}}]^a [\mathrm{MT^{-3}}]^{b}[\mathrm{T}^{-1}]^c. \end{align} Equate the exponents of M and L in above equation to get \begin{align} & a+b=0,\\ &1=-3a. \end{align} Solve these equations to get $b=1/3$.