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Non-Rationalised Science NCERT Notes and Solutions (Class 6th to 10th)
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Non-Rationalised Science NCERT Notes and Solutions (Class 11th)
Physics Chemistry Biology
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Physics Chemistry Biology

Class 11th (Physics) Chapters
1. Physical World 2. Units And Measurements 3. Motion In A Straight Line
4. Motion In A Plane 5. Laws Of Motion 6. Work, Energy And Power
7. System Of Particles And Rotational Motion 8. Gravitation 9. Mechanical Properties Of Solids
10. Mechanical Properties Of Fluids 11. Thermal Properties Of Matter 12. Thermodynamics
13. Kinetic Theory 14. Oscillations 15. Waves



Chapter 2 Units And Measurement

The chapter on Units and Measurement serves as the fundamental foundation for all quantitative sciences. It establishes that the measurement of any physical quantity involves a comparison with an internationally accepted reference standard known as a unit. The modern standard is the Système Internationale d’ Unites (SI), which is based on seven base units: metre ($m$), kilogram ($kg$), second ($s$), ampere ($A$), kelvin ($K$), mole ($mol$), and candela ($cd$). Beyond these, derived units are formed through various combinations of base units, while plane angles and solid angles are measured in radians and steradians, respectively. The chapter highlights that SI units are preferred globally due to their decimal-based structure, which simplifies conversions.

Practical measurement techniques range from direct methods using vernier callipers or screw gauges to indirect methods like the parallax method for measuring astronomical distances and the oleic acid film method for estimating molecular sizes. A critical distinction is made between Accuracy (how close a measurement is to the true value) and Precision (the resolution of the measuring instrument). Every measurement is subject to errors, which are classified as Systematic (instrumental, procedural, or personal) and Random. The chapter provides a mathematical framework for calculating absolute, relative, and percentage errors, emphasizing that when quantities are combined, their uncertainties propagate according to specific arithmetic rules.

Finally, the chapter introduces Significant Figures to ensure that reported measurements reflect the reliability and precision of the data. To describe the nature of physical quantities, Dimensions ($[M], [L], [T]$, etc.) are used. The Principle of Homogeneity of Dimensions is a vital tool, stating that only physical quantities with identical dimensions can be added or subtracted. This principle allows scientists to check the consistency of equations and deduce relationships among different physical variables. While dimensional analysis is a powerful analytical tool, it is limited by its inability to determine dimensionless constants, requiring experimental observation for complete theoretical accuracy.



Introduction

The study of physics is based on the measurement of physical quantities. Measurement of any physical quantity involves the comparison of that quantity with a certain basic, arbitrarily chosen, internationally accepted reference standard called a unit.

The result of a measurement is always expressed by a numerical measure (a number) followed by the unit. For example, if the length of a room in an Indian house is measured to be $5$ metres, the number $5$ is the numerical measure and 'metre' is the unit.


Fundamental and Derived Quantities

In nature, there are a large number of physical quantities, but they are all interrelated. Therefore, we do not need a separate unique unit for every single quantity. These quantities are categorized into two types:

1. Fundamental (Base) Quantities

The physical quantities which are independent of each other and cannot be further simplified are called fundamental or base quantities. The units defined for these quantities are known as fundamental or base units. Examples include length, mass, and time.

2. Derived Quantities

All other physical quantities whose units can be expressed as combinations of the fundamental units are called derived quantities. Their units are known as derived units. For example, Speed is derived from length and time.

A complete set of these units, including both the base units and derived units, is known as a system of units.


Comparison of Base and Derived Units

The following table illustrates the distinction between the two types of units with examples.

Type of Quantity Example Quantity Standard Unit (SI)
Fundamental Length Metre ($m$)
Fundamental Mass Kilogram ($kg$)
Derived Area (Length $\times$ Breadth) Metre$^2$ ($m^2$)
Derived Speed (Distance / Time) Metre / Second ($m/s$)
Total Base Quantities 7 Fixed by SI


The International System of Units

Before the global adoption of a single system, scientists across the world relied on different localized systems. The most notable among these were the CGS system, the FPS system (often called the British system), and the MKS system. These systems primarily focused on three base quantities: length, mass, and time.

System Name CGS FPS MKS
Length Unit centimetre ($cm$) foot ($ft$) metre ($m$)
Mass Unit gram ($g$) pound ($lb$) kilogram ($kg$)
Time Unit second ($s$) second ($s$) second ($s$)

The Advent of SI Units

The Système Internationale d’ Unites (SI) was established to provide a universal language for science and trade. Managed by the International Bureau of Weights and Measures (BIPM), it was significantly revised in November 2018 to define units based on fundamental physical constants rather than physical prototypes. This ensures that the units are stable over time and can be reproduced anywhere in the universe.

Advantages of the SI System

1. Decimal Nature: SI is a metric system. Multiples and sub-multiples are related by powers of $10$, making calculations straightforward.

2. Universal Acceptance: It is used in all scientific, technical, and industrial work globally, including all research conducted by ISRO and DRDO in India.

3. Coherence: All derived units are obtained by simple multiplication or division of base units without introducing numerical factors.


The Seven SI Base Units

The following table provides the complete list of SI base quantities, their names, symbols, and the rigorous scientific definitions adopted recently.

Base Quantity Unit Name Symbol Definition and Fixed Constant
Length metre m Defined by the fixed numerical value of the speed of light in vacuum $c$ to be $299,792,458$ when expressed in $\text{m s}^{-1}$.
Mass kilogram kg Defined by the fixed numerical value of the Planck constant $h$ to be $6.62607015 \times 10^{-34}$ when expressed in $\text{J s}$ (or $\text{kg m}^2 \text{s}^{-1}$).
Time second s Defined by the fixed numerical value of the caesium frequency $\Delta \nu_{Cs}$ (ground-state hyperfine transition) to be $9,192,631,770$ when expressed in $\text{Hz}$.
Electric current ampere A Defined by the fixed numerical value of the elementary charge $e$ to be $1.602176634 \times 10^{-19}$ when expressed in unit $\text{C}$ (or $\text{A s}$).
Thermodynamic Temperature kelvin K Defined by the fixed numerical value of the Boltzmann constant $k$ to be $1.380649 \times 10^{-23}$ when expressed in unit $\text{J K}^{-1}$.
Amount of substance mole mol One mole contains exactly $6.02214076 \times 10^{23}$ elementary entities (Avogadro constant $N_A$).
Luminous intensity candela cd Defined by the luminous efficacy of monochromatic radiation of frequency $540 \times 10^{12} \text{ Hz}$ to be $683 \text{ lm W}^{-1}$.

Mathematical Logic Behind the Definitions

The modern SI system is unique because it defines units by "fixing" the value of nature's constants. This ensures that the units remain invariant and can be reproduced in any laboratory, whether in New Delhi or on the Moon.

1. The Metre and Speed of Light

The metre is no longer a distance between two marks on a bar. It is derived from the constant speed of light ($c$).

$$1 \text{ metre} = \left( \frac{c}{299,792,458} \right) \text{ seconds}$$

2. The Kilogram and Planck Constant

The kilogram is now linked to the Planck constant ($h$). The relation between energy ($E$), mass ($m$), and frequency ($\nu$) is given by Einstein's and Planck's formulas:

$$E = mc^2 \quad \text{and} \quad E = h\nu$$

$$mc^2 = h\nu \implies m = \frac{h\nu}{c^2}$$

By fixing $h$ and $c$, the unit of mass (kg) is precisely determined without the need for a physical weight.

3. The Mole and Avogadro Number

The Mole is used to measure the amount of substance. It is strictly numerical. If you have $6.02214076 \times 10^{23}$ atoms of Gold, you have exactly $1 \text{ mole}$ of Gold.


Supplementary Units: Angles

In addition to the seven fundamental base units, the SI system includes two supplementary units used to measure angles. These units are unique because they represent ratios of similar physical quantities (length/length or area/area$^2$), making them dimensionless. However, they are assigned units to distinguish them from pure numbers.

1. Plane Angle ($d\theta$)

A Plane Angle is a two-dimensional angle formed at the centre of a circle by an arc. It is defined as the ratio of the length of the arc ($ds$) to the radius ($r$) of the circle.

Mathematical Formula and Unit

The plane angle $d\theta$ is expressed as:

$$d\theta = \frac{ds}{r}$$

The SI unit for a plane angle is the radian, denoted by the symbol rad.

Derivation of Total Angle and Conversions

For a complete circle, the total arc length is equal to the circumference, which is $2\pi r$. Therefore, the total plane angle subtended by a circle at its centre is:

$$\theta_{total} = \frac{2\pi r}{r} = 2\pi \text{ rad}$$

Since a full circle is also $360^{\circ}$, we can derive the following conversion factors:

1. $2\pi \text{ rad} = 360^{\circ}$

2. $\pi \text{ rad} = 180^{\circ}$

3. $1 \text{ rad} = \frac{180}{\pi} \approx 57.3^{\circ}$

Angle in Degrees ($^{\circ}$) Angle in Radians (rad)
$30^{\circ}$ $\frac{\pi}{6}$
$45^{\circ}$ $\frac{\pi}{4}$
$60^{\circ}$ $\frac{\pi}{3}$
$90^{\circ}$ $\frac{\pi}{2}$
$180^{\circ}$ $\pi$
$360^{\circ}$ $2\pi$

2. Solid Angle ($d\Omega$)

A Solid Angle is the three-dimensional equivalent of a plane angle. It is the angle subtended at a point (the apex) by a portion of a spherical surface. It is defined as the ratio of the intercepted area ($dA$) of the spherical surface to the square of its radius ($r^2$).

Mathematical Formula and Unit

The solid angle $d\Omega$ is expressed as:

$$d\Omega = \frac{dA}{r^2}$$

The SI unit for a solid angle is the steradian, denoted by the symbol sr.

Derivation of Total Solid Angle for a Sphere

For a complete sphere, the total surface area is $4\pi r^2$. Therefore, the total solid angle subtended by a sphere at its centre is:

$$\Omega_{total} = \frac{4\pi r^2}{r^2} = 4\pi \text{ sr}$$


The Dimensionless Nature of Angles

Although radians and steradians are units, the quantities they measure have no dimensions. This is because they are calculated as the ratio of quantities with identical dimensions.

For Plane Angle:

$$\text{Dimension of } \theta = \frac{[\text{Length}]}{[\text{Length}]} = [L^0]$$

For Solid Angle:

$$\text{Dimension of } \Omega = \frac{[\text{Area}]}{[\text{Length}]^2} = \frac{[L^2]}{[L^2]} = [L^0]$$

In dimensional analysis, these are represented as $[M^0 L^0 T^0]$.


Visual representation of Plane Angle and Solid Angle

Units Retained for General Use

While the SI system is the international standard for scientific and industrial work, several non-SI units remain in frequent use across the globe and within the Indian domestic context. These units are retained because they are deeply embedded in daily life, commerce, and specific fields of study like medicine, agriculture, and astronomy. For convenience, their values are precisely defined in terms of SI base units.

Comprehensive List of Non-SI Units

The following table provides the complete list of units that are outside the SI system but are officially recognized for continued use. This includes units for time, angle, mass, area, and pressure.

Physical Quantity Unit Name Symbol Value in SI Units
Time minute min $60 \text{ s}$
Time hour h $60 \text{ min} = 3600 \text{ s}$
Time day d $24 \text{ h} = 86400 \text{ s}$
Time year y $365.25 \text{ d} \approx 3.156 \times 10^7 \text{ s}$
Angle degree $^{\circ}$ $1^{\circ} = (\pi/180) \text{ rad}$
Volume litre L $1 \text{ dm}^3 = 10^{-3} \text{ m}^3$
Mass tonne t $10^3 \text{ kg} = 1000 \text{ kg}$
Mass quintal q $100 \text{ kg}$
Mass carat c $200 \text{ mg} = 2 \times 10^{-4} \text{ kg}$
Pressure bar bar $0.1 \text{ MPa} = 10^5 \text{ Pa}$
Pressure standard atmosphere atm $101325 \text{ Pa} = 1.013 \times 10^5 \text{ Pa}$
Area hectare ha $1 \text{ hm}^2 = 10^4 \text{ m}^2$
Area are a $1 \text{ dam}^2 = 10^2 \text{ m}^2$
Radioactivity curie Ci $3.7 \times 10^{10} \text{ s}^{-1}$

Key Categories

I. Units of Time

The SI unit for time is the second. However, for longer durations, we use minutes, hours, and days. The value of a Year is typically taken as $365.25$ days to account for leap years in long-term calculations.

$$\text{1 Year} \approx 365.25 \times 24 \times 60 \times 60 \text{ s} \approx 3.156 \times 10^7 \text{ s}$$

II. Units of Area (Agriculture)

In Indian agriculture and land records, the Hectare is a common unit. It represents the area of a square with sides of $100$ metres.

$$\text{1 Hectare} = 100 \text{ m} \times 100 \text{ m} = 10^4 \text{ m}^2$$

Similarly, the Are represents a square of $10$ metres side length ($100 \text{ m}^2$).

III. Units of Mass (Trade)

For heavy goods like grains and steel in Indian markets, Quintal ($100 \text{ kg}$) and Metric Tonne ($1000 \text{ kg}$) are used. In the jewellery industry, the Carat is used for measuring gemstones.

$$\text{1 Carat} = 200 \text{ mg} = 0.2 \text{ grams}$$


Derived Units and Scientific Notation

While the seven base units form the foundation of the SI system, most physical quantities we encounter are derived quantities. These are expressed as combinations of the base units. To handle extremely large or small values, the SI system employs a standard set of prefixes and specific guidelines for symbolic representation.

SI Derived Units

Derived units are obtained by performing mathematical operations (multiplication or division) on the base units. Some derived units are expressed directly in terms of base units, while others have been given special names in honour of famous scientists.

A. Units Expressed in Base Units

The following vertical table lists common physical quantities and how their units are derived from the seven fundamental pillars.

Physical Quantity SI Unit Name Symbol
Area square metre $m^2$
Volume cubic metre $m^3$
Speed, velocity metre per second $m/s$ or $m \, s^{-1}$
Acceleration metre per second square $m/s^2$ or $m \, s^{-2}$
Density kilogram per cubic metre $kg/m^3$ or $kg \, m^{-3}$
Momentum kilogram metre per second $kg \, m \, s^{-1}$
Moment of inertia kilogram square metre $kg \, m^2$
B. Units with Special Names

Certain quantities are used so frequently that their complex base-unit combinations are replaced by single symbols. For example, the unit of Force ($kg \, m/s^2$) is called the Newton ($N$).

Quantity Name Symbol In Base Units
Frequency hertz $Hz$ $s^{-1}$
Force newton $N$ $kg \, m \, s^{-2}$
Pressure, Stress pascal $Pa$ $kg \, m^{-1} s^{-2}$
Energy, Work joule $J$ $kg \, m^2 s^{-2}$
Power watt $W$ $kg \, m^2 s^{-3}$
Electric Charge coulomb $C$ $A \, s$
Electric Potential volt $V$ $kg \, m^2 s^{-3} A^{-1}$
Resistance ohm $\Omega$ $kg \, m^2 s^{-3} A^{-2}$

Mathematical Derivations of Derived Units

To find the unit of any derived quantity, we simply substitute the units of the base quantities into the defining formula.

1. Derivation of the Joule (Energy)

Energy or Work is defined as Force $\times$ Displacement.

$$\text{Unit of Work} = \text{Unit of Force} \times \text{Unit of Displacement}$$

$$\text{Unit of Work} = (kg \, m/s^2) \times m$$

$$\text{Unit of Work} = kg \, m^2 s^{-2}$$

This composite unit is defined as $1$ Joule ($J$).

2. Derivation of the Pascal (Pressure)

Pressure is defined as Force per unit Area.

$$\text{Unit of Pressure} = \frac{\text{Force}}{\text{Area}}$$

$$\text{Unit of Pressure} = \frac{kg \, m \, s^{-2}}{m^2}$$

$$\text{Unit of Pressure} = kg \, m^{-1} s^{-2}$$

This is defined as $1$ Pascal ($Pa$).


Common SI Prefixes

When measuring quantities like the distance between Indian cities (very large) or the size of a virus (very small), using standard units is inconvenient. Prefixes are used to indicate multiples or sub-multiples of $10$.

Factor Prefix Symbol Factor Prefix Symbol
$10^{12}$ Tera $T$ $10^{-3}$ milli $m$
$10^{9}$ Giga $G$ $10^{-6}$ micro $\mu$
$10^{6}$ Mega $M$ $10^{-9}$ nano $n$
$10^{3}$ kilo $k$ $10^{-12}$ pico $p$
$10^{2}$ hecto $h$ $10^{-15}$ femto $f$

Guidelines for Symbols

To ensure international consistency, strict rules must be followed when writing symbols for units and physical quantities.

Rules for Physical Quantities

1. Italics: Symbols for physical quantities are printed in italics (e.g., $v$ for velocity, $m$ for mass).

2. Vectors: Vectors are printed in bold upright type or indicated with an arrow (e.g., $\vec{F}$).

3. Chemical Elements: Symbols for elements like $Ca$ or $Fe$ are written in Roman upright type and never followed by a full stop.

Rules for SI Units

1. Capitalization: Unit names (like newton, joule) are never capitalized. Unit symbols are capitalized only if named after a scientist (e.g., $N$ for newton, $W$ for watt), with the exception of $L$ for litre.

2. Plurals: Symbols remain unaltered in the plural. We write $25 \, cm$, not $25 \, cms$.

3. Punctuation: Do not use a full stop at the end of a unit symbol unless it is at the end of a sentence.

4. Prefix Placement: Prefixes are attached directly to the unit without a space (e.g., $km$, $ns$).

5. The Mass Exception: Prefixes are always attached to 'gram', never to 'kilogram'. For example, $10^{-6} \, kg$ is written as $1 \, mg$ (milligram), not $1 \, \mu kg$.



Measurement of Length

The measurement of length involves determining the distance between two points in space. In physics, the range of lengths varies from the extremely minute (sub-atomic levels) to the incredibly vast (astronomical levels). Because of this vast range, a single instrument cannot be used for all measurements. We categorize these into Direct Methods and Indirect Methods.


1. Direct Methods of Measurement

The measurement of length is a fundamental task in physics. Depending on the magnitude of the length and the accuracy required, we choose different instruments. Direct methods involve the physical use of an instrument to read the value against a standard scale.

The most important concept in direct measurement is the Least Count (LC). It is the smallest value that can be measured accurately with a given instrument. A smaller least count implies higher precision.

1. Metre Scale

The Metre Scale is the most basic instrument used in Indian schools and laboratories. It is typically a wooden or plastic strip of $1 \text{ m}$ length, graduated in centimetres and millimetres.

A standard laboratory metre scale showing cm and mm divisions

2. Vernier Callipers

When we need to measure the thickness of a small object or the internal/external diameter of a tube with greater precision, we use the Vernier Callipers. It overcomes the limitation of the metre scale where values between two millimetre marks cannot be read accurately.

Construction and Principle

It consists of two scales:

  1. Main Scale (MS): Fixed scale graduated in $mm$.
  2. Vernier Scale (VS): A sliding scale that moves over the main scale.
Derivation of Least Count (LC)

The Least Count is the difference between the value of one Main Scale Division ($MSD$) and one Vernier Scale Division ($VSD$).

$LC = 1 \text{ MSD} - 1 \text{ VSD}$

Usually, $n$ divisions on the Vernier Scale coincide with $(n-1)$ divisions on the Main Scale.

$n \text{ VSD} = (n-1) \text{ MSD}$

$1 \text{ VSD} = \left( \frac{n-1}{n} \right) \text{ MSD}$

Substituting this back into the $LC$ formula:

$LC = 1 \text{ MSD} - \left( \frac{n-1}{n} \right) \text{ MSD}$

$LC = \left( 1 - \frac{n-1}{n} \right) \text{ MSD} = \frac{1}{n} \text{ MSD}$

Standard Case: In most Indian labs, the smallest $MSD = 1 \text{ mm}$ and the Vernier scale has $10$ divisions.

$LC = \frac{1 \text{ mm}}{10} = 0.1 \text{ mm} = 0.01 \text{ cm} = 10^{-4} \text{ m}$.

Diagram of Vernier Callipers identifying main scale, vernier scale, and jaws

3. Screw Gauge

The Screw Gauge works on the principle of a screw rotating in a nut. It is used for extremely precise measurements, such as the diameter of a thin copper wire or the thickness of a glass slide.

Key Terms and Derivation

The Least Count is derived as the ratio of the pitch to the total number of divisions on the circular scale.

$LC = \frac{\text{Pitch of the Screw}}{\text{Total number of divisions on Circular Scale}}$

Calculation: If the pitch is $1 \text{ mm}$ and there are $100$ divisions on the circular scale:

$LC = \frac{1 \text{ mm}}{100} = 0.01 \text{ mm} = 0.001 \text{ cm} = 10^{-5} \text{ m}$.

Detailed diagram of a Screw Gauge showing spindle, sleeve, and thimble

4. Spherometer

A Spherometer works on the same principle as the screw gauge. It consists of a metallic frame with three fixed legs and a central moving screw. It is primarily used to measure the radius of curvature of spherical surfaces like lenses or mirrors.

The $LC$ is calculated identically to the screw gauge:

$LC = \frac{\text{Pitch}}{\text{Number of circular divisions}}$


Summary Table of Direct Instruments

Instrument Least Count ($m$) Application
Metre Scale $10^{-3} \text{ m}$ General cloth/paper measurement
Vernier Callipers $10^{-4} \text{ m}$ Internal/External diameters
Screw Gauge $10^{-5} \text{ m}$ Diameter of thin wires
Spherometer $10^{-5} \text{ m}$ Curvature of lenses

2. Indirect Methods for Large Distances

For distances that are beyond the reach of physical scales, such as the distance to a mountain peak, a planet, or a star, we use Indirect Methods.

The Concept of Parallax

Diagram illustrating parallax with two eyes viewing a pencil against a wall

Mathematical Derivation for Distance ($D$)

To find the distance $D$ of a planet $S$, follow these steps:

  1. Observe the planet from two different observatories $A$ and $B$ on the Earth, separated by a distance (basis) $b$.
  2. Measure the angle $\angle ASB = \theta$ between the two directions along which the planet is viewed. This $\theta$ is called the parallactic angle.
  3. Since the planet is very far away, $D$ is very large compared to $b$. Therefore, the ratio $b/D$ is very small, and $\theta$ is also very small.
  4. For such small angles, we consider $AB$ as an arc of a circle with centre $S$ and radius $D$.
  5. From the definition of an angle in radians:

    $\text{Angle } (\theta) = \frac{\text{Arc Length } (b)}{\text{Radius } (D)}$

  6. Rearranging to find the distance $D$:

    $D = \frac{b}{\theta}$

Important Note: In the above formula, the angle $\theta$ must be expressed in radians.

Diagram of Parallax method showing Earth observatories A and B, planet S, distance D and angle theta

Determination of Angular Size and Linear Diameter

Angle Conversions for Calculations

In physics and astronomy, angles are treated as ratios of lengths. To perform calculations using the parallax formula $D = \frac{b}{\theta}$, the angle must be in the SI unit, which is the Radian ($rad$).

The fundamental relationship between degrees and radians is based on a full circle:

1. Calculating $1$ Degree ($1^\circ$) in Radians

To find the value of $1^\circ$, we use the basic proportionality:

  1. We have the relation: $180^\circ = \pi \ rad$.
  2. $1^\circ = \frac{\pi}{180} \ rad$.
  3. Substituting the value of $\pi \approx 3.14159$:

    $1^\circ = \frac{3.14159}{180} \ rad$

  4. Result: $1^\circ \approx 1.745 \times 10^{-2} \ rad$.
2. Calculating $1$ Minute of arc ($1'$) in Radians

A degree is further divided into $60$ smaller parts called minutes.

  1. Relation: $1^\circ = 60'$.
  2. Therefore, $1' = \frac{1^\circ}{60}$.
  3. Using the radian value of $1^\circ$ calculated above:

    $1' = \frac{1.745 \times 10^{-2}}{60} \ rad$

  4. Result: $1' \approx 2.908 \times 10^{-4} \ rad$ (commonly rounded to $2.91 \times 10^{-4} \ rad$).
3. Calculating $1$ Second of arc ($1''$) in Radians

A minute is further divided into $60$ smaller parts called seconds.

  1. Relation: $1' = 60''$.
  2. Therefore, $1'' = \frac{1'}{60}$.
  3. Using the radian value of $1'$ calculated above:

    $1'' = \frac{2.908 \times 10^{-4}}{60} \ rad$

  4. Result: $1'' \approx 4.847 \times 10^{-6} \ rad$ (commonly rounded to $4.85 \times 10^{-6} \ rad$).

Summary of Angle Conversions

This table is essential for solving problems in Chapter 2 of Class 11 Physics (Units and Measurement):

Angle Unit Symbol Value in Radians
Degree $1^\circ$ $1.745 \times 10^{-2} \ rad$
Minute $1'$ $2.91 \times 10^{-4} \ rad$
Second $1''$ $4.85 \times 10^{-6} \ rad$

Example 1. A man wishes to estimate the distance of a nearby tower $C$ from him. He stands at a point $A$ in front of the tower and spots a very distant object $O$ in line with $AC$. He then walks perpendicular to $AC$ up to $B$, a distance of $100 \text{ m}$, and looks at $O$ and $C$ again. Since $O$ is very distant, the direction $BO$ is practically the same as $AO$; but he finds the line of sight of $C$ shifted from the original line of sight by an angle $\theta = 40^\circ$. Estimate the distance of the tower $C$ from his original position $A$.

Answer:

From the given information, we can represent the situation as a right-angled triangle $ABC$, where the angle at $A$ is $90^\circ$ because the man walks perpendicular to $AC$.

Right angled triangle showing points A, B and C for tower distance estimation

We have the parallax angle $\theta = 40^\circ$ and the distance $AB = 100 \text{ m}$.

In $\Delta ABC$:

$\tan \theta = \frac{AB}{AC}$

Rearranging the formula to find the distance $AC$:

$AC = \frac{AB}{\tan \theta}$

$AC = \frac{100 \text{ m}}{\tan 40^\circ}$

Using the value $\tan 40^\circ \approx 0.8391$:

$AC = \frac{100}{0.8391} \approx 119.17 \text{ m}$

The estimated distance of the tower $C$ from the original position $A$ is $119 \text{ m}$.


Example 2. The Moon is observed from two diametrically opposite points $A$ and $B$ on Earth. The angle $\theta$ subtended at the Moon by the two directions of observation is $1^\circ 54'$. Given the diameter of the Earth to be about $1.276 \times 10^7 \text{ m}$, compute the distance of the Moon from the Earth.

Answer:

Step 1: Convert the angle into radians.

We are given $\theta = 1^\circ 54' = (60 + 54)' = 114'$

Since $1' = 2.91 \times 10^{-4} \text{ rad}$:

$\theta = 114 \times 2.91 \times 10^{-4} \text{ rad} \approx 3.32 \times 10^{-2} \text{ rad}$

Step 2: Use the parallax formula to find distance $D$.

Parallax method diagram showing Earth diameter as basis and Moon as the distant object

Basis ($b$) = Diameter of Earth = $1.276 \times 10^7 \text{ m}$

The distance $D$ is given by:

$D = \frac{b}{\theta}$

$D = \frac{1.276 \times 10^7 \text{ m}}{3.32 \times 10^{-2}}$

$D \approx 3.84 \times 10^8 \text{ m}$

The distance of the Moon from the Earth is $3.84 \times 10^8 \text{ m}$.


Example 3. The Sun’s angular diameter is measured to be $1920''$. The distance $D$ of the Sun from the Earth is $1.496 \times 10^{11} \text{ m}$. What is the linear diameter of the Sun?

Answer:

Step 1: Convert the angular diameter into radians.

We are given angular diameter $\alpha = 1920''$.

Since $1'' = 4.85 \times 10^{-6} \text{ rad}$:

$\alpha = 1920 \times 4.85 \times 10^{-6} \text{ rad} = 9.31 \times 10^{-3} \text{ rad}$

Step 2: Calculate the linear diameter $d$.

Diagram showing angular diameter alpha subtended by linear diameter d at distance D

The relationship between linear diameter ($d$), distance ($D$), and angular size ($\alpha$) is:

$d = \alpha \times D$

$d = (9.31 \times 10^{-3}) \times (1.496 \times 10^{11} \text{ m})$

$d \approx 1.39 \times 10^9 \text{ m}$

The linear diameter of the Sun is $1.39 \times 10^9 \text{ m}$.


Example 4. An Indian astronomer observes the Moon from two points on Earth separated by $1.276 \times 10^7 \text{ m}$. The parallax angle $\theta$ is found to be $1^\circ 54'$. Calculate the distance of the Moon from the Earth.

Answer:

Step 1: Convert the parallax angle to radians.

$\theta = 1^\circ 54' = 114'$

Using the conversion $1' = 2.91 \times 10^{-4} \text{ rad}$:

$\theta = 114 \times 2.91 \times 10^{-4} \text{ rad} = 3.32 \times 10^{-2} \text{ rad}$

Step 2: Calculate the distance using the parallax formula.

Distance $D = \frac{b}{\theta}$

$D = \frac{1.276 \times 10^7 \text{ m}}{3.32 \times 10^{-2}}$

$D = 3.84 \times 10^8 \text{ m}$

The distance of the Moon from the Earth is $3.84 \times 10^8 \text{ m}$.


Example 5. The angular diameter of the Sun is $1920''$. If the distance to the Sun is $1.496 \times 10^{11} \text{ m}$, find its linear diameter.

Answer:

Step 1: Convert angular diameter to radians.

$\alpha = 1920''$

Using the conversion $1'' = 4.85 \times 10^{-6} \text{ rad}$:

$\alpha = 1920 \times 4.85 \times 10^{-6} \text{ rad} = 9.31 \times 10^{-3} \text{ rad}$

Step 2: Calculate linear diameter.

Linear Diameter $d = \alpha \times D$

$d = (9.31 \times 10^{-3}) \times (1.496 \times 10^{11} \text{ m})$

$d = 1.39 \times 10^9 \text{ m}$

The linear diameter of the Sun is $1.39 \times 10^9 \text{ m}$.


3. Estimation of Very Small Distances

To measure extremely small sizes, such as those of a molecule ($10^{-8} \text{ m}$ to $10^{-10} \text{ m}$), we cannot use ordinary tools or even standard optical microscopes. This requires indirect chemical methods or high-resolution electronic instruments.

Limitations of Optical Microscopes

Diagram showing light waves passing around a small molecule illustrating diffraction limits

Advanced Microscopic Techniques

To overcome the resolution barriers of visible light, scientists use alternative beams with much smaller wavelengths.

1. Electron Microscopes
2. Tunnelling Microscopy

Example 6. An electron microscope has a resolution of $0.6 \text{ Å}$. Calculate how many such objects (atoms or molecules) would fit linearly within the span of the smallest wavelength of visible light, which is approximately $4000 \text{ Å}$.

Answer:

1. Given Data:

  • Resolution of the Electron Microscope ($d$) $= 0.6 \text{ Å}$
  • Smallest wavelength of visible light ($\lambda$) $= 4000 \text{ Å}$

2. Concept:

To find the number of objects ($N$) that fit within the wavelength, we divide the wavelength of light by the resolution of the microscope.

Comparison scale showing thousands of small units fitting inside one large wave cycle

3. Calculation:

$$N = \frac{\lambda}{d}$$

$$N = \frac{4000 \text{ Å}}{0.6 \text{ Å}}$$

$$N = \frac{40000}{6}$$

$$N \approx 6666.67$$

4. Result:

Approximately $6,667$ objects of size $0.6 \text{ Å}$ can fit linearly within a single wavelength of the smallest visible light. This high number illustrates why visible light is unable to "see" or resolve individual atoms; the light wave is simply too large and "skips" over these tiny particles.


The Oleic Acid Method

To estimate the size of a molecule, which is typically in the range of $10^{-8} \text{ m}$ to $10^{-10} \text{ m}$, we cannot use direct measurement. Instead, we use a chemical method involving Oleic Acid to form a mono-molecular layer.

Experimental Procedure

The estimation involves the following steps to create a extremely thin film of known concentration:

  1. First Solution Preparation:
    • Take $1 \text{ cm}^3$ of pure oleic acid.
    • Dissolve it in alcohol to prepare a solution of $20 \text{ cm}^3$.
  2. Second Dilution:
    • Take $1 \text{ cm}^3$ of the above solution.
    • Dilute it again with alcohol to make a final volume of $20 \text{ cm}^3$.
    • Concentration: The final concentration of oleic acid is now $\frac{1}{20 \times 20} = \frac{1}{400} \text{ cm}^3$ of oleic acid per $\text{ cm}^3$ of solution.
  3. Film Formation:
    • Take a large trough filled with water and lightly sprinkle lycopodium powder on its surface.
    • Add $n$ drops of the prepared solution to the water surface.
    • The oleic acid spreads quickly, and the alcohol evaporates, leaving a thin circular film of pure oleic acid.
  4. Measurement:
    • Quickly measure the diameter of the circular film.
    • Calculate the Area ($A$) of the film.
Process of forming a monomolecular layer of oleic acid on water surface

Mathematical Derivation

We assume that the film formed on the water surface is exactly one molecule thick. This thickness ($t$) is treated as the diameter of the oleic acid molecule.

Step 1: Calculate Total Volume of Solution

If we drop $n$ drops of the solution into the trough, and the volume of each drop is $V$:

$\text{Total Volume of solution} = nV \text{ cm}^3$

Step 2: Calculate Volume of Pure Oleic Acid

Since the concentration is $\frac{1}{400}$:

$\text{Volume of Oleic Acid in the film} = \frac{nV}{400} \text{ cm}^3$

Step 3: Calculate Thickness (Molecular Diameter)

The film spreads over an area $A$. The relationship between Volume, Area, and Thickness is:

$\text{Volume} = \text{Area} \times \text{Thickness}$

$t = \frac{\text{Volume of Oleic Acid}}{\text{Area}}$

Substituting the volume from Step 2:

$t = \frac{nV}{400A} \text{ cm}$

By using this formula, the thickness $t$ is found to be approximately $10^{-9} \text{ m}$.


Example 7. A student drops $10$ drops of oleic acid solution (concentration $1/400$) on water. The volume of each drop is $0.02 \text{ cm}^3$. The film spreads to form a circle with a radius of $10 \text{ cm}$. Calculate the thickness of the film.

Answer:

1. Given Data:

  • Number of drops ($n$) $= 10$
  • Volume of each drop ($V$) $= 0.02 \text{ cm}^3$
  • Radius of film ($r$) $= 10 \text{ cm}$
  • Concentration $= \frac{1}{400}$

2. Calculate Area ($A$):

$A = \pi r^2 = 3.14 \times (10)^2 = 314 \text{ cm}^2$

3. Calculate Thickness ($t$):

$t = \frac{nV}{400A}$

$t = \frac{10 \times 0.02}{400 \times 314}$

$t = \frac{0.2}{125600} \text{ cm}$

$t \approx 1.59 \times 10^{-6} \text{ cm}$

$t \approx 1.59 \times 10^{-8} \text{ m}$


Example 8. If the molecular size of a substance is known to be $10^{-9} \text{ m}$ and a student wants to form a film of area $100 \text{ cm}^2$ using a solution of concentration $1/400$, how many drops are required? (Assume volume of 1 drop $= 0.01 \text{ cm}^3$).

Answer:

1. Given Data:

  • $t = 10^{-9} \text{ m} = 10^{-7} \text{ cm}$
  • $A = 100 \text{ cm}^2$
  • $V = 0.01 \text{ cm}^3$
  • Concentration $= 1/400$

2. Rearrange formula to find $n$:

$t = \frac{nV}{400A} \implies n = \frac{t \times 400 \times A}{V}$

3. Calculation:

$n = \frac{10^{-7} \times 400 \times 100}{0.01}$

$n = \frac{4 \times 10^{-3}}{10^{-2}} = 0.4$

Note: Practically, this means even a fraction of a drop would cover that area for this molecular thickness.


4. Range of Lengths

The universe exhibits a staggering range of lengths, spanning from the incredibly small sub-atomic particles to the vast extent of the observable universe. In Physics, we deal with lengths as small as $10^{-14} \text{ m}$ (the size of a tiny nucleus) to as large as $10^{26} \text{ m}$ (the extent of the observable universe).

Because these numbers are either extremely small or extremely large, we use Scientific Notation (powers of 10) and certain Special Units to make measurements and calculations more convenient.

Special Units for Small (Microscopic) Lengths

For atomic and nuclear scales, the metre is too large a unit. We use the following:

Special Units for Large (Astronomical) Lengths

For distances between planets and stars, the following units are used:

1. Astronomical Unit (AU)

It is the average distance of the centre of the Sun from the centre of the Earth.

$1 \text{ AU} = 1.496 \times 10^{11} \text{ m}$

2. Light Year (ly)

It is the distance that light travels in vacuum in one year with a velocity of $3 \times 10^8 \text{ m s}^{-1}$.

Derivation:

$1 \text{ light year} = \text{Speed of light} \times \text{Time (1 year in seconds)}$

$1 \text{ ly} = (3 \times 10^8 \text{ m/s}) \times (365.25 \times 24 \times 60 \times 60 \text{ s})$

$1 \text{ ly} = 9.46 \times 10^{15} \text{ m}$

3. Parsec (pc)

Parsec (Parallactic Second) is the distance at which the average radius of the Earth’s orbit ($1 \text{ AU}$) subtends an angle of 1 arc second.

Derivation:

Using the formula $\theta = \frac{l}{r} \implies r = \frac{l}{\theta}$

Where $l = 1 \text{ AU}$ and $\theta = 1''$ (converted to radians).

$1 \text{ pc} = \frac{1.496 \times 10^{11} \text{ m}}{4.85 \times 10^{-6} \text{ rad}}$

$1 \text{ pc} \approx 3.08 \times 10^{16} \text{ m}$

Comparison Table of Units

Unit Name Fermi ($f$) Angstrom (Å) Astron. Unit (AU) Light Year (ly) Parsec (pc)
Value in Metres $10^{-15} \text{ m}$ $10^{-10} \text{ m}$ $1.496 \times 10^{11} \text{ m}$ $9.46 \times 10^{15} \text{ m}$ $3.08 \times 10^{16} \text{ m}$

Range and Order of Lengths

The following table illustrates the wide spectrum of sizes and distances that are studied in Physics. Note that the ratio of the longest and shortest lengths in our universe is approximately $10^{41}$.

Size of object or distance Length (m)
Size of a proton $10^{-15}$
Size of atomic nucleus $10^{-14}$
Size of hydrogen atom $10^{-10}$
Length of typical virus $10^{-8}$
Wavelength of light $10^{-7}$
Size of red blood corpuscle $10^{-5}$
Thickness of a paper $10^{-4}$
Height of Mount Everest above sea level $10^{4}$
Radius of the Earth $10^{7}$
Distance of Moon from the Earth $10^{8}$
Distance of the Sun from the Earth $10^{11}$
Distance of Pluto from the Sun $10^{13}$
Size of our galaxy $10^{21}$
Distance to Andromeda galaxy $10^{22}$
Distance to the boundary of observable universe $10^{26}$

Example 9. If the size of a nucleus ($10^{-15} \text{ m}$) is scaled up to the tip of a sharp pin ($10^{-5} \text{ m}$), what roughly is the size of an atom ($10^{-10} \text{ m}$)?

Answer:

To find the scaled-up size, we first determine the Scaling Factor.

$\text{Scaling Factor} = \frac{\text{Scaled-up size of nucleus}}{\text{Actual size of nucleus}}$

$\text{Scaling Factor} = \frac{10^{-5} \text{ m}}{10^{-15} \text{ m}} = 10^{10}$

Now, we apply this factor to the actual size of the atom:

$\text{Scaled-up size of atom} = \text{Actual size of atom} \times \text{Scaling Factor}$

$\text{Scaled-up size of atom} = 10^{-10} \text{ m} \times 10^{10} = 1 \text{ m}$

Thus, if a nucleus is scaled to the size of a pin-tip, the atom would be as large as a sphere with a radius of 1 metre.


Example 10. How many astronomical units (AU) are there in 1 parsec?

Answer:

We know that:

$1 \text{ pc} = 3.08 \times 10^{16} \text{ m}$

$1 \text{ AU} = 1.496 \times 10^{11} \text{ m}$

To find the number of AU in 1 parsec, we divide the value of 1 parsec by 1 AU:

$\text{Number of AU} = \frac{3.08 \times 10^{16}}{1.496 \times 10^{11}}$

$\text{Number of AU} \approx 2,05,882$

So, 1 parsec is approximately $2.06 \times 10^5 \text{ AU}$.



Measurement of Mass

Mass is a fundamental and basic property of matter. Unlike weight, mass does not depend on the temperature, pressure, or the location of the object in space. For instance, the mass of an object remains the same whether it is on the Earth, the Moon, or in deep space.


SI Unit of Mass

The kilogram ($kg$) is the internationally accepted base unit of mass. Historically, it was defined by a physical prototype—a cylinder made of platinum-iridium alloy kept at the International Bureau of Weights and Measures in France. However, physical objects can accumulate dust or lose atoms over time, leading to slight variations in measurement.

The Modern Definition (Post-2018)

Mass is a basic property of any object that does not change with location. The standard unit to measure mass is the Kilogram ($kg$).

Before 2018, the kilogram was defined by a physical metal cylinder kept in France. However, physical objects can change over time (due to dust or wear). To make the measurement permanent and universal, scientists now use a constant of nature called the Planck Constant ($h$).

Key Constant Values

Derivation of Mass

The new definition links mass to energy. We use two famous equations from physics to understand how mass is calculated from the Planck constant:

  1. Quantum Energy Equation: $E = h\nu$ (where $\nu$ is frequency)
  2. Mass-Energy Equation: $E = mc^2$

Since both represent energy ($E$), we can put them together:

$mc^2 = h\nu$

To find the mass ($m$), we rearrange the formula:

$m = \frac{h\nu}{c^2}$

Conclusion: This formula shows that if we know the fixed value of $h$ and the speed of light $c$, we can define the kilogram ($m$) using the Second (linked to frequency $\nu$) and the Metre (linked to $c$).

Dimensional Consistency

To check if the units are correct, we look at the dimensions:

When we put them in the formula $m = \frac{h\nu}{c^2}$:

$\text{Mass Unit} = \frac{(\text{kg m}^2 \text{ s}^{-1}) \times (\text{s}^{-1})}{\text{m}^2 \text{ s}^{-2}}$

The $\text{m}^2$ and $\text{s}^{-2}$ cancel out, leaving only kilogram ($kg$).


Example 1. If the Planck constant $h$ has a value of $6.6 \times 10^{-34} \text{ kg m}^2 \text{ s}^{-1}$ and the speed of light $c$ is $3 \times 10^8 \text{ m/s}$, show that the units of $\frac{h}{c^2}$ combined with a frequency $\nu$ (measured in $\text{s}^{-1}$) will result in the unit of mass ($kg$).

Answer:

We need to perform a dimensional check on the formula $m = \frac{h\nu}{c^2}$.

1. Dimensions of the Numerator:

Unit of $h = kg \cdot m^2 \cdot s^{-1}$

Unit of $\nu = s^{-1}$

Product $(h \cdot \nu) = (kg \cdot m^2 \cdot s^{-1}) \times (s^{-1}) = kg \cdot m^2 \cdot s^{-2}$

2. Dimensions of the Denominator:

Unit of $c = m \cdot s^{-1}$

Unit of $c^2 = m^2 \cdot s^{-2}$

3. Final Division:

$\text{Resulting Unit} = \frac{kg \cdot m^2 \cdot s^{-2}}{m^2 \cdot s^{-2}}$

The terms $m^2$ and $s^{-2}$ cancel out from the numerator and denominator.

$\text{Remaining Unit} = kg$.

This confirms that mass is successfully defined using the constants $h$ and $c$ along with time-based frequency.


Unified Atomic Mass Unit ($u$)

When dealing with atoms and molecules, using the kilogram is impractical because the numbers are too small. For example, the mass of a hydrogen atom is approximately $0.00000000000000000000000000167 \text{ kg}$. To simplify this, scientists established the Unified Atomic Mass Unit.

Definition

One unified atomic mass unit ($1 \text{ u}$) is defined as exactly one-twelfth ($\frac{1}{12}$) of the mass of an unexcited atom of the Carbon-12 isotope ($_{6}^{12}C$), including the mass of its electrons.

Mathematical Expression and Derivation

The relationship is given by the formula:

$1 \text{ u} = \frac{\text{Mass of one atom of } ^{12}C}{12}$

To find the value in kilograms, we use Avogadro's number ($N_A \approx 6.022 \times 10^{23} \text{ mol}^{-1}$):

Mass of $1 \text{ mole}$ of $^{12}C = 12 \text{ g} = 0.012 \text{ kg}$

Mass of $1 \text{ atom}$ of $^{12}C = \frac{0.012}{6.022 \times 10^{23}} \text{ kg}$

$1 \text{ u} = \frac{1}{12} \times \left( \frac{0.012}{6.022 \times 10^{23}} \right) \text{ kg}$

Result: $1 \text{ u} \approx 1.66 \times 10^{-27} \text{ kg}$


Methods of Measuring Mass

The technique used to measure mass depends entirely on the scale of the object being measured. We categorize these into three main methods:

1. Common Balance (Macroscopic Scale)

For everyday objects like vegetables, grains, or metals, we use a Common Balance (Beam Balance). This is frequently seen in Indian grocery shops and markets. It works by comparing the unknown mass with a known standard mass using the principle of moments.

2. Gravitational Method (Large Scale)

To measure the mass of planets, moons, and stars, we cannot use a balance. Instead, we use Newton’s Law of Gravitation. By observing the orbital period ($T$) and the radius of the orbit ($R$) of a satellite or planet, the mass ($M$) of the central body can be calculated.

The gravitational force provides the centripetal force:

$G \frac{Mm}{R^2} = \frac{mv^2}{R}$

From this, the mass $M$ of the heavy celestial body is derived as:

$M = \frac{v^2 R}{G}$

3. Mass Spectrograph (Atomic Scale)

For sub-atomic particles and atoms, a Mass Spectrograph is used. In this device, a charged particle is made to move through uniform electric and magnetic fields. The particle follows a curved trajectory.


Comparison of Measurement Techniques

Scale Example Object Instrument/Method
Small (Atomic) Protons, Electrons, Atoms Mass Spectrograph
Medium (Everyday) Rice, Gold, Books Common Balance / Electronic Scale
Large (Celestial) Earth, Sun, Jupiter Gravitational Method (Newton's Laws)

Range and Order of Masses

The range of masses encountered in the universe is incredibly vast, spanning over $85$ orders of magnitude. The table below lists the typical masses of various objects:

Object Mass ($kg$)
Electron $10^{-30}$
Proton $10^{-27}$
Uranium atom $10^{-25}$
Red blood cell $10^{-13}$
Dust particle $10^{-9}$
Rain drop $10^{-6}$
Mosquito $10^{-5}$
Grape $10^{-3}$
Human $10^{2}$
Automobile $10^{3}$
Boeing 747 aircraft $10^{8}$
Moon $10^{23}$
Earth $10^{25}$
Sun $10^{30}$
Milky way galaxy $10^{41}$
Observable Universe $10^{55}$


Measurement of Time

To measure any time interval, we require a clock. In modern science, time measurement has reached extraordinary levels of precision by moving from mechanical vibrations to atomic vibrations. The current international standard for time is based on the Atomic Clock.


The Atomic Standard of Time (Caesium Clock)

The Caesium Atomic Clock uses the vibrations of atoms as a "pendulum." In a standard wristwatch, a quartz crystal vibrates at $32,768$ times per second. In contrast, a Caesium atom vibrates billions of times per second, making it significantly more stable and accurate.

Mechanism of Vibration

The "vibrations" in an atomic clock actually refer to the transitions between two energy states of an atom. Specifically, we use the Hyperfine Levels of the ground state of the Caesium-133 atom. When the atom absorbs a specific frequency of microwave radiation, it jumps between these levels. The frequency of this radiation is so constant that it serves as the ultimate "tick" of the universal clock.

Simplified diagram of a Caesium atomic clock showing the transition between two energy levels

Standard Definition of a Second

The Second ($s$) is defined by taking the fixed numerical value of the Caesium frequency, $\Delta\nu_{Cs}$, which is the unperturbed ground-state hyperfine transition frequency of the Caesium-133 atom.

The definition is mathematically stated as:

$\Delta\nu_{Cs} = 9,192,631,770 \text{ Hz}$

This means that exactly $9,192,631,770$ cycles of this radiation must pass for one second to be completed.


Time Maintenance and IST in India

In India, the National Physical Laboratory (NPL) in New Delhi is the custodian of the nation's time. It maintains the Indian Standard Time (IST) using an ensemble of four Caesium atomic clocks.


Linking Time to Length

Because time can be measured with such immense accuracy ($1$ part in $10^{15}$), whereas other quantities like length cannot, the scientific community decided to define Length in terms of Time. This was done by fixing the Speed of Light ($c$).

The Metre is now defined as the distance light travels in vacuum in a time interval of:

$t = \frac{1}{299,792,458} \text{ seconds}$

This ensures that as long as our atomic clocks are accurate, our definition of the metre remains perfectly constant everywhere in the universe.


Range and Order of Time Intervals

The events in our universe occur over a massive range of time, from the fleeting life of an unstable particle to the vast age of the universe itself. The ratio between the longest and shortest time intervals is approximately $10^{41}$.

Event Time Interval ($s$)
Life-span of most unstable particle $10^{-24}$
Time required for light to cross a nuclear distance $10^{-22}$
Period of x-rays $10^{-19}$
Period of atomic vibrations $10^{-15}$
Period of light wave $10^{-15}$
Life time of an excited state of an atom $10^{-8}$
Period of radio wave $10^{-6}$
Period of a sound wave $10^{-3}$
Wink of eye $10^{-1}$
Time between successive human heart beats $10^{0}$
Travel time for light from Moon to the Earth $10^{0}$
Travel time for light from the Sun to the Earth $10^{2}$
Time period of a satellite $10^{4}$
Rotation period of the Earth $10^{5}$
Rotation and revolution periods of the Moon $10^{6}$
Revolution period of the Earth $10^{7}$
Travel time for light from nearest star $10^{8}$
Average human life-span $10^{9}$
Age of Egyptian pyramids $10^{11}$
Time since dinosaurs became extinct $10^{15}$
Age of the universe $10^{17}$

Cosmic Ratios and Curious Coincidences

When we examine the tables for the range of Length, Time, and Mass, a fascinating and mysterious pattern emerges. Scientists have discovered that certain ratios between the largest and smallest scales of our universe result in the same numerical order of magnitude: $10^{41}$.

Analysis of the "Large Number" Coincidence

By comparing the extremes of the universe, we can derive the following mathematical ratios based on the data provided in the previous sections:

1. Ratio of Lengths ($R_L$)

The ratio of the longest length (distance to the boundary of the observable universe) to the shortest length (size of a proton) is calculated as:

$R_L = \frac{10^{26} \text{ m}}{10^{-15} \text{ m}} = 10^{41}$

2. Ratio of Time Intervals ($R_T$)

The ratio of the longest time interval (age of the universe) to the shortest time interval (life-span of the most unstable particle) is calculated as:

$R_T = \frac{10^{17} \text{ s}}{10^{-24} \text{ s}} \approx 10^{41}$

3. Ratio of Masses ($R_M$)

The ratio of the largest mass (observable universe) to the smallest mass (electron) is approximately the square of this "magic number":

$R_M = \frac{10^{55} \text{ kg}}{10^{-30} \text{ kg}} \approx 10^{85}$

Interestingly, $(10^{41})^2 = 10^{82}$, which is very close to the order of magnitude of the mass ratio of the universe.

Summary of Cosmic Ratios

The following table summarizes these coincidences which have led many physicists, including Paul Dirac, to speculate about deeper laws of nature.

Physical Quantity Calculated Ratio Order of Magnitude
Length (Largest/Smallest) $10^{26} / 10^{-15}$ $10^{41}$
Time (Longest/Shortest) $10^{17} / 10^{-24}$ $10^{41}$
Mass (Largest/Smallest) $10^{55} / 10^{-30}$ $\approx (10^{41})^2$

Whether these ratios are purely accidental or point toward a deeper physical symmetry in the fabric of the universe remains one of the greatest mysteries in modern physics.

Infographic showing the scale of the universe from a proton to the observable boundary

Example 1. If a Caesium atomic clock at NPL New Delhi has an uncertainty of $1$ part in $10^{15}$, calculate the maximum error (in seconds) it might accumulate over a period of $10^8$ seconds.

Answer:

Step 1: Identify the uncertainty.

Relative Uncertainty $= \frac{1}{10^{15}} = 10^{-15}$.

Step 2: Calculate the error for the given time interval.

Total time ($t$) $= 10^8 \ s$.

$\text{Error} = \text{Uncertainty} \times \text{Total time}$

$\text{Error} = 10^{-15} \times 10^{8} \ s$

$\text{Error} = 10^{-7} \ s$.

The clock would accumulate a maximum error of only $0.0000001 \text{ seconds}$ over this long period.


Example 2. Light takes approximately $500 \ s$ to travel from the Sun to the Earth. Given the speed of light is $3 \times 10^8 \text{ m/s}$, calculate the distance between the Sun and the Earth.

Answer:

1. Given Data:

  • Time ($t$) $= 500 \ s$.
  • Speed ($v$) $= 3 \times 10^8 \text{ m/s}$.

2. Formula:

$\text{Distance} = \text{Speed} \times \text{Time}$

3. Calculation:

$\text{Distance} = (3 \times 10^8 \text{ m/s}) \times 500 \ s$

$\text{Distance} = 1500 \times 10^8 \text{ m}$

$\text{Distance} = 1.5 \times 10^{11} \text{ m}$.

This distance is approximately $1 \text{ Astronomical Unit (AU)}$.


Example 3. Based on the above tables, calculate the ratio of the radius of the Earth to the thickness of a paper, and the ratio of the age of the universe to the time taken for light to cross a nuclear distance. Do these ratios also relate to the large number coincidence?

Answer:

1. Length Ratio:

Radius of Earth $= 10^7 \text{ m}$

Thickness of paper $= 10^{-4} \text{ m}$

$\text{Ratio} = \frac{10^7}{10^{-4}} = 10^{11}$

2. Time Ratio:

Age of universe $= 10^{17} \text{ s}$

Time for light to cross nuclear distance $= 10^{-22} \text{ s}$

$\text{Ratio} = \frac{10^{17}}{10^{-22}} = 10^{39}$

Conclusion: The second ratio ($10^{39}$) is very close to the Large Number Coincidence of $10^{41}$, while the first ratio ($10^{11}$) represents a smaller macroscopic scale within our local environment.



Accuracy, Precision of Instruments and Errors in Measurement

Measurement is the cornerstone of all experimental science and technology. However, the result of every measurement by any measuring instrument contains some degree of uncertainty. This uncertainty is technically called an error. To understand the quality of a measurement, we must analyze two fundamental concepts: Accuracy and Precision.


Accuracy and Precision

While often used interchangeably in daily life, these terms have distinct meanings in scientific measurement:

1. Accuracy: It is a measure of how close the measured value is to the true value of the quantity. Accuracy depends on the calibration of the instrument and the removal of systematic biases.

2. Precision: It tells us to what resolution or limit the quantity is measured. Precision is determined by the least count of the instrument. A measurement can be very precise but inaccurate if the instrument is not correctly calibrated.

Illustrative Example

Suppose the true value of a certain length is exactly $3.678 \text{ cm}$.

In this case, the first measurement is more accurate (closer to $3.678$) but less precise. The second measurement is more precise (higher resolution) but less accurate.

Target board analogy showing high accuracy/low precision vs low accuracy/high precision

Classification of Errors

Errors in measurement are generally classified into two broad categories based on their nature and source: Systematic Errors and Random Errors.

1. Systematic Errors

These are errors that tend to be in one direction, either consistently positive or consistently negative. Since the cause is often known, these errors can be minimized or corrected. Major sources include:


2. Random Errors

Random errors occur irregularly and vary in both sign and size. They are unpredictable and arise due to:

Because these errors are random, they can be minimized by taking a large number of observations and calculating their Arithmetic Mean. The likelihood of overestimating the value becomes equal to the likelihood of underestimating it, causing the errors to cancel out on average.


3. Least Count Error

The Least Count is the smallest value that can be measured by a measuring instrument. The uncertainty associated with this limit is called the Least Count Error.

We can reduce this error by using instruments of higher precision or by improving experimental techniques.


Mathematical Analysis of Errors

In scientific experiments, since the "true value" of a physical quantity is often unknown, we use statistical methods to estimate it and determine the reliability of our measurements. This process involves calculating the average of multiple readings and analyzing the deviations from that average.

1. Arithmetic Mean ($a_{mean}$)

When we perform an experiment multiple times, random errors cause the readings to vary. By taking the arithmetic mean, we assume that overestimations and underestimations will cancel each other out, providing us with a value closest to the true value.

If $a_1, a_2, a_3, ..., a_n$ are the values obtained in $n$ measurements, then the arithmetic mean is calculated as:

$a_{mean} = \frac{a_1 + a_2 + a_3 + ... + a_n}{n}$

In summation notation, this is expressed as:

$a_{mean} = \sum\limits_{i=1}^{n} \frac{a_i}{n}$

2. Absolute Error ($|\Delta a|$)

The Absolute Error is the magnitude of the difference between the true value (taken as the arithmetic mean) and the individual measured value. It tells us how much a specific reading "deviated" from the average.

The absolute errors for individual readings are calculated as follows:

$\Delta a_1 = a_1 - a_{mean}$

$\Delta a_2 = a_2 - a_{mean}$

... and so on up to $\Delta a_n = a_n - a_{mean}$.

Note: While the calculated $\Delta a_i$ can be positive or negative, the absolute error is always considered as the magnitude, denoted by $|\Delta a_i|$, which is always positive.

3. Mean Absolute Error ($\Delta a_{mean}$)

To find the overall uncertainty in our set of measurements, we take the arithmetic mean of all the individual absolute errors. This is called the Mean Absolute Error.

The formula for Mean Absolute Error is:

$\Delta a_{mean} = \frac{|\Delta a_1| + |\Delta a_2| + |\Delta a_3| + ... + |\Delta a_n|}{n}$

Or in summation form:

$\Delta a_{mean} = \sum\limits_{i=1}^{n} \frac{|\Delta a_i|}{n}$

The final result of a measurement is usually reported as: $a = a_{mean} \pm \Delta a_{mean}$. This implies that the true value is likely to lie between $(a_{mean} + \Delta a_{mean})$ and $(a_{mean} - \Delta a_{mean})$.

4. Relative Error and Percentage Error

Absolute error does not always give a clear picture of the "seriousness" of the error. For example, an error of $1 \text{ cm}$ is huge if you are measuring a pen, but negligible if you are measuring a national highway. To solve this, we use Relative and Percentage errors.

Relative Error

It is the ratio of the mean absolute error to the mean value of the quantity measured.

$\text{Relative Error} = \frac{\Delta a_{mean}}{a_{mean}}$

Percentage Error ($\delta a$)

When the relative error is expressed in percent, it is called the Percentage Error.

$\delta a = \left( \frac{\Delta a_{mean}}{a_{mean}} \right) \times 100 \%$


Summary Table of Error Analysis

Term Mathematical Definition Purpose
Arithmetic Mean $\sum \frac{a_i}{n}$ Acts as the 'True Value'
Absolute Error $|a_i - a_{mean}|$ Individual deviation
Mean Absolute Error $\sum \frac{|\Delta a_i|}{n}$ Average uncertainty
Relative Error $\frac{\Delta a_{mean}}{a_{mean}}$ Normalized error
Percentage Error $\text{Relative Error} \times 100$ Standardized error in %

Example 1. An Indian student measuring the diameter of a thin wire using a screw gauge recorded the following five readings: $2.04 \text{ mm}$, $2.06 \text{ mm}$, $2.08 \text{ mm}$, $2.03 \text{ mm}$, and $2.04 \text{ mm}$. Calculate the (i) Mean Diameter, (ii) Absolute error in each measurement, (iii) Mean absolute error, and (iv) Percentage error.

Answer:

(i) Mean Diameter ($d_{mean}$):

$d_{mean} = \frac{2.04 + 2.06 + 2.08 + 2.03 + 2.04}{5} = \frac{10.25}{5} = 2.05 \text{ mm}$

(ii) Absolute Errors ($\Delta d_i = d_i - d_{mean}$):

$|\Delta d_1| = |2.04 - 2.05| = 0.01 \text{ mm}$

$|\Delta d_2| = |2.06 - 2.05| = 0.01 \text{ mm}$

$|\Delta d_3| = |2.08 - 2.05| = 0.03 \text{ mm}$

$|\Delta d_4| = |2.03 - 2.05| = 0.02 \text{ mm}$

$|\Delta d_5| = |2.04 - 2.05| = 0.01 \text{ mm}$

(iii) Mean Absolute Error ($\Delta d_{mean}$):

$\Delta d_{mean} = \frac{0.01 + 0.01 + 0.03 + 0.02 + 0.01}{5} = \frac{0.08}{5} = 0.016 \text{ mm}$

Rounding off to the precision of the instrument: $\Delta d_{mean} \approx 0.02 \text{ mm}$.

(iv) Percentage Error ($\delta d$):

$\delta d = \left( \frac{\Delta d_{mean}}{d_{mean}} \right) \times 100 = \left( \frac{0.016}{2.05} \right) \times 100 \approx 0.78 \%$

The reported measurement is $(2.05 \pm 0.02) \text{ mm}$ with a $0.78 \%$ error.


Example 2. In a physics lab, the refractive index of glass was measured to be $1.54, 1.53, 1.44, 1.54, 1.56$ and $1.45$. Calculate the mean value, absolute error, and relative error.

Answer:

Step 1: Mean Refractive Index ($\mu_{mean}$)

$\mu_{mean} = \frac{1.54 + 1.53 + 1.44 + 1.54 + 1.56 + 1.45}{6} = \frac{9.06}{6} = 1.51$

Step 2: Absolute Errors ($|\Delta \mu_i|$)

$|\Delta \mu_1| = |1.54 - 1.51| = 0.03$

$|\Delta \mu_2| = |1.53 - 1.51| = 0.02$

$|\Delta \mu_3| = |1.44 - 1.51| = 0.07$

$|\Delta \mu_4| = |1.54 - 1.51| = 0.03$

$|\Delta \mu_5| = |1.56 - 1.51| = 0.05$

$|\Delta \mu_6| = |1.45 - 1.51| = 0.06$

Step 3: Mean Absolute Error ($\Delta \mu_{mean}$)

$\Delta \mu_{mean} = \frac{0.03 + 0.02 + 0.07 + 0.03 + 0.05 + 0.06}{6} = \frac{0.26}{6} \approx 0.043$

Step 4: Relative Error

$\text{Relative Error} = \frac{\Delta \mu_{mean}}{\mu_{mean}} = \frac{0.043}{1.51} \approx 0.028$

The refractive index of glass is $1.51 \pm 0.04$.


Example 3. Two masses are measured as $m_1 = (25.4 \pm 0.1) \text{ g}$ and $m_2 = (15.2 \pm 0.2) \text{ g}$. Find the sum of the masses and the absolute error in the sum.

Answer:

This is a case of Combination of Errors (Summation).

1. Sum of the Mean Values:

$M = m_1 + m_2 = 25.4 + 15.2 = 40.6 \text{ g}$

2. Sum of the Absolute Errors:

As per the rule of summation, $\Delta Z = \Delta A + \Delta B$.

$\Delta M = \Delta m_1 + \Delta m_2$

$\Delta M = 0.1 + 0.2 = 0.3 \text{ g}$

Illustration showing how error bars from two separate measurements combine into a larger error bar for the sum

3. Result:

The total mass is $(40.6 \pm 0.3) \text{ g}$.


Example 4. The period of oscillation of a simple pendulum is $T = 2\pi\sqrt{L/g}$. If the measured length $L = 10 \text{ cm}$ with $1 \text{ mm}$ accuracy and the time $T$ for $1$ oscillation is $2.0 \text{ s}$ with $0.1 \text{ s}$ resolution, find the percentage error in the determination of acceleration due to gravity ($g$).

Answer:

Step 1: Express the formula for $g$.

$T = 2\pi\sqrt{\frac{L}{g}} \implies T^2 = 4\pi^2 \frac{L}{g} \implies g = \frac{4\pi^2 L}{T^2}$

Step 2: Formula for Relative Error.

Since $4\pi^2$ is a constant, it has no error. Using the product and power rule:

$\frac{\Delta g}{g} = \frac{\Delta L}{L} + 2 \frac{\Delta T}{T}$

Step 3: Plug in the values.

$L = 10 \text{ cm} = 100 \text{ mm}, \Delta L = 1 \text{ mm}$

$T = 2.0 \text{ s}, \Delta T = 0.1 \text{ s}$

$\frac{\Delta g}{g} = \left( \frac{1}{100} \right) + 2 \left( \frac{0.1}{2.0} \right)$

$\frac{\Delta g}{g} = 0.01 + 0.1 = 0.11$

Step 4: Calculate Percentage Error.

$\delta g = 0.11 \times 100\% = 11\%$

The percentage error in calculating $g$ is $11 \%$.


Combination of Errors

When we perform an experiment, the final result often depends on several different measured quantities. Since each measurement has its own uncertainty, these errors "propagate" or combine to affect the final calculated value. Understanding how these errors combine is essential for reporting accurate scientific results.

1. Error of a Sum or a Difference

The fundamental principle in error analysis is that we always calculate the Maximum Possible Error. This ensures that the true value of the quantity is guaranteed to lie within the range we specify. This is why, even when two quantities are subtracted, their absolute errors are added, never subtracted.

A. Error of a Sum

Suppose two physical quantities $A$ and $B$ have measured values $(A \pm \Delta A)$ and $(B \pm \Delta B)$. Let $Z$ be their sum:

$Z = A + B$

Derivation:

  1. Including the errors, the value of $Z$ becomes:

    $Z \pm \Delta Z = (A \pm \Delta A) + (B \pm \Delta B)$

  2. Rearranging the terms on the right-hand side:

    $Z \pm \Delta Z = (A + B) \pm \Delta A \pm \Delta B$

  3. Since we know that $Z = A + B$, the term $(A + B)$ cancels out from both sides:

    $\pm \Delta Z = \pm \Delta A \pm \Delta B$

  4. The four possible values for $\Delta Z$ are $(+\Delta A + \Delta B)$, $(+\Delta A - \Delta B)$, $(-\Delta A + \Delta B)$, and $(-\Delta A - \Delta B)$.
  5. The maximum absolute error is therefore:

    $\Delta Z = \Delta A + \Delta B$

B. Error of a Difference

Suppose we want to find the error in the difference of two quantities $Z = A - B$.

Derivation:

  1. Including the errors, the value of $Z$ becomes:

    $Z \pm \Delta Z = (A \pm \Delta A) - (B \pm \Delta B)$

  2. Opening the brackets:

    $Z \pm \Delta Z = (A - B) \pm \Delta A \mp \Delta B$

  3. Since $Z = A - B$, the equation simplifies to:

    $\pm \Delta Z = \pm \Delta A \mp \Delta B$

  4. To find the most pessimistic or maximum possible error, we take the sum of the magnitudes of the errors. We do not subtract the errors because a smaller error would lead us to be overconfident in our result.
  5. Thus, the maximum absolute error is:

    $\Delta Z = \Delta A + \Delta B$

Summary Rule for Sum and Difference

Rule: 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.

Mathematical Operation Measured Result ($Z$) Maximum Absolute Error ($\Delta Z$)
Addition $Z = A + B$ $\Delta A + \Delta B$
Subtraction $Z = A - B$ $\Delta A + \Delta B$

Example 5. Two steel rods used in a construction project in Mumbai have lengths $L_1 = (15.4 \pm 0.2) \text{ cm}$ and $L_2 = (10.1 \pm 0.1) \text{ cm}$. Find the total length of the two rods joined together and the uncertainty in the measurement.

Answer:

1. Calculate the total mean length:

$L = L_1 + L_2 = 15.4 \text{ cm} + 10.1 \text{ cm} = 25.5 \text{ cm}$

2. Calculate the maximum absolute error:

$\Delta L = \Delta L_1 + \Delta L_2$

$\Delta L = 0.2 \text{ cm} + 0.1 \text{ cm} = 0.3 \text{ cm}$

3. Result:

The total length is $(25.5 \pm 0.3) \text{ cm}$.


Example 6. A student weighs two chemical samples in a lab. The mass of the first sample is $M_1 = (2.53 \pm 0.01) \text{ g}$ and the mass of the second is $M_2 = (1.42 \pm 0.01) \text{ g}$. Calculate the difference in their masses and the absolute error.

Answer:

1. Calculate the mean difference:

$M = M_1 - M_2 = 2.53 \text{ g} - 1.42 \text{ g} = 1.11 \text{ g}$

2. Calculate the maximum absolute error:

Even though we are subtracting the masses, the errors must be added.

$\Delta M = \Delta M_1 + \Delta M_2$

$\Delta M = 0.01 \text{ g} + 0.01 \text{ g} = 0.02 \text{ g}$

3. Result:

The difference in mass is $(1.11 \pm 0.02) \text{ g}$.


Example 7. The initial and final readings of a water meter in an Indian household are $(342.6 \pm 0.2) \text{ Litres}$ and $(450.8 \pm 0.3) \text{ Litres}$. Calculate the total water consumed and the uncertainty in this value.

Answer:

1. Mean value of consumption:

$V = V_{final} - V_{initial} = 450.8 - 342.6 = 108.2 \text{ Litres}$

2. Absolute error in consumption:

$\Delta V = \Delta V_{final} + \Delta V_{initial}$

$\Delta V = 0.3 + 0.2 = 0.5 \text{ Litres}$

3. Final Result:

Total water consumed is $(108.2 \pm 0.5) \text{ Litres}$.

2. Error of a Product or a Quotient

When physical quantities are related by multiplication or division, the errors do not combine as simple additions of absolute values. Instead, it is the Relative Error (or Fractional Error) that propagates. This section explains how uncertainties are calculated for derived quantities like area, density, or speed.

1. Error of a Product

Suppose a physical quantity $Z$ is the product of two independent measured quantities $A$ and $B$. Let the measured values be $(A \pm \Delta A)$ and $(B \pm \Delta B)$, where $\Delta A$ and $\Delta B$ are their respective absolute errors.

Mathematical Derivation

We start with the relationship:

$Z = AB$

Including the uncertainties, the measured value of $Z$ becomes:

$Z \pm \Delta Z = (A \pm \Delta A)(B \pm \Delta B)$

Expanding the brackets on the right-hand side:

$Z \pm \Delta Z = AB \pm A\Delta B \pm B\Delta A \pm \Delta A\Delta B$

To find the relative error, we divide the left side by $Z$ and the right side by $AB$ (since $Z = AB$):

$\frac{Z \pm \Delta Z}{Z} = \frac{AB \pm A\Delta B \pm B\Delta A \pm \Delta A\Delta B}{AB}$

$1 \pm \frac{\Delta Z}{Z} = 1 \pm \frac{\Delta B}{B} \pm \frac{\Delta A}{A} \pm \frac{\Delta A\Delta B}{AB}$

In scientific measurements, the absolute errors $\Delta A$ and $\Delta B$ are very small compared to the actual values $A$ and $B$. Therefore, their product $\Delta A\Delta B$ is negligibly small and can be ignored. After canceling $1$ from both sides, we get:

$\pm \frac{\Delta Z}{Z} = \pm \frac{\Delta A}{A} \pm \frac{\Delta B}{B}$

For the Maximum Relative Error, we sum the individual relative errors:

$\frac{\Delta Z}{Z} = \frac{\Delta A}{A} + \frac{\Delta B}{B}$

2. Error of a Quotient

The same rule applies when quantities are divided. If $Z = A/B$, the maximum relative error is still the sum of the relative errors of $A$ and $B$.

Logic using Logarithmic Differentiation

A more sophisticated way to derive this for any power or quotient is by using natural logarithms:

$ln \ Z = ln \ A - ln \ B$

Differentiating both sides:

$\frac{dZ}{Z} = \frac{dA}{A} - \frac{dB}{B}$

To obtain the maximum possible error, we take the absolute values and add them:

$\frac{\Delta Z}{Z} = \frac{\Delta A}{A} + \frac{\Delta B}{B}$

The General Rule

Rule: When two quantities are multiplied or divided, the relative error in the result is the sum of the relative errors in the multipliers or dividers.

Relationship Result ($Z$) Maximum Relative Error ($\frac{\Delta Z}{Z}$)
Product $Z = A \times B$ $\frac{\Delta A}{A} + \frac{\Delta B}{B}$
Quotient $Z = A / B$ $\frac{\Delta A}{A} + \frac{\Delta B}{B}$

Example 8. The length and breadth of a rectangular plot in an Indian housing society are measured as $L = (50.0 \pm 0.5) \text{ m}$ and $B = (20.0 \pm 0.2) \text{ m}$. Calculate the area of the plot and the percentage error in the area.

Answer:

Step 1: Calculate the mean Area ($A$).

$A = L \times B = 50.0 \times 20.0 = 1000 \text{ m}^2$

Step 2: Calculate the relative error in Area.

$\frac{\Delta A}{A} = \frac{\Delta L}{L} + \frac{\Delta B}{B}$

$\frac{\Delta A}{A} = \frac{0.5}{50.0} + \frac{0.2}{20.0}$

$\frac{\Delta A}{A} = 0.01 + 0.01 = 0.02$

Step 3: Calculate the Percentage Error.

Percentage Error $= 0.02 \times 100\% = 2\%$

Step 4: Express the result.

Absolute error $\Delta A = 0.02 \times 1000 = 20 \text{ m}^2$

Final Area $= (1000 \pm 20) \text{ m}^2$.


Example 9. A scientist in an Indian lab measures the mass of a metal block as $M = (2.50 \pm 0.05) \text{ kg}$ and its volume as $V = (0.20 \pm 0.01) \text{ m}^3$. Find the density ($\rho$) and its percentage error.

Answer:

Step 1: Calculate Mean Density.

$\rho = \frac{M}{V} = \frac{2.50}{0.20} = 12.5 \text{ kg/m}^3$

Step 2: Calculate Relative Errors.

$\text{Relative error in Mass } = \frac{0.05}{2.50} = 0.02$

$\text{Relative error in Volume } = \frac{0.01}{0.20} = 0.05$

Step 3: Total Percentage Error.

$\text{Percentage error in Density } = (0.02 + 0.05) \times 100\% = 7\%$

The density is $12.5 \text{ kg/m}^3$ with an uncertainty of $7\%$.


Example 10. The length and breadth of a rectangular field are measured with percentage errors of $2\%$ and $3\%$ respectively. Calculate the total percentage error in the measurement of the area of the field.

Answer:

Step 1: Identify the relationship between the quantities.

The area ($A$) of a rectangular field is given by the product of its length ($L$) and breadth ($B$):

$A = L \times B$

Step 2: Apply the rule for combination of errors in a product.

According to the rule of propagation of errors, when two quantities are multiplied, the relative error in the result is the sum of the relative errors of the individual quantities. This can be expressed as:

$\frac{\Delta A}{A} = \frac{\Delta L}{L} + \frac{\Delta B}{B}$

Step 3: Convert the relative error to percentage error.

Multiplying the entire equation by $100$:

$\left( \frac{\Delta A}{A} \times 100 \right) = \left( \frac{\Delta L}{L} \times 100 \right) + \left( \frac{\Delta B}{B} \times 100 \right)$

$\text{Percentage Error in Area} = \text{Percentage Error in Length} \ $$ + \text{Percentage Error in Breadth}$

Step 4: Calculation.

Given:

  • Percentage error in length $= 2\%$
  • Percentage error in breadth $= 3\%$

$\text{Total Percentage Error in Area} = 2\% + 3\% = 5\%$

The total percentage error in the measurement of the area of the field is $5\%$.

3. Error in Case of a Power

When a physical quantity is raised to a mathematical power, the relative error in that quantity is multiplied by the value of the power. This is a very important rule because it shows that quantities with higher exponents contribute more significantly to the total uncertainty of the result.

1. Derivation for a Simple Power

Suppose a physical quantity $Z$ depends on a measured quantity $A$ such that:

$Z = A^2$

We can write this expression as a product of the same quantity:

$Z = A \times A$

Applying the Product Rule of errors (where relative errors are added):

$\frac{\Delta Z}{Z} = \frac{\Delta A}{A} + \frac{\Delta A}{A}$

$\frac{\Delta Z}{Z} = 2 \frac{\Delta A}{A}$

Hence, the relative error in $A^2$ is twice the relative error in $A$.

2. General Dimensional Equation Rule

In a more complex scenario where a quantity $Z$ depends on multiple variables raised to different powers, such as:

$Z = \frac{A^p B^q}{C^r}$

The maximum relative error in $Z$ is obtained by summing the relative errors of each component multiplied by their respective powers. Note that even if a quantity is in the denominator (like $C^r$), its error is added because we always seek the maximum possible uncertainty.

The General Formula:

$\frac{\Delta Z}{Z} = p \left( \frac{\Delta A}{A} \right) + q \left( \frac{\Delta B}{B} \right) + r \left( \frac{\Delta C}{C} \right)$

Rule: The relative error in a physical quantity raised to the power $k$ is the $k$ times the relative error in the individual quantity.

Summary of Power Rule Multipliers
Quantity Form Power ($k$) Relative Error Contribution
$A^2$ $2$ $2 \left( \frac{\Delta A}{A} \right)$
$A^3$ $3$ $3 \left( \frac{\Delta A}{A} \right)$
$\sqrt{A}$ $1/2$ $\frac{1}{2} \left( \frac{\Delta A}{A} \right)$
$\frac{1}{A}$ $1$ $1 \left( \frac{\Delta A}{A} \right)$
$A^{3/2}$ $1.5$ $1.5 \left( \frac{\Delta A}{A} \right)$

Example 11. Find the relative error in $Z$, if $Z = \frac{A^4 B^{1/3}}{C D^{3/2}}$. If the percentage errors in the measurement of $A, B, C$ and $D$ are $1\%, 3\%, 2\%$ and $4\%$ respectively, calculate the total percentage error in $Z$.

Answer:

Step 1: Apply the General Rule for Errors in Powers

According to the rule of propagation of errors for powers, the relative error in $Z$ is the sum of the relative errors of individual quantities multiplied by their respective exponents. For $Z = \frac{A^4 B^{1/3}}{C D^{3/2}}$, we have:

$$\frac{\Delta Z}{Z} = 4 \left( \frac{\Delta A}{A} \right) + \frac{1}{3} \left( \frac{\Delta B}{B} \right) + 1 \left( \frac{\Delta C}{C} \right) + \frac{3}{2} \left( \frac{\Delta D}{D} \right)$$

Step 2: Substitute the Given Percentage Errors

The percentage error is simply the relative error multiplied by $100$. We can substitute the given values directly into the equation:

  • Percentage error in $A = 1\%$
  • Percentage error in $B = 3\%$
  • Percentage error in $C = 2\%$
  • Percentage error in $D = 4\%$

$\text{Percentage error in } Z = [4 \times (1\%)] + [\frac{1}{3} \times (3\%)] + [1 \times (2\%)] \ $$ + [\frac{3}{2} \times (4\%)]$

Step 3: Perform the Calculation

$$\text{Percentage error in } Z = 4\% + 1\% + 2\% + 6\% = 13\%$$

The total percentage error in the physical quantity $Z$ is $13\%$.


Example 12. In a laboratory experiment, the volume of a sphere is calculated using the formula $V = \frac{4}{3}\pi r^3$. If the error in the measurement of the radius $r$ is $2\%$, find the percentage error in the calculated volume.

Answer:

1. Identification of Constants and Variables:

The formula for the volume of a sphere is $V = \frac{4}{3}\pi r^3$. In this expression, the term $\frac{4}{3}\pi$ is a numerical constant. Constants do not have any uncertainty or error associated with them.

2. Application of the Power Rule:

The error in volume $V$ depends solely on the error in the radius $r$ raised to the power of $3$. Applying the rule for errors in powers:

$$\frac{\Delta V}{V} = 3 \times \frac{\Delta r}{r}$$

3. Final Calculation:

Given that the percentage error in $r$ is $2\%$:

$$\text{Percentage error in } V = 3 \times 2\% = 6\%$$

Thus, the percentage error in the volume of the sphere is $6\%$.


Example 13. The period of oscillation of a simple pendulum is $T = 2\pi \sqrt{L/g}$. The measured value of $L$ is $20.0 \text{ cm}$ known to $1 \text{ mm}$ accuracy and the time for $100$ oscillations is found to be $90 \text{ s}$ using a wrist watch of $1 \text{ s}$ resolution. Calculate the percentage error in the determination of $g$.

Answer:

Step 1: Rearrange the Formula to Isolate $g$

Starting from $T = 2\pi \sqrt{\frac{L}{g}}$, square both sides to get $T^2 = 4\pi^2 \frac{L}{g}$. Rearranging gives:

$$g = \frac{4\pi^2 L}{T^2}$$

A simple pendulum diagram showing length L and a bob

Step 2: Express the Relative Error

Since $4\pi^2$ is a constant, the relative error in $g$ is the sum of the relative error in $L$ and twice the relative error in $T$:

$$\frac{\Delta g}{g} = \frac{\Delta L}{L} + 2 \left( \frac{\Delta T}{T} \right)$$

Step 3: Identify the Given Values and Accuracies

  • Measured length $L = 20.0 \text{ cm}$
  • Accuracy (Absolute Error) in length $\Delta L = 1 \text{ mm} = 0.1 \text{ cm}$
  • Total time for 100 oscillations $t = 90 \text{ s}$
  • Resolution (Absolute Error) in time $\Delta t = 1 \text{ s}$

Note: The relative error in the time period $T$ is the same as the relative error in the total time $t$ ($\frac{\Delta T}{T} = \frac{\Delta t}{t}$).

Step 4: Substitution and Calculation

$$\frac{\Delta g}{g} = \frac{0.1}{20.0} + 2 \times \left( \frac{1}{90} \right)$$

$$\frac{\Delta g}{g} = 0.005 + 0.0222 = 0.0272$$

Step 5: Convert to Percentage Error

$$\text{Percentage error in } g = 0.0272 \times 100\% = 2.72\%$$

Rounding off to appropriate significant figures, the final accuracy in the determination of $g$ is approximately $3\%$.


Visualization of Power Impact

Quantities with higher powers in a formula contribute more to the final error. For instance, in the formula for $g$ above, an error in time ($T$) has double the impact of an error in length ($L$) because $T$ is squared.

A graph showing how relative error increases more sharply for quantities with higher powers

Summary Table of Error Rules

The following table summarizes how absolute and relative errors combine during various arithmetic operations.

Mathematical Operation Result ($Z$) Error Rule (Maximum Error)
Addition $Z = A + B$ $\Delta Z = \Delta A + \Delta B$ (Absolute errors add)
Subtraction $Z = A - B$ $\Delta Z = \Delta A + \Delta B$ (Absolute errors add)
Multiplication $Z = A \times B$ $\frac{\Delta Z}{Z} = \frac{\Delta A}{A} + \frac{\Delta B}{B}$ (Relative errors add)
Division $Z = A / B$ $\frac{\Delta Z}{Z} = \frac{\Delta A}{A} + \frac{\Delta B}{B}$ (Relative errors add)
Power $Z = A^k$ $\frac{\Delta Z}{Z} = k \left( \frac{\Delta A}{A} \right)$ (Relative error multiplied by power)

Practical Challenges in Measurement

One of the most fascinating challenges in measurement is dealing with the fractal nature of objects. A common example is the measurement of international boundaries and coastlines.

Illustration of a coastline being measured with different scales showing length increase

Least Count of Measuring Instruments

The Least Count is the smallest value that can be measured by a measuring instrument. All readings or measured values are reliable only up to this limit. Using instruments with a smaller least count increases the precision of the measurement.

Standard Least Counts in Laboratory
Instrument Least Count (in cm)
Metre Scale $0.1 \text{ cm}$
Vernier Callipers $0.01 \text{ cm}$
Screw Gauge $0.001 \text{ cm}$
Spherometer $0.001 \text{ cm}$

Detailed Analysis of Error Propagation

In many experiments, like finding the acceleration due to gravity ($g$) using a simple pendulum, we combine multiple measurements (Length and Time). The error in each measurement contributes to the total error in the result.



Significant Figures



Dimensions of Physical Quantities



Dimensional Formulae and Dimensional Equations



Dimensional Analysis and Its Applications



Exercises



Question 2.1. Fill in the blanks

(a) The volume of a cube of side 1 cm is equal to .....$m^3$

(b) The surface area of a solid cylinder of radius 2.0 cm and height 10.0 cm is equal to ...$(mm)^2$

(c) A vehicle moving with a speed of 18 km h$^{–1}$ covers....m in 1 s

(d) The relative density of lead is 11.3. Its density is ....g cm$^{–3}$ or ....kg m$^{–3}$.

Answer:

Question 2.2. Fill in the blanks by suitable conversion of units

(a) 1 kg m$^2$ s$^{–2}$ = ....g cm$^2$ s$^{–2}$

(b) 1 m = ..... ly

(c) 3.0 m s$^{–2}$ = .... km h$^{–2}$

(d) G = 6.67 × 10$^{–11}$ N m$^2$ (kg)$^{–2}$ = .... (cm)$^3$ s$^{–2}$ g$^{–1}$.

Answer:

Question 2.3. A calorie is a unit of heat (energy in transit) and it equals about 4.2 J where 1J = 1 kg m$^2$ s$^{–2}$. Suppose we employ a system of units in which the unit of mass equals $\alpha$ kg, the unit of length equals $\beta$ m, the unit of time is $\gamma$ s. Show that a calorie has a magnitude 4.2 $\alpha^{–1} \beta^{–2} \gamma^2$ in terms of the new units.

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Question 2.4. Explain this statement clearly :

“To call a dimensional quantity ‘large’ or ‘small’ is meaningless without specifying a standard for comparison”. In view of this, reframe the following statements wherever necessary :

(a) atoms are very small objects

(b) a jet plane moves with great speed

(c) the mass of Jupiter is very large

(d) the air inside this room contains a large number of molecules

(e) a proton is much more massive than an electron

(f) the speed of sound is much smaller than the speed of light.

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Question 2.5. A new unit of length is chosen such that the speed of light in vacuum is unity. What is the distance between the Sun and the Earth in terms of the new unit if light takes 8 min and 20 s to cover this distance ?

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Question 2.6. Which of the following is the most precise device for measuring length :

(a) a vernier callipers with 20 divisions on the sliding scale

(b) a screw gauge of pitch 1 mm and 100 divisions on the circular scale

(c) an optical instrument that can measure length to within a wavelength of light ?

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Question 2.7. A student measures the thickness of a human hair by looking at it through a microscope of magnification 100. He makes 20 observations and finds that the average width of the hair in the field of view of the microscope is 3.5 mm. What is the estimate on the thickness of hair ?

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Question 2.8. Answer the following :

(a)You are given a thread and a metre scale. How will you estimate the diameter of the thread ?

(b)A screw gauge has a pitch of 1.0 mm and 200 divisions on the circular scale. Do you think it is possible to increase the accuracy of the screw gauge arbitrarily by increasing the number of divisions on the circular scale ?

(c) The mean diameter of a thin brass rod is to be measured by vernier callipers. Why is a set of 100 measurements of the diameter expected to yield a more reliable estimate than a set of 5 measurements only ?

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Question 2.9. The photograph of a house occupies an area of 1.75 cm$^2$ on a 35 mm slide. The slide is projected on to a screen, and the area of the house on the screen is 1.55 m$^2$. What is the linear magnification of the projector-screen arrangement.

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Question 2.10. State the number of significant figures in the following :

(a) 0.007 m$^2$

(b) 2.64 × 10$^{24}$ kg

(c) 0.2370 g cm$^{–3}$

(d) 6.320 J

(e) 6.032 N m$^{–2}$

(f) 0.0006032 m$^2$

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Question 2.11. The length, breadth and thickness of a rectangular sheet of metal are 4.234 m, 1.005 m, and 2.01 cm respectively. Give the area and volume of the sheet to correct significant figures.

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Question 2.12. The mass of a box measured by a grocer’s balance is 2.30 kg. Two gold pieces of masses 20.15 g and 20.17 g are added to the box. What is (a) the total mass of the box, (b) the difference in the masses of the pieces to correct significant figures ?

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Question 2.13. A physical quantity P is related to four observables a, b, c and d as follows :

$P = a^3 b^2 / (\sqrt{c} d)$

The percentage errors of measurement in a, b, c and d are 1%, 3%, 4% and 2%, respectively. What is the percentage error in the quantity P ? If the value of P calculated using the above relation turns out to be 3.763, to what value should you round off the result ?

Answer:

Question 2.14. A book with many printing errors contains four different formulas for the displacement y of a particle undergoing a certain periodic motion :

(a) $y = a \sin(2\pi t/T)$

(b) $y = a \sin(vt)$

(c) $y = (a/T) \sin(t/a)$

(d) $y = (a\sqrt{2}) (\sin(2\pi t/T) + \cos(2\pi t/T))$

(a = maximum displacement of the particle, v = speed of the particle. T = time-period of motion). Rule out the wrong formulas on dimensional grounds.

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Question 2.15. A famous relation in physics relates ‘moving mass’ m to the ‘rest mass’ m$_0$ of a particle in terms of its speed v and the speed of light, c. (This relation first arose as a consequence of special relativity due to Albert Einstein). A boy recalls the relation almost correctly but forgets where to put the constant c. He writes :

$m = \frac{m_0}{(1 - v^2)^{1/2}}$

Guess where to put the missing c.

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Question 2.16. The unit of length convenient on the atomic scale is known as an angstrom and is denoted by Å: 1 Å = 10$^{–10}$ m. The size of a hydrogen atom is about 0.5 Å. What is the total atomic volume in m$^3$ of a mole of hydrogen atoms ?

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Question 2.17. One mole of an ideal gas at standard temperature and pressure occupies 22.4 L (molar volume). What is the ratio of molar volume to the atomic volume of a mole of hydrogen ? (Take the size of hydrogen molecule to be about 1 Å). Why is this ratio so large ?

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Question 2.18. Explain this common observation clearly : If you look out of the window of a fast moving train, the nearby trees, houses etc. seem to move rapidly in a direction opposite to the train’s motion, but the distant objects (hill tops, the Moon, the stars etc.) seem to be stationary. (In fact, since you are aware that you are moving, these distant objects seem to move with you).

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Question 2.19. The principle of ‘parallax’ in section 2.3.1 is used in the determination of distances of very distant stars. The baseline AB is the line joining the Earth’s two locations six months apart in its orbit around the Sun. That is, the baseline is about the diameter of the Earth’s orbit ≈ 3 × 10$^{11}$m. However, even the nearest stars are so distant that with such a long baseline, they show parallax only of the order of 1” (second) of arc or so. A parsec is a convenient unit of length on the astronomical scale. It is the distance of an object that will show a parallax of 1” (second of arc) from opposite ends of a baseline equal to the distance from the Earth to the Sun. How much is a parsec in terms of metres ?

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Question 2.20. The nearest star to our solar system is 4.29 light years away. How much is this distance in terms of parsecs? How much parallax would this star (named Alpha Centauri) show when viewed from two locations of the Earth six months apart in its orbit around the Sun ?

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Question 2.21. Precise measurements of physical quantities are a need of science. For example, to ascertain the speed of an aircraft, one must have an accurate method to find its positions at closely separated instants of time. This was the actual motivation behind the discovery of radar in World War II. Think of different examples in modern science where precise measurements of length, time, mass etc. are needed. Also, wherever you can, give a quantitative idea of the precision needed.

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Question 2.22. Just as precise measurements are necessary in science, it is equally important to be able to make rough estimates of quantities using rudimentary ideas and common observations. Think of ways by which you can estimate the following (where an estimate is difficult to obtain, try to get an upper bound on the quantity) :

(a) the total mass of rain-bearing clouds over India during the Monsoon

(b) the mass of an elephant

(c) the wind speed during a storm

(d) the number of strands of hair on your head

(e) the number of air molecules in your classroom.

Answer:

Question 2.23. The Sun is a hot plasma (ionized matter) with its inner core at a temperature exceeding 10$^7$ K, and its outer surface at a temperature of about 6000 K. At these high temperatures, no substance remains in a solid or liquid phase. In what range do you expect the mass density of the Sun to be, in the range of densities of solids and liquids or gases ? Check if your guess is correct from the following data : mass of the Sun = 2.0 × 10$^{30}$ kg, radius of the Sun = 7.0 × 10$^8$ m.

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Question 2.24. When the planet Jupiter is at a distance of 824.7 million kilometers from the Earth, its angular diameter is measured to be 35.72" of arc. Calculate the diameter of Jupiter.

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Additional Exercises

Question 2.25. A man walking briskly in rain with speed v must slant his umbrella forward making an angle $\theta$ with the vertical. A student derives the following relation between $\theta$ and v : tan $\theta$ = v and checks that the relation has a correct limit: as $v \rightarrow 0, \theta \rightarrow 0$, as expected. (We are assuming there is no strong wind and that the rain falls vertically for a stationary man). Do you think this relation can be correct ? If not, guess the correct relation.

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Question 2.26. It is claimed that two cesium clocks, if allowed to run for 100 years, free from any disturbance, may differ by only about 0.02 s. What does this imply for the accuracy of the standard cesium clock in measuring a time-interval of 1 s ?

Answer:

Question 2.27. Estimate the average mass density of a sodium atom assuming its size to be about 2.5 Å. (Use the known values of Avogadro’s number and the atomic mass of sodium). Compare it with the mass density of sodium in its crystalline phase : 970 kg m$^{–3}$. Are the two densities of the same order of magnitude ? If so, why ?

Answer:

Question 2.28. The unit of length convenient on the nuclear scale is a fermi : 1 f = 10$^{–15}$ m. Nuclear sizes obey roughly the following empirical relation :

$r = r_0 A^{1/3}$

where r is the radius of the nucleus, A its mass number, and r$_0$ is a constant equal to about, 1.2 f. Show that the rule implies that nuclear mass density is nearly constant for different nuclei. Estimate the mass density of sodium nucleus. Compare it with the average mass density of a sodium atom obtained in Exercise. 2.27.

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Question 2.29. A LASER is a source of very intense, monochromatic, and unidirectional beam of light. These properties of a laser light can be exploited to measure long distances. The distance of the Moon from the Earth has been already determined very precisely using a laser as a source of light. A laser light beamed at the Moon takes 2.56 s to return after reflection at the Moon’s surface. How much is the radius of the lunar orbit around the Earth ?

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Question 2.30. A SONAR (sound navigation and ranging) uses ultrasonic waves to detect and locate objects under water. In a submarine equipped with a SONAR, the time delay between generation of a probe wave and the reception of its echo after reflection from an enemy submarine is found to be 77.0 s. What is the distance of the enemy submarine? (Speed of sound in water = 1450 m s$^{–1}$).

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Question 2.31. The farthest objects in our Universe discovered by modern astronomers are so distant that light emitted by them takes billions of years to reach the Earth. These objects (known as quasars) have many puzzling features, which have not yet been satisfactorily explained. What is the distance in km of a quasar from which light takes 3.0 billion years to reach us ?

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Question 2.32. It is a well known fact that during a total solar eclipse the disk of the moon almost completely covers the disk of the Sun. From this fact and from the information you can gather from examples 2.3 and 2.4, determine the approximate diameter of the moon.

Answer:

Question 2.33. A great physicist of this century (P.A.M. Dirac) loved playing with numerical values of Fundamental constants of nature. This led him to an interesting observation. Dirac found that from the basic constants of atomic physics (c, e, mass of electron, mass of proton) and the gravitational constant G, he could arrive at a number with the dimension of time. Further, it was a very large number, its magnitude being close to the present estimate on the age of the universe (~15 billion years). From the table of fundamental constants in this book, try to see if you too can construct this number (or any other interesting number you can think of ). If its coincidence with the age of the universe were significant, what would this imply for the constancy of fundamental constants?

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