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Chapter 2 Units And Measurement
Introduction
Measurement is a fundamental process in physics and science, involving the comparison of a physical quantity with a predefined, internationally agreed-upon reference standard known as a unit.
The result of a measurement is always expressed as a numerical value accompanied by the appropriate unit (e.g., 5 meters, 10 kilograms).
While there are numerous physical quantities, they are interrelated, meaning we only need a limited number of fundamental units to describe all others.
Units associated with the basic or fundamental quantities are called fundamental or base units.
Units for all other physical quantities, which can be expressed as combinations of base units, are called derived units.
A system of units refers to a complete set encompassing both base units and derived units.
The International System Of Units
Historically, different countries used different systems of units. Common examples included:
- CGS System: Based on centimetre (length), gram (mass), and second (time).
- FPS System (British System): Based on foot (length), pound (mass), and second (time).
- MKS System: Based on metre (length), kilogram (mass), and second (time).
The internationally accepted system currently is the Système Internationale d’ Unites, abbreviated as SI.
- It was developed by the Bureau International des Poids et Measures (BIPM).
- It was formally established in 1971 and has undergone revisions, including significant changes in 2018 based on fundamental physical constants.
- SI units are used worldwide in scientific, technical, industrial, and commercial fields.
- A key advantage is its decimal nature, simplifying conversions.
The SI system is built upon seven base units:
Base Quantity | Name (Unit) | Symbol | Brief Concept of Definition |
---|---|---|---|
Length | metre | m | Defined based on the speed of light in vacuum ($c$). |
Mass | kilogram | kg | Defined based on the Planck constant ($h$). |
Time | second | s | Defined based on the frequency of radiation from a Caesium-133 atom ($\Delta \nu_{Cs}$). |
Electric Current | ampere | A | Defined based on the elementary charge ($e$). |
Thermodynamic Temperature | kelvin | K | Defined based on the Boltzmann constant ($k$). |
Amount of Substance | mole | mol | Defined based on the Avogadro constant ($N_A$) ($6.02214076 \times 10^{23}$ entities). |
Luminous Intensity | candela | cd | Defined based on the luminous efficacy of monochromatic radiation ($K_{cd}$). |
The definitions of the base units link them to fundamental constants of nature, providing a stable and universal basis for measurement.
Besides the seven base units, SI includes two supplementary units, which are dimensionless:
- Radian (rad): Unit for plane angle ($d\theta$). Defined as the ratio of the arc length ($ds$) subtending the angle to the radius ($r$) of the circle ($d\theta = ds/r$).
- Steradian (sr): Unit for solid angle ($d\Omega$). Defined as the ratio of the area ($dA$) intercepted on a spherical surface to the square of its radius ($r$) ($d\Omega = dA/r^2$).
Units for other physical quantities are derived units, formed by combining the base units algebraically (e.g., unit of speed is m/s, unit of force is kg m/s$^2$ which is called Newton, N).
Some derived units have special names (e.g., Newton, Joule, Watt, Volt, Pascal, etc.).
SI also uses standard prefixes (like kilo-, milli-, micro-, nano-) to denote multiples and sub-multiples of units, simplifying the representation of very large or very small values (e.g., 1 km = $10^3$ m, 1 mm = $10^{-3}$ m).
General guidelines exist for the proper use of symbols for physical quantities, units, and prefixes to ensure clarity and consistency in scientific communication.
Measurement Of Length
Measuring length can be done using various methods depending on the scale of the distance.
Direct Methods: Involve using measuring instruments directly:
- Metre Scale: Typically measures lengths from $10^{-3}$ m (millimeter) to $10^2$ m (hundreds of meters).
- Vernier Callipers: Provides greater accuracy, measuring lengths up to $10^{-4}$ m (0.1 mm).
- Screw Gauge / Spherometer: Offer even higher precision for small lengths, measuring up to $10^{-5}$ m (0.01 mm).
For lengths outside these ranges (very large or very small), indirect methods are employed.
Measurement Of Large Distances
Large distances like the distance to a planet or a star cannot be measured directly. The parallax method is a common indirect technique.
Parallax: The apparent shift in the position of an object when viewed from two different points. The distance between the two observation points is called the basis.
To measure the distance ($D$) of a distant object (like a planet $S$) from Earth:
- Observe the object from two widely separated points ($A$ and $B$) on Earth simultaneously. The distance between $A$ and $B$ is the basis ($b = AB$).
- Measure the angle ($\theta$) between the two lines of sight ($AS$ and $BS$). This angle is the parallax angle or parallactic angle.
- Since the object is very far ($D >> b$), $\theta$ is very small. The distance $AB$ can be approximated as an arc of a circle with radius $D$ centered at $S$.
- The relationship between the basis, distance, and angle (in radians) is given by: $$b = D \theta$$
- Therefore, the distance is calculated as: $$D = \frac{b}{\theta}$$ (Note: $\theta$ must be in radians).
Once the distance ($D$) to a planet is known, its diameter ($d$) can be determined by measuring its angular size ($\alpha$).
- Observe the planet through a telescope from one location on Earth.
- Measure the angle ($\alpha$) subtended by two diametrically opposite points of the planet at the observation point. This is the angular size ($\alpha$).
- The relationship between diameter, distance, and angular size (in radians) is: $$\alpha = \frac{d}{D}$$
- The diameter is then calculated as: $$d = \alpha D$$
Example 2.1. Calculate the angle of (a) $1^\circ$ (degree) (b) $1'$ (minute of arc or arcmin) and (c) $1''$ (second of arc or arc second) in radians. Use $360^\circ = 2\pi \text{ rad}$, $1^\circ = 60'$ and $1' = 60''$.
Answer:
(a) We know $360^\circ = 2\pi$ radians.
So, $1^\circ = \frac{2\pi}{360} \text{ rad} = \frac{\pi}{180} \text{ rad}$.
Using $\pi \approx 3.14159$, $1^\circ \approx \frac{3.14159}{180} \text{ rad} \approx 0.01745 \text{ rad} = 1.745 \times 10^{-2} \text{ rad}$.
(b) We have $1^\circ = 60'$. From (a), $1^\circ = 1.745 \times 10^{-2}$ rad.
So, $60' = 1.745 \times 10^{-2}$ rad.
$1' = \frac{1.745 \times 10^{-2}}{60} \text{ rad} \approx 0.02908 \times 10^{-2} \text{ rad} = 2.908 \times 10^{-4} \text{ rad}$.
Rounding off to two significant figures, $1' \approx 2.91 \times 10^{-4} \text{ rad}$.
(c) We have $1' = 60''$. From (b), $1' = 2.908 \times 10^{-4}$ rad.
So, $60'' = 2.908 \times 10^{-4}$ rad.
$1'' = \frac{2.908 \times 10^{-4}}{60} \text{ rad} \approx 0.04847 \times 10^{-4} \text{ rad} = 4.847 \times 10^{-6} \text{ rad}$.
Rounding off to two significant figures, $1'' \approx 4.85 \times 10^{-6} \text{ rad}$.
Example 2.2. A man wishes to estimate the distance of a nearby tower from him. He stands at a point A in front of the tower C and spots a very distant object O in line with AC. He then walks perpendicular to AC up to B, a distance of 100 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 $\theta = 40^\circ$ ($\theta$ is known as ‘parallax’) estimate the distance of the tower C from his original position A.
Answer:
Let the distance of the tower C from point A be AC. The man moves from A to B, such that AB = 100 m and AB is perpendicular to AC.
The distant object O is in line with AC when viewed from A. When viewed from B, the line of sight to C is shifted by $\theta = 40^\circ$ relative to the line of sight to O (which is effectively the same direction as AO). This means the angle $\angle ABC = \theta = 40^\circ$.
In the right-angled triangle $\triangle ABC$, we have:
$$\tan \theta = \frac{AB}{AC}$$We are given $AB = 100$ m and $\theta = 40^\circ$.
$$AC = \frac{AB}{\tan \theta} = \frac{100 \text{ m}}{\tan 40^\circ}$$Using the value $\tan 40^\circ \approx 0.8391$,
$$AC = \frac{100}{0.8391} \text{ m} \approx 119.17 \text{ m}$$Rounding to appropriate significant figures (based on 100 m, assuming 2 or 3 sig figs), the distance of the tower C from A is approximately 119 m.
Example 2.3. 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 $1.276 \times 10^7$ m, compute the distance of the moon from the Earth.
Answer:
The two diametrically opposite points A and B on Earth serve as the observation points. The distance between them is the diameter of the Earth, which is the basis $b = 1.276 \times 10^7$ m.
The parallax angle subtended at the moon is $\theta = 1^\circ 54'$.
First, convert $\theta$ to radians:
$$1^\circ 54' = 1^\circ + 54'$$We know $1^\circ = 60'$, so $54' = \frac{54}{60}^\circ = 0.9^\circ$.
Thus, $\theta = 1^\circ + 0.9^\circ = 1.9^\circ$.
Now, convert degrees to radians using the conversion from Example 2.1(a): $1^\circ \approx 1.745 \times 10^{-2}$ rad.
$$\theta = 1.9^\circ \times \frac{\pi}{180} \text{ rad} \approx 1.9 \times 1.745 \times 10^{-2} \text{ rad} \approx 3.3155 \times 10^{-2} \text{ rad}$$Alternatively, convert directly to arc seconds and then to radians:
$$1^\circ 54' = (1 \times 60 + 54)' = (60 + 54)' = 114'$$ $$114' = (114 \times 60)'' = 6840''$$Using the conversion from Example 2.1(c): $1'' \approx 4.85 \times 10^{-6}$ rad.
$$\theta = 6840'' \times 4.85 \times 10^{-6} \text{ rad} \approx 33174 \times 10^{-6} \text{ rad} = 3.3174 \times 10^{-2} \text{ rad}$$Let's use the value from the text's calculation which is $3.32 \times 10^{-2}$ rad.
Using the parallax formula $D = \frac{b}{\theta}$, where $b = 1.276 \times 10^7$ m and $\theta = 3.32 \times 10^{-2}$ rad:
$$D = \frac{1.276 \times 10^7 \text{ m}}{3.32 \times 10^{-2}} = \frac{1.276}{3.32} \times 10^{7 - (-2)} \text{ m} = \frac{1.276}{3.32} \times 10^9 \text{ m}$$ $$D \approx 0.3843 \times 10^9 \text{ m} = 3.843 \times 10^8 \text{ m}$$Rounding to three significant figures (based on $1.276 \times 10^7$ m and the calculated $\theta$), the distance of the moon from the Earth is approximately $3.84 \times 10^8$ m.
Example 2.4. The Sun’s angular diameter is measured to be $1920''$. The distance D of the Sun from the Earth is $1.496 \times 10^{11}$ m. What is the diameter of the Sun ?
Answer:
The angular diameter of the Sun is given as $\alpha = 1920''$.
The distance of the Sun from the Earth is $D = 1.496 \times 10^{11}$ m.
First, convert the angular diameter from arc seconds to radians using the conversion from Example 2.1(c): $1'' \approx 4.85 \times 10^{-6}$ rad.
$$\alpha = 1920'' \times 4.85 \times 10^{-6} \text{ rad} = 9312 \times 10^{-6} \text{ rad} = 9.312 \times 10^{-3} \text{ rad}$$Let the diameter of the Sun be $d$. The relationship between diameter, distance, and angular size (in radians) is $d = \alpha D$.
$$d = (9.312 \times 10^{-3} \text{ rad}) \times (1.496 \times 10^{11} \text{ m})$$ $$d = (9.312 \times 1.496) \times 10^{-3 + 11} \text{ m} = 13.93 \times 10^8 \text{ m}$$Rounding to three significant figures (based on $1.496 \times 10^{11}$ m), the diameter of the Sun is approximately $1.39 \times 10^9$ m.
Estimation Of Very Small Distances: Size Of A Molecule
Measuring extremely small sizes, like those of molecules (ranging from $10^{-8}$ m to $10^{-10}$ m), requires specialized indirect techniques as conventional instruments like screw gauges or even optical microscopes have limitations.
Limitations of Microscopes:
- Optical Microscope: Limited by the wave nature of visible light. Its resolution cannot be better than the wavelength of light (typically $4000 \text{ Å}$ to $7000 \text{ Å}$, where $1 \text{ Å} = 10^{-10} \text{ m}$). It cannot resolve objects smaller than this range.
- Electron Microscope: Uses electron beams, which have much smaller wavelengths than visible light. While offering better resolution (down to $\sim 0.6 \text{ Å}$), it is still limited by the wave nature of electrons.
- Tunnelling Microscopy: Offers resolution better than $1 \text{ Å}$ and can image individual atoms and molecules.
A simple method to estimate the size of a molecule is using oleic acid, which forms a thin, mono-molecular layer on a water surface.
The procedure involves:
- Preparing a dilute solution of oleic acid in alcohol (e.g., 1 cm$^3$ of oleic acid in 20 cm$^3$ alcohol, then 1 cm$^3$ of this solution diluted to 20 cm$^3$). The final concentration is $\frac{1}{20 \times 20} = \frac{1}{400}$ cm$^3$ of oleic acid per cm$^3$ of solution.
- Sprinkling lycopodium powder on the surface of water in a large trough. The powder helps to visualize the extent of the oleic acid film.
- Carefully placing a known number of drops ($n$) of the prepared oleic acid solution onto the water surface.
- The alcohol evaporates, and the oleic acid spreads to form a thin, roughly circular film of molecular thickness on the water surface, pushing the lycopodium powder outwards.
- Measuring the diameter of the circular film to calculate its area ($A$).
- Estimating the volume ($V$) of a single drop of the solution beforehand (e.g., by counting how many drops make 1 cm$^3$).
- Calculating the total volume of oleic acid in the $n$ drops: $$\text{Volume of oleic acid} = n \times V \times \text{Concentration} = n \times V \times \frac{1}{400} \text{ cm}^3$$
- Assuming the film is a single layer of molecules, its thickness ($t$) is the volume of oleic acid divided by the area of the film: $$t = \frac{\text{Volume of oleic acid}}{\text{Area of film}} = \frac{nV/400}{A} \text{ cm}$$
This calculated thickness ($t$) provides an estimate for the size or diameter of an oleic acid molecule, which typically comes out to be around $10^{-9}$ m.
Example 2.5. If the size of a nucleus (in the range of $10^{-15}$ to $10^{-14}$ m) is scaled up to the tip of a sharp pin, what roughly is the size of an atom ? Assume tip of the pin to be in the range $10^{-5}$m to $10^{-4}$m.
Answer:
Let's take typical values for the size of a nucleus and the size of an atom:
- Typical size of a nucleus $\sim 10^{-14}$ m.
- Typical size of an atom $\sim 10^{-10}$ m.
The ratio of the size of an atom to the size of a nucleus is $\frac{10^{-10} \text{ m}}{10^{-14} \text{ m}} = 10^{-10 - (-14)} = 10^4$. This means an atom is roughly $10^4$ times larger than its nucleus.
Now, let's consider the given scaling scenario:
- Size of a nucleus is scaled up to the size of the tip of a sharp pin.
- Assume the size of the pin tip is in the range $10^{-5}$ m to $10^{-4}$ m. Let's take a value in this range, say $\sim 10^{-4}$ m.
The scaling factor used for the nucleus is the ratio of the pin tip size to the actual nucleus size:
$$\text{Scaling Factor} = \frac{\text{Size of pin tip}}{\text{Size of nucleus}} \approx \frac{10^{-4} \text{ m}}{10^{-14} \text{ m}} = 10^{10}$$This means the nucleus's size is being increased by a factor of $10^{10}$. To find the scaled size of the atom, we must apply the same scaling factor to the actual size of the atom:
$$\text{Scaled size of atom} = \text{Actual size of atom} \times \text{Scaling Factor}$$ $$\text{Scaled size of atom} \approx 10^{-10} \text{ m} \times 10^{10} = 10^{(-10+10)} \text{ m} = 10^0 \text{ m} = 1 \text{ m}$$Thus, if a nucleus were scaled up to the size of a pin tip (around $10^{-4}$ m), the atom itself would be scaled up to roughly the size of 1 meter. This illustrates how much larger an atom is compared to its nucleus; the nucleus is like a tiny pin tip at the center of a sphere about a meter in radius representing the atom.
Range Of Lengths
The sizes of objects and distances in the universe vary over an extraordinarily wide spectrum.
The range spans from the size of the smallest known particles (like the diameter of a proton, $\sim 10^{-15}$ m) to the estimated extent of the observable universe ($\sim 10^{26}$ m).
Special units are used for convenience at very small or very large scales:
- fermi (f): $1 \text{ f} = 10^{-15} \text{ m}$ (typical nuclear size scale).
- angstrom (Å): $1 \text{ Å} = 10^{-10} \text{ m}$ (typical atomic size scale).
- astronomical unit (AU): $1 \text{ AU} = 1.496 \times 10^{11} \text{ m}$ (average distance between the Earth and the Sun).
- light year (ly): $1 \text{ ly} = 9.46 \times 10^{15} \text{ m}$ (distance light travels in vacuum in one year).
- parsec (pc): $1 \text{ pc} = 3.08 \times 10^{16} \text{ m}$. Defined as the distance at which the average radius of Earth's orbit (1 AU) subtends a parallax angle of one arc second ($1''$). ($1 \text{ pc} = \frac{1 \text{ AU}}{1'' \text{ in radians}}$).
Measurement Of Mass
Mass is an intrinsic property of matter, representing the amount of substance. It remains constant regardless of temperature, pressure, or location in space.
The SI unit of mass is the kilogram (kg). It is defined based on the fixed numerical value of the Planck constant ($h$).
For dealing with the masses of atoms and molecules, the kilogram is an inconveniently large unit. A more suitable standard unit is the unified atomic mass unit (u).
- $1 \text{ u}$ is defined as exactly one-twelfth ($1/12$) of the mass of an atom of the carbon-12 isotope ($^{12}\text{C}$), including the mass of its electrons.
- $1 \text{ u} \approx 1.66 \times 10^{-27} \text{ kg}$.
Methods for measuring mass depend on the scale:
- Common Balance: Used for measuring the mass of everyday objects (like in shops), comparing an unknown mass with standard masses.
- Gravitational Method: Used for large masses like planets and stars, based on Newton's law of gravitation (related to how they interact gravitationally).
- Mass Spectrograph: Used for measuring the masses of atomic and sub-atomic particles. It works by analyzing the trajectory of charged particles in electric and magnetic fields, where the radius of the path is proportional to the particle's mass.
Range Of Masses
Similar to length, the masses of objects in the universe cover a vast range.
The spectrum extends from the tiny mass of an electron ($\sim 10^{-30}$ kg) to the immense mass of the known observable universe ($\sim 10^{55}$ kg).
This ratio of largest to smallest mass ($\sim 10^{85}$) is the square of the ratio observed for length and time scales ($\sim (10^{41})^2$).
Measurement Of Time
Measuring time intervals requires a device that exhibits a regular, repeating phenomenon – a clock.
The modern standard for time measurement is based on an atomic standard.
- This standard uses the precise, periodic vibrations of radiation corresponding to a specific transition in a Cesium-133 atom.
- This principle is used in caesium clocks (atomic clocks), which serve as national standards for time and frequency.
The SI unit of time, the second (s), is defined based on this atomic standard:
- One second is the duration of 9,192,631,770 periods of the radiation corresponding to the transition between the two hyperfine levels of the ground state of the Caesium-133 atom.
Cesium atomic clocks are extremely accurate and provide highly stable timekeeping. For example, they can maintain an uncertainty of $\pm 1 \times 10^{-15}$ parts, meaning they lose or gain no more than about 32 microseconds in one year.
The high accuracy of time measurement has led to the definition of the SI unit of length (the metre) being based on the speed of light and the second.
The time intervals associated with phenomena in the universe also span a tremendous range, from extremely short lifetimes of unstable particles ($\sim 10^{-24}$ s) to the age of the universe ($\sim 10^{18}$ s).
The ratio of the longest to the shortest time intervals is approximately $10^{42}$, which is of the same order of magnitude as the ratio found for length scales ($\sim 10^{40}$). This numerical coincidence across different physical quantities is noteworthy.
Accuracy, Precision Of Instruments And Errors In Measurement
All experimental science and technology rely on measurement. However, every measurement made with any instrument contains some level of uncertainty, referred to as error.
It's important to distinguish between accuracy and precision:
- Accuracy: Describes how close a measured value is to the true or actual value of the quantity. A highly accurate measurement is very near the true value.
- Precision: Refers to the resolution or the limit to which a quantity is measured. It indicates how repeatable the measurements are, i.e., how close multiple measurements are to each other, regardless of their proximity to the true value.
A measurement can be precise but not accurate, or accurate but not precise, or neither, or both.
Due to errors, every measurement is inherently approximate. Errors in measurement can be broadly categorized into:
- Systematic Errors:
- Random Errors:
Systematic Errors:
- These errors tend to occur consistently in one direction (either always making the measurement too high or always too low).
- Sources of systematic errors include:
- Instrumental Errors: Due to imperfect design, faulty calibration (e.g., thermometer reading $104^\circ\text{C}$ at water's boiling point), or zero error (e.g., vernier scale not aligning at zero, worn-off scale ends).
- Imperfection in Experimental Technique or Procedure: Results from flaws in the experimental setup or method (e.g., measuring body temperature under the armpit gives a lower reading than actual), or external conditions affecting the measurement (temperature changes, air currents).
- Personal Errors: Arise from the observer's habits, biases, lack of proper setup, or carelessness (e.g., parallax error from consistently viewing a scale from an angle).
- Systematic errors can often be minimised by improving techniques, using better instruments, and correcting for known biases. They can sometimes be estimated and accounted for.
Random Errors:
- These errors occur irregularly and unpredictably, varying in both sign (positive or negative) and magnitude.
- Sources include:
- Unpredictable fluctuations in experimental conditions (e.g., sudden voltage changes, mechanical vibrations).
- Unbiased personal errors (slight variations in how an observer takes a reading each time).
- Random errors cannot be eliminated entirely but can be reduced by repeating measurements multiple times and taking the average.
Least Count Error:
- This error is associated with the resolution or the smallest value that can be measured by an instrument (its least count).
- All measurements have an uncertainty related to the least count (e.g., a metre scale with 1 mm divisions has a least count error of $\pm 0.5$ mm or $\pm 1$ mm depending on convention/estimation).
- Least count error contributes to both systematic (if there's a zero error) and random errors.
- It can be reduced by using instruments with higher resolution or by averaging repeated measurements.
Absolute Error, Relative Error And Percentage Error
To quantify errors and represent the uncertainty in a measurement, we use different error metrics.
Suppose a quantity is measured $n$ times, yielding values $a_1, a_2, a_3, ..., a_n$.
- Mean Value: The best estimate of the true value is typically the arithmetic mean of the measurements: $$a_{mean} = \frac{a_1 + a_2 + ... + a_n}{n} = \frac{1}{n}\sum_{i=1}^{n} a_i$$
- Absolute Error ($\Delta a_i$): The difference between an individual measurement and the mean value: $$\Delta a_i = a_i - a_{mean}$$ These individual errors can be positive or negative. The magnitude, $|\Delta a_i|$, is the absolute error.
- Mean Absolute Error ($\Delta a_{mean}$): The average of the magnitudes of the absolute errors from all measurements: $$\Delta a_{mean} = \frac{|\Delta a_1| + |\Delta a_2| + ... + |\Delta a_n|}{n} = \frac{1}{n}\sum_{i=1}^{n} |\Delta a_i|$$ This represents the overall uncertainty in the measurements.
The result of a measurement is conventionally expressed as $a = a_{mean} \pm \Delta a_{mean}$. This indicates that the true value of the quantity is likely to lie within the range $[a_{mean} - \Delta a_{mean}, a_{mean} + \Delta a_{mean}]$.
Other ways to express error include relative error and percentage error:
- Relative Error: The ratio of the mean absolute error to the mean value: $$\text{Relative Error} = \frac{\Delta a_{mean}}{a_{mean}}$$ This is a dimensionless quantity.
- Percentage Error ($\delta a$): The relative error expressed as a percentage: $$\text{Percentage Error} = \frac{\Delta a_{mean}}{a_{mean}} \times 100\%$$
Example 2.6. Two clocks are being tested against a standard clock located in a national laboratory. At 12:00:00 noon by the standard clock, the readings of the two clocks are :
Clock 1 | Clock 2 | |
---|---|---|
Monday | 12:00:05 | 10:15:06 |
Tuesday | 12:01:15 | 10:14:59 |
Wednesday | 11:59:08 | 10:15:18 |
Thursday | 12:01:50 | 10:15:07 |
Friday | 11:59:15 | 10:14:53 |
Saturday | 12:01:30 | 10:15:24 |
Sunday | 12:01:19 | 10:15:11 |
If you are doing an experiment that requires precision time interval measurements, which of the two clocks will you prefer ?
Answer:
To assess which clock is better for precise time interval measurements, we should look at the consistency of its readings, not just how close its average reading is to the standard time. A consistent clock (high precision) is preferable because any constant offset (zero error) can be corrected. An inconsistent clock (low precision) will have larger variations in measurements, making precise time intervals difficult.
Let's analyze the spread of readings for each clock over the seven days:
- Clock 1: Readings range from 11:59:08 (minimum) to 12:01:50 (maximum).
- Minimum reading is 52 seconds before 12:00:00 (11:59:08).
- Maximum reading is 1 minute and 50 seconds after 12:00:00 (12:01:50), which is $60 + 50 = 110$ seconds after.
- Clock 2: All readings are around 10:15.
- Minimum reading is 10:14:53.
- Maximum reading is 10:15:24.
Clock 1's readings are sometimes before and sometimes after the standard time, with a large total spread of 162 seconds. Clock 2's readings are consistently around 10:15, which is far from the standard 12:00:00 noon, but its readings are much closer to each other, with a total spread of only 31 seconds.
Clock 2 has a large systematic error (it's about 1 hour 45 minutes behind), but its readings are much more consistent (higher precision) over the days compared to Clock 1. Clock 1's readings are closer to the standard time on average (possibly higher accuracy in terms of mean reading relative to the target time), but they vary significantly day-to-day (lower precision).
For measuring precise time intervals, consistency (precision) is more important than having an average reading close to a specific target time (accuracy with respect to a reference point). A systematic error like the one in Clock 2 can be easily corrected for (e.g., by simply noting that 12:00:00 standard time corresponds to 10:15:something on Clock 2). The variation in time measurements (precision) is the limiting factor for interval measurements.
Therefore, for an experiment requiring precise time interval measurements, you would prefer Clock 2 because it is more precise, despite its significant offset (zero error) from the standard time.
Example 2.7. We measure the period of oscillation of a simple pendulum. In successive measurements, the readings turn out to be 2.63 s, 2.56 s, 2.42 s, 2.71s and 2.80 s. Calculate the absolute errors, relative error or percentage error.
Answer:
The successive measurements of the period of oscillation are $T_1 = 2.63 \text{ s}$, $T_2 = 2.56 \text{ s}$, $T_3 = 2.42 \text{ s}$, $T_4 = 2.71 \text{ s}$, and $T_5 = 2.80 \text{ s}$. The number of measurements is $n=5$.
First, calculate the mean period ($T_{mean}$), which is the best estimate of the true period:
$$T_{mean} = \frac{T_1 + T_2 + T_3 + T_4 + T_5}{5}$$ $$T_{mean} = \frac{2.63 + 2.56 + 2.42 + 2.71 + 2.80}{5} \text{ s}$$ $$T_{mean} = \frac{13.12}{5} \text{ s} = 2.624 \text{ s}$$Since the original measurements are given to two decimal places (0.01 s resolution), the mean should also be rounded to two decimal places for reporting the final result with error. So, $T_{mean} \approx 2.62 \text{ s}$.
Next, calculate the absolute error for each measurement relative to the mean value ($2.62$ s):
- $\Delta T_1 = T_1 - T_{mean} = 2.63 \text{ s} - 2.62 \text{ s} = +0.01 \text{ s}$
- $\Delta T_2 = T_2 - T_{mean} = 2.56 \text{ s} - 2.62 \text{ s} = -0.06 \text{ s}$
- $\Delta T_3 = T_3 - T_{mean} = 2.42 \text{ s} - 2.62 \text{ s} = -0.20 \text{ s}$
- $\Delta T_4 = T_4 - T_{mean} = 2.71 \text{ s} - 2.62 \text{ s} = +0.09 \text{ s}$
- $\Delta T_5 = T_5 - T_{mean} = 2.80 \text{ s} - 2.62 \text{ s} = +0.18 \text{ s}$
The magnitudes of these absolute errors are $|+0.01| = 0.01$, $|-0.06| = 0.06$, $|-0.20| = 0.20$, $|+0.09| = 0.09$, $|+0.18| = 0.18$ s.
Now, calculate the mean absolute error ($\Delta T_{mean}$):
$$\Delta T_{mean} = \frac{|\Delta T_1| + |\Delta T_2| + |\Delta T_3| + |\Delta T_4| + |\Delta T_5|}{5}$$ $$\Delta T_{mean} = \frac{0.01 + 0.06 + 0.20 + 0.09 + 0.18}{5} \text{ s}$$ $$\Delta T_{mean} = \frac{0.54}{5} \text{ s} = 0.108 \text{ s}$$The mean absolute error is 0.108 s. When reporting the error, it should have the same number of decimal places as the uncertain digit in the mean value. The mean value $2.62$ s has uncertainty in the hundredths place. However, the error 0.108 s affects the tenths place significantly. It is conventional to round the mean absolute error to one significant figure or to the same decimal place as the least precise reading allows. Here, the mean absolute error $0.108 \text{ s}$ rounds to $0.1 \text{ s}$. The mean period is then reported as $T = (2.6 \pm 0.1) \text{ s}$.
Finally, calculate the relative error and percentage error:
Relative Error $= \frac{\Delta T_{mean}}{T_{mean}} = \frac{0.11 \text{ s}}{2.62 \text{ s}}$ (Using the less rounded mean absolute error for calculation before final rounding)
Relative Error $\approx 0.04198$
Percentage Error $= \text{Relative Error} \times 100\% \approx 0.04198 \times 100\% \approx 4.198\%$
Rounding the percentage error, we can report it as approximately 4% or 4.2%.
Using $\Delta T_{mean} = 0.1$ and $T_{mean} = 2.6$: Percentage Error $= \frac{0.1}{2.6} \times 100\% \approx 3.846\%$ or $\approx 4\%$.
Using the values from the text's calculation ($T_{mean}=2.6$, $\Delta T_{mean}=0.1$): Percentage Error $= \frac{0.1}{2.6} \times 100\% \approx 3.846\%$. Rounding to 1 significant figure for the relative error/percentage error (since $\Delta T_{mean}$ has 1 sig fig when rounded to 0.1), we get $\approx 4\%$.
Summary of Results:
- Mean Period: $T_{mean} \approx 2.62 \text{ s}$ (using all data) or $2.6 \text{ s}$ (considering error propagation).
- Absolute Errors: $+0.01, -0.06, -0.20, +0.09, +0.18$ s.
- Mean Absolute Error: $\Delta T_{mean} \approx 0.11 \text{ s}$ or $0.1 \text{ s}$.
- Reported Period with error: $(2.6 \pm 0.1) \text{ s}$.
- Relative Error: $\approx \frac{0.1}{2.6} \approx 0.0385$.
- Percentage Error: $\approx 3.85\%$ or $\approx 4\%$.
Combination Of Errors
When calculations involve multiple measured quantities, each with its own error, it's necessary to determine how these individual errors combine to affect the final result. This is known as error propagation.
Rules for error combination:
(a) Error of a Sum or a Difference:
- If $Z = A + B$ or $Z = A - B$, and the measured values are $A \pm \Delta A$ and $B \pm \Delta B$ (where $\Delta A$ and $\Delta B$ are absolute errors).
- The maximum possible absolute error in $Z$ is the sum of the absolute errors: $$\Delta Z = \Delta A + \Delta B$$
- Rule: When quantities are added or subtracted, the absolute error in the result is the sum of the absolute errors in the individual quantities.
Example 2.8. The temperatures of two bodies measured by a thermometer are $t_1 = 20^\circ\text{C} \pm 0.5^\circ\text{C}$ and $t_2 = 50^\circ\text{C} \pm 0.5^\circ\text{C}$. Calculate the temperature difference and the error therein.
Answer:
The temperature difference is $\Delta t = t_2 - t_1$.
Mean difference $= 50^\circ\text{C} - 20^\circ\text{C} = 30^\circ\text{C}$.
The absolute error in $t_1$ is $\Delta t_1 = 0.5^\circ\text{C}$.
The absolute error in $t_2$ is $\Delta t_2 = 0.5^\circ\text{C}$.
For a difference, the absolute errors add up to give the maximum possible error in the result.
Error in difference $\Delta(\Delta t) = \Delta t_1 + \Delta t_2 = 0.5^\circ\text{C} + 0.5^\circ\text{C} = 1.0^\circ\text{C}$.
The temperature difference is $(30 \pm 1.0)^\circ\text{C}$.
(b) Error of a Product or a Quotient:
- If $Z = AB$ or $Z = A/B$, and the measured values are $A \pm \Delta A$ and $B \pm \Delta B$.
- Assuming $\Delta A$ and $\Delta B$ are small compared to A and B, the maximum relative error in $Z$ is the sum of the relative errors in $A$ and $B$: $$\frac{\Delta Z}{Z} = \frac{\Delta A}{A} + \frac{\Delta B}{B}$$
- This rule applies to both multiplication and division.
- Rule: When quantities are multiplied or divided, the relative error in the result is the sum of the relative errors in the quantities being combined.
Example 2.9. The resistance $R = V/I$ where $V = (100 \pm 5)\text{V}$ and $I = (10 \pm 0.2)\text{A}$. Find the percentage error in R.
Answer:
The resistance $R$ is obtained by dividing voltage ($V$) by current ($I$).
The value of R using the mean values is $R = \frac{100 \text{ V}}{10 \text{ A}} = 10 \, \Omega$.
The voltage measurement is $V = (100 \pm 5)\text{V}$. The absolute error is $\Delta V = 5$ V. The relative error in V is $\frac{\Delta V}{V} = \frac{5}{100}$.
The percentage error in V is $\frac{5}{100} \times 100\% = 5\%$.
The current measurement is $I = (10 \pm 0.2)\text{A}$. The absolute error is $\Delta I = 0.2$ A. The relative error in I is $\frac{\Delta I}{I} = \frac{0.2}{10}$.
The percentage error in I is $\frac{0.2}{10} \times 100\% = 2\%$.
For division, the relative errors add up. The relative error in R is $\frac{\Delta R}{R} = \frac{\Delta V}{V} + \frac{\Delta I}{I}$.
The percentage error in R is $\left(\frac{\Delta R}{R} \times 100\%\right) = \left(\frac{\Delta V}{V} \times 100\%\right) + \left(\frac{\Delta I}{I} \times 100\%\right)$.
Percentage error in R = $5\% + 2\% = 7\%$.
Example 2.10. Two resistors of resistances $R_1 = 100\pm3$ ohm and $R_2 = 200 \pm 4$ ohm are connected (a) in series, (b) in parallel. Find the equivalent resistance of the (a) series combination, (b) parallel combination. Use for (a) the relation $R = R_1 + R_2$, and for (b) $\frac{1}{R'} = \frac{1}{R_1} + \frac{1}{R_2}$.
Answer:
(a) Series Combination:
For resistors in series, the equivalent resistance is $R = R_1 + R_2$.
Mean equivalent resistance $= R_1 + R_2 = 100 \, \Omega + 200 \, \Omega = 300 \, \Omega$.
The absolute error in a sum is the sum of absolute errors: $\Delta R = \Delta R_1 + \Delta R_2$.
Absolute error in equivalent resistance $= 3 \, \Omega + 4 \, \Omega = 7 \, \Omega$.
The equivalent resistance in series is $(300 \pm 7) \, \Omega$.
(b) Parallel Combination:
For resistors in parallel, the reciprocal of the equivalent resistance is the sum of the reciprocals: $\frac{1}{R'} = \frac{1}{R_1} + \frac{1}{R_2}$.
First, calculate the mean equivalent resistance $R'$.
$$\frac{1}{R'} = \frac{1}{100 \, \Omega} + \frac{1}{200 \, \Omega} = \frac{2 + 1}{200 \, \Omega} = \frac{3}{200 \, \Omega}$$ $$R' = \frac{200}{3} \, \Omega \approx 66.666... \, \Omega$$For error calculation, it's easier to work with relative errors for products and quotients. Consider the reciprocal relation. Differentiating the relation $\frac{1}{R'} = \frac{1}{R_1} + \frac{1}{R_2}$ with respect to R' etc. and treating differentials as small errors:
$$-\frac{1}{(R')^2} \Delta R' = -\frac{1}{R_1^2} \Delta R_1 - \frac{1}{R_2^2} \Delta R_2$$Taking magnitudes for maximum error:
$$\frac{\Delta R'}{(R')^2} = \frac{\Delta R_1}{R_1^2} + \frac{\Delta R_2}{R_2^2}$$ $$\Delta R' = (R')^2 \left(\frac{\Delta R_1}{R_1^2} + \frac{\Delta R_2}{R_2^2}\right)$$Using mean values: $R_1=100, \Delta R_1=3$, $R_2=200, \Delta R_2=4$, $R' = 200/3 \approx 66.7$.
$$\Delta R' = \left(\frac{200}{3}\right)^2 \left(\frac{3}{100^2} + \frac{4}{200^2}\right)$$ $$\Delta R' = \frac{40000}{9} \left(\frac{3}{10000} + \frac{4}{40000}\right)$$ $$\Delta R' = \frac{40000}{9} \left(\frac{12}{40000} + \frac{4}{40000}\right) = \frac{40000}{9} \left(\frac{16}{40000}\right) = \frac{16}{9} \approx 1.77...$$Rounding to two significant figures (since 66.7 has three, but the errors are 3 and 4, implying uncertainty starts earlier), or matching decimal places with calculated R' mean (66.7), let's round $\Delta R'$ to $1.8$.
The equivalent resistance in parallel is $(66.7 \pm 1.8) \, \Omega$.
(c) Error in case of a Measured Quantity Raised to a Power:
- If $Z = A^k$, where $k$ is a constant and the measured value is $A \pm \Delta A$.
- The maximum relative error in $Z$ is the relative error in $A$ multiplied by the power $k$: $$\frac{\Delta Z}{Z} = |k| \frac{\Delta A}{A}$$
- For a more general expression like $Z = \frac{A^p B^q}{C^r}$: $$\frac{\Delta Z}{Z} = p \frac{\Delta A}{A} + q \frac{\Delta B}{B} + r \frac{\Delta C}{C}$$ (Here we take the magnitudes of the powers and add the relative errors).
- Rule: The relative error in a quantity raised to a power is that power times the relative error in the original quantity. For combined operations involving powers, the maximum relative error in the result is the sum of the relative errors of each factor, multiplied by their respective powers.
Example 2.11. Find the relative error in Z, if $Z = \frac{A^4 B^{1/3}}{C D^{3/2}}$.
Answer:
Using the rule for the error in a quantity raised to a power and for combined multiplication/division:
The relative error in Z is given by:
$$\frac{\Delta Z}{Z} = 4 \frac{\Delta A}{A} + \frac{1}{3} \frac{\Delta B}{B} + 1 \frac{\Delta C}{C} + \frac{3}{2} \frac{\Delta D}{D}$$Note that even though C and D are in the denominator, their powers are treated as positive in the sum of relative errors for calculating the maximum possible error.
Example 2.12. The period of oscillation of a simple pendulum is $T = 2\pi \sqrt{\frac{L}{g}}$. Measured value of L is 20.0 cm known to 1 mm accuracy and time for 100 oscillations of the pendulum is found to be 90 s using a wrist watch of 1 s resolution. What is the accuracy in the determination of g ?
Answer:
The formula for the period of a simple pendulum is $T = 2\pi \sqrt{\frac{L}{g}}$. We want to find the accuracy in the determination of $g$. Let's rearrange the formula to express $g$ in terms of $T$ and $L$:
$$T^2 = (2\pi)^2 \frac{L}{g}$$ $$g = \frac{4\pi^2 L}{T^2}$$We are given the measured length $L = 20.0 \text{ cm}$ and its accuracy is 1 mm. So, $\Delta L = 1 \text{ mm} = 0.1 \text{ cm}$. The value of L has 3 significant figures (20.0). The percentage error in L is $\frac{\Delta L}{L} \times 100\% = \frac{0.1 \text{ cm}}{20.0 \text{ cm}} \times 100\% = 0.5\%$.
We are given the total time for 100 oscillations is $t = 90 \text{ s}$ with a wrist watch resolution of 1 s. This means $\Delta t = 1 \text{ s}$. The period of one oscillation is $T = \frac{t}{n}$, where $n = 100$.
Mean period $T = \frac{90 \text{ s}}{100} = 0.9 \text{ s}$.
The error in the period $\Delta T$ can be related to the error in the total time $\Delta t$: $\Delta T = \frac{\Delta t}{n}$.
So, $\Delta T = \frac{1 \text{ s}}{100} = 0.01 \text{ s}$.
The period measurement is $(0.9 \pm 0.01) \text{ s}$. The value 0.9 s has 2 significant figures (if we consider the error 0.01 s, implying uncertainty in the hundredths place of a single period, or in the tenths place of the total time). Let's use the given data's apparent precision.
The relative error in T is $\frac{\Delta T}{T} = \frac{0.01 \text{ s}}{0.9 \text{ s}}$.
The percentage error in T is $\frac{0.01}{0.9} \times 100\% \approx 1.11\%$.
Now consider the formula $g = \frac{4\pi^2 L}{T^2}$. The constant $4\pi^2$ is exact and has no error.
Using the rule for combining errors in products/quotients and powers, the relative error in $g$ is:
$$\frac{\Delta g}{g} = \frac{\Delta L}{L} + 2 \frac{\Delta T}{T}$$In terms of percentage errors:
$$\left(\frac{\Delta g}{g} \times 100\%\right) = \left(\frac{\Delta L}{L} \times 100\%\right) + 2 \left(\frac{\Delta T}{T} \times 100\%\right)$$Percentage error in $L = 0.5\%$.
Percentage error in $T = \frac{0.01}{0.9} \times 100\% \approx 1.11\%$.
Percentage error in $g = 0.5\% + 2 \times (1.11\%) = 0.5\% + 2.22\% = 2.72\%$.
Rounding to 3 significant figures (based on L's precision), the accuracy in the determination of $g$ is approximately $2.72\%$. Or rounding based on the overall precision of inputs, it's around 3%.
The text's answer uses $\frac{\Delta L}{L} + 2 \frac{\Delta T}{T}$ calculation first: $\frac{0.1}{20.0} + 2 \frac{0.01}{0.9} = 0.005 + 2(0.0111...) = 0.005 + 0.0222... = 0.0272...$. Then $\times 100\% \approx 2.72\%$.
Final Answer based on textbook phrasing/rounding: Percentage error in $g \approx 3\%$.
Significant Figures
Since every measurement involves some error or uncertainty, the reported result should reflect the precision of the measurement. Significant figures (or significant digits) are used for this purpose.
The significant figures in a measured value consist of all the digits that are known reliably (certain) plus the first digit that is uncertain or estimated.
For example, if a length is measured as 287.5 cm, the digits 2, 8, and 7 are certain, while 5 is the first uncertain digit. This measurement has four significant figures.
Reporting more digits than are significant is misleading, as it suggests a higher level of precision than was actually achieved.
The number of significant figures in a value indicates the precision of the measurement and is related to the least count of the measuring instrument.
An important rule is that changing the unit of measurement does not change the number of significant figures.
Rules for determining the number of significant figures in a number:
- All non-zero digits are significant. (e.g., 287.5 has 4 sig figs; 123 has 3 sig figs).
- Zeros between two non-zero digits are significant. (e.g., 2.008 has 4 sig figs; 101 has 3 sig figs).
- Leading zeros (zeros to the left of the first non-zero digit) are NOT significant. These zeros only indicate the position of the decimal point. (e.g., $\underline{0}.\underline{00}237$ has 3 sig figs; $\underline{0}.5$ has 1 sig fig).
- Trailing zeros (zeros at the end of a number):
- In a number WITHOUT a decimal point, trailing zeros are generally NOT significant. They may just be placeholders for the magnitude. (e.g., 12300 has 3 sig figs; 500 has 1 sig fig).
- In a number WITH a decimal point, trailing zeros ARE significant. They indicate the precision of the measurement. (e.g., 3.500 has 4 sig figs; 0.06900 has 4 sig figs; 20.0 has 3 sig figs).
- The zero conventionally placed to the left of the decimal point for a number less than 1 (like 0.1250) is never significant; it's just for readability.
Scientific Notation ($a \times 10^b$) helps resolve ambiguity about trailing zeros in numbers without a decimal point.
- In this form, $a$ is a number between 1 and 10, and $b$ is an integer power of 10.
- All digits shown in the number 'a' are considered significant.
- The power of 10 ($10^b$) is irrelevant to the number of significant figures.
- Example: 12300 can be written as $1.23 \times 10^4$ (3 sig figs), $1.230 \times 10^4$ (4 sig figs), or $1.2300 \times 10^4$ (5 sig figs) depending on its known precision.
- The exponent $b$ in scientific notation is called the order of magnitude of the quantity.
Exact Numbers: Quantities that are counted or defined (like the '2' in $2\pi r$) are considered to have an infinite number of significant figures. This means they do not limit the number of significant figures in a calculation result.
Rules For Arithmetic Operations With Significant Figures
When performing calculations with measured values, the result's precision is limited by the least precise input value. The rules for handling significant figures in arithmetic operations are:
(1) In Multiplication or Division:
- The final result should have the same number of significant figures as the input number with the least number of significant figures.
- Example: Calculating density = mass / volume. If mass = 4.237 g (4 sig figs) and volume = 2.51 cm$^3$ (3 sig figs), the result should have 3 significant figures. $4.237 / 2.51 \approx 1.688$. Rounded to 3 sig figs, density = 1.69 g/cm$^3$.
- Example: Calculating light year distance. If speed of light = $3 \times 10^8$ m/s (1 sig fig) and seconds in a year = $3.156 \times 10^7$ s (4 sig figs), the result should have 1 significant figure. $3 \times 10^8 \times 3.156 \times 10^7 \approx 9.468 \times 10^{15}$. Rounded to 1 sig fig, light year $\approx 9 \times 10^{15}$ m (though in practice, usually more sig figs are used for constants).
(2) In Addition or Subtraction:
- The final result should retain the same number of decimal places as the input number with the least number of decimal places.
- Example: Summing masses 436.32 g (2 decimal places), 227.2 g (1 decimal place), and 0.301 g (3 decimal places). The least precise is 227.2 g with 1 decimal place. $$436.32$$ $$227.2\underline{0}$$ $$\underline{0.301}$$ $$663.821$$ The sum is 663.821 g. Rounding to 1 decimal place, the result is 663.8 g.
- Example: Difference in length 0.307 m (3 decimal places) - 0.304 m (3 decimal places) = 0.003 m (3 decimal places). This result has 1 significant figure, but the rule is based on decimal places for addition/subtraction.
Rounding Off The Uncertain Digits
When a calculation produces more digits than are significant, the result must be rounded off to the appropriate number of significant figures.
Standard rules for rounding:
- If the digit to be dropped is greater than 5, increase the preceding digit by one. (e.g., 2.746 rounded to 3 sig figs becomes 2.75).
- If the digit to be dropped is less than 5, leave the preceding digit unchanged. (e.g., 2.743 rounded to 3 sig figs becomes 2.74).
- If the digit to be dropped is exactly 5:
- If the preceding digit is even, leave it unchanged. (e.g., 2.745 rounded to 3 sig figs becomes 2.74).
- If the preceding digit is odd, increase it by one. (e.g., 2.735 rounded to 3 sig figs becomes 2.74).
In multi-step calculations, it is good practice to keep at least one extra digit beyond the significant figures in intermediate steps to avoid accumulation of rounding errors, and only round the final result to the correct number of significant figures.
Example 2.13. Each side of a cube is measured to be 7.203 m. What are the total surface area and the volume of the cube to appropriate significant figures?
Answer:
The measured side length of the cube is $s = 7.203$ m.
The number 7.203 has four significant figures.
Any calculation based on this measurement should also be reported with four significant figures.
Total Surface Area:
The formula for the surface area of a cube is $A = 6s^2$.
$$A = 6 \times (7.203 \text{ m})^2 = 6 \times 51.883209 \text{ m}^2$$ $$A = 311.299254 \text{ m}^2$$Since the original measurement has 4 significant figures, the result should be rounded to 4 significant figures.
The fourth significant digit is the first '2' in 311.2. The digit following it is '9', which is greater than 5. So we round up the preceding digit.
$$A \approx 311.3 \text{ m}^2$$Volume:
The formula for the volume of a cube is $V = s^3$.
$$V = (7.203 \text{ m})^3 = 373.714754209 \text{ m}^3$$Since the original measurement has 4 significant figures, the result should be rounded to 4 significant figures.
The fourth significant digit is the first '7' in 373.7. The digit following it is '1', which is less than 5. So we leave the preceding digit unchanged.
$$V \approx 373.7 \text{ m}^3$$Example 2.14. 5.74 g of a substance occupies 1.2 cm$^3$. Express its density by keeping the significant figures in view.
Answer:
The mass of the substance is $m = 5.74$ g.
The mass has three significant figures (5, 7, 4).
The volume occupied is $V = 1.2$ cm$^3$.
The volume has two significant figures (1, 2).
Density ($\rho$) is given by the formula $\rho = \frac{m}{V}$.
$$\rho = \frac{5.74 \text{ g}}{1.2 \text{ cm}^3} \approx 4.7833... \text{ g/cm}^3$$In multiplication or division, the result should have the same number of significant figures as the input value with the least number of significant figures.
Mass (5.74 g) has 3 significant figures.
Volume (1.2 cm$^3$) has 2 significant figures.
The least number of significant figures is 2.
Therefore, the result for density should be rounded to 2 significant figures.
The first two significant digits in 4.7833... are 4 and 7. The digit following 7 is 8, which is greater than 5. So, we round up the preceding digit (7) by one.
$$\rho \approx 4.8 \text{ g/cm}^3$$Rules For Determining The Uncertainty In The Results Of Arithmatic Calculations
Beyond simply counting significant figures, we can also determine the actual uncertainty (error) in the result of calculations involving measured quantities with known uncertainties.
These rules are essentially a re-application or illustration of the error combination rules (Section 2.6.2) in the context of reporting the final result with significant figures and uncertainty.
(1) Uncertainty in Product (Example):
- If length $l = 16.2 \pm 0.1$ cm (16.2 cm is mean value, 0.1 cm is absolute error) and breadth $b = 10.1 \pm 0.1$ cm.
- $l$ has 3 significant figures; $\Delta l = 0.1$ cm. Relative error in $l = \frac{0.1}{16.2} \times 100\% \approx 0.6\%$.
- $b$ has 3 significant figures; $\Delta b = 0.1$ cm. Relative error in $b = \frac{0.1}{10.1} \times 100\% \approx 1.0\%$.
- Area $A = l \times b = 16.2 \text{ cm} \times 10.1 \text{ cm} = 163.62 \text{ cm}^2$.
- Relative error in Area $\frac{\Delta A}{A} = \frac{\Delta l}{l} + \frac{\Delta b}{b}$.
- Percentage error in Area = Percentage error in $l$ + Percentage error in $b \approx 0.6\% + 1.0\% = 1.6\%$.
- Absolute error in Area $\Delta A = A \times (\text{Relative error in A}) = 163.62 \times (1.6/100) \approx 163.62 \times 0.016 \approx 2.618 \text{ cm}^2$.
- Rounding the absolute error to one or two significant figures, say 3 cm$^2$.
- The area is reported as $163.62 \pm 2.618 \text{ cm}^2$. Given the uncertainty $\pm 2.618$, the digit '3' in 163 is uncertain. The result should be rounded to reflect this. 163.62 rounded to the nearest integer (because the error is around 3) is 164.
- The area is reported as $(164 \pm 3) \text{ cm}^2$. This shows the mean value rounded appropriately and the absolute uncertainty.
(2) Uncertainty in Subtraction (Impact on Sig Figs):
- If subtracting values with the same number of significant figures, the result may have fewer significant figures.
- Example: $12.9$ g (3 sig figs, 1 decimal place) - $7.06$ g (3 sig figs, 2 decimal places).
- Using the rule for addition/subtraction (least decimal places), the result should have 1 decimal place. $$12.9$$ $$-\underline{7.06}$$ $$5.84$$
Rounding 5.84 to 1 decimal place gives 5.8 g. This result (5.8) has only 2 significant figures, less than the original numbers.
(3) Relative Error Dependence:
- The relative error for a number with $n$ significant figures depends not just on $n$, but also on the value of the number itself.
- Example:
- Measurement 1: 1.02 g. Absolute error $\pm 0.01$ g (uncertainty in the last digit). Relative error = $\frac{0.01}{1.02} \times 100\% \approx 0.98\% \approx 1\%$.
- Measurement 2: 9.89 g. Absolute error $\pm 0.01$ g. Relative error = $\frac{0.01}{9.89} \times 100\% \approx 0.101\% \approx 0.1\%$.
- Both measurements have 3 significant figures and the same absolute error magnitude ($\pm 0.01$), but their relative errors are different.
General Advice for Complex Calculations:
- In intermediate steps of multi-step calculations, carry at least one more significant figure than required for the final result (based on the least precise input). This prevents rounding errors from accumulating and affecting the final precision.
Dimensions Of Physical Quantities
The fundamental nature of a physical quantity, independent of the specific system of units used for measurement, is described by its dimensions.
All physical quantities can be expressed in terms of a combination of the seven fundamental or base quantities defined in SI.
These seven base quantities are considered the seven dimensions of the physical world. They are denoted using square brackets [ ]:
- Length: [L]
- Mass: [M]
- Time: [T]
- Electric Current: [A]
- Thermodynamic Temperature: [K]
- Amount of Substance: [mol]
- Luminous Intensity: [cd]
The dimensions of a physical quantity are the powers (exponents) to which the base quantities are raised to represent that quantity.
For example, in mechanics, quantities can often be expressed using only the dimensions of length, mass, and time ([L], [M], [T]).
- Volume: Defined by Length $\times$ Breadth $\times$ Height. Dimensions are $[L] \times [L] \times [L] = [L]^3$. In terms of all base dimensions: $[M^0 L^3 T^0 A^0 K^0 mol^0 cd^0]$. We typically only show the non-zero dimensions.
- Force: Defined by Mass $\times$ Acceleration. Acceleration is (Length / Time$^2$). So, Force has dimensions $[M] \times \frac{[L]}{[T]^2} = [MLT^{-2}]$. This means force has one dimension in mass, one dimension in length, and minus two dimensions in time.
- Velocity or Speed: Defined by Distance / Time. Dimensions are $\frac{[L]}{[T]} = [LT^{-1}]$.
When considering dimensions, we focus on the fundamental nature or 'quality' of the physical quantity, not its magnitude or whether it is a vector or scalar. Quantities of the same type (like speed, velocity, initial velocity) have the same dimensions.
Dimensional Formulae And Dimensional Equations
A dimensional formula is an expression that shows which base quantities represent the dimensions of a physical quantity and to what powers they are raised.
- Dimensional formula of Volume: $[M^0 L^3 T^0]$
- Dimensional formula of Speed/Velocity: $[M^0 L T^{-1}]$
- Dimensional formula of Acceleration: $[M^0 L T^{-2}]$
- Dimensional formula of Mass Density: $[M L^{-3} T^0]$
A dimensional equation is formed by equating a physical quantity symbol (placed within square brackets) to its dimensional formula.
- Dimensional equation of Volume: $[V] = [M^0 L^3 T^0]$
- Dimensional equation of Speed/Velocity: $[v] = [M^0 L T^{-1}]$
- Dimensional equation of Force: $[F] = [M L T^{-2}]$
- Dimensional equation of Mass Density: $[\rho] = [M L^{-3} T^0]$
Dimensional formulae and equations can be derived from the defining equations or relationships between physical quantities.
Dimensional Analysis And Its Applications
Understanding the dimensions of physical quantities is crucial because it leads to a powerful tool called dimensional analysis.
A fundamental principle of dimensional analysis is the Principle of Homogeneity of Dimensions: Physical quantities can only be added to or subtracted from other quantities if they have the same dimensions. Similarly, the terms on both sides of a valid equation must have the same dimensions.
Just as units can be treated as algebraic symbols (cancelled in numerator/denominator), dimensions can also be manipulated algebraically.
Dimensional analysis has several key applications:
- Checking the dimensional consistency (correctness) of equations.
- Deducing possible relationships between physical quantities (within limitations).
Checking The Dimensional Consistency Of Equations
To check if an equation is dimensionally consistent (homogeneous), verify that the dimensions of every term on both sides of the equation are the same.
If the dimensions of terms in an equation are not the same, the equation is definitely wrong.
If the dimensions are the same, the equation is dimensionally consistent. However, a dimensionally consistent equation is not necessarily correct. Dimensional analysis cannot verify dimensionless constants (like 1/2, $2\pi$) or the correctness of trigonometric, logarithmic, or exponential functions.
Important considerations:
- Arguments of dimensionless functions (like $\sin(\theta)$, $\log(x)$, $e^y$) must be dimensionless. For $\sin(2\pi t/T)$, the argument $2\pi t/T$ must be dimensionless. Dimension of $t$ is [T], dimension of $T$ is [T], $2\pi$ is dimensionless. $[T]/[T]$ is dimensionless.
- Pure numbers or ratios of similar physical quantities (like angle = arc length/radius, refractive index = speed/speed) are dimensionless.
Example: Check the dimensional consistency of the equation of motion: $x = x_0 + v_0 t + \frac{1}{2}at^2$, where $x, x_0$ are positions (length), $v_0$ is initial velocity, $a$ is acceleration, and $t$ is time.
- Dimension of the left side term $[x]$: $[L]$
- Dimension of the first term on the right side $[x_0]$: $[L]$
- Dimension of the second term on the right side $[v_0 t]$: $[LT^{-1}] \times [T] = [L]$
- Dimension of the third term on the right side $[\frac{1}{2}at^2]$: The constant $\frac{1}{2}$ is dimensionless. Dimension of $a$ is $[LT^{-2}]$, dimension of $t^2$ is $[T^2]$. So, $[LT^{-2}] \times [T^2] = [L]$.
Since every term has the dimension of length [L], the equation is dimensionally consistent.
Example 2.15. Let us consider an equation $\frac{1}{2} m v^2 = m g h$ where m is the mass of the body, v its velocity, g is the acceleration due to gravity and h is the height. Check whether this equation is dimensionally correct.
Answer:
The equation is $\frac{1}{2} m v^2 = m g h$.
We need to check the dimensions of the terms on the left side (LHS) and the right side (RHS).
Dimensions of LHS: $[\frac{1}{2} m v^2]$
- The constant $\frac{1}{2}$ is dimensionless.
- Dimension of mass $[m]$ is $[M]$.
- Dimension of velocity $[v]$ is $[LT^{-1}]$, so dimension of $[v^2]$ is $[LT^{-1}]^2 = [L^2 T^{-2}]$.
Dimension of LHS $= [M] \times [L^2 T^{-2}] = [ML^2 T^{-2}]$.
Dimensions of RHS: $[m g h]$
- Dimension of mass $[m]$ is $[M]$.
- Dimension of acceleration due to gravity $[g]$ (which is an acceleration) is $[LT^{-2}]$.
- Dimension of height $[h]$ (which is a length) is $[L]$.
Dimension of RHS $= [M] \times [LT^{-2}] \times [L] = [M L^{1+1} T^{-2}] = [ML^2 T^{-2}]$.
Since the dimensions of the LHS ($[ML^2 T^{-2}]$) are equal to the dimensions of the RHS ($[ML^2 T^{-2}]$), the equation $\frac{1}{2} m v^2 = m g h$ is dimensionally correct. (This equation relates kinetic energy to potential energy, and energy has dimensions $[ML^2 T^{-2}]$).
Example 2.16. The SI unit of energy is J = kg m$^2$ s$^{-2}$; that of speed v is m s$^{-1}$ and of acceleration a is m s$^{-2}$. Which of the formulae for kinetic energy (K) given below can you rule out on the basis of dimensional arguments (m stands for the mass of the body) :
(a) $K = m^2 v^3$
(b) $K = (1/2)mv^2$
(c) $K = ma$
(d) $K = (3/16)mv^2$
(e) $K = (1/2)mv^2 + ma$
Answer:
The SI unit of energy is kg m$^2$ s$^{-2}$. In terms of base dimensions, this is $[M][L]^2[T]^{-2} = [ML^2 T^{-2}]$. So, the dimensions of kinetic energy $[K]$ must be $[ML^2 T^{-2}]$.
We will check the dimensions of the RHS for each given formula:
(a) $K = m^2 v^3$
- Dimension of $m^2$: $[M]^2 = [M^2]$.
- Dimension of $v^3$: $[LT^{-1}]^3 = [L^3 T^{-3}]$.
Dimension of RHS $= [M^2] [L^3 T^{-3}] = [M^2 L^3 T^{-3}]$.
This does not match $[ML^2 T^{-2}]$. So, formula (a) is ruled out.
(b) $K = (1/2)mv^2$
- The constant $1/2$ is dimensionless.
- Dimension of $m$: $[M]$.
- Dimension of $v^2$: $[LT^{-1}]^2 = [L^2 T^{-2}]$.
Dimension of RHS $= [M] [L^2 T^{-2}] = [ML^2 T^{-2}]$.
This matches $[ML^2 T^{-2}]$. Formula (b) is dimensionally consistent, but not necessarily correct based on dimensions alone.
(c) $K = ma$
- Dimension of $m$: $[M]$.
- Dimension of $a$: $[LT^{-2}]$.
Dimension of RHS $= [M] [LT^{-2}] = [MLT^{-2}]$.
This does not match $[ML^2 T^{-2}]$. So, formula (c) is ruled out. (This is the dimension of force).
(d) $K = (3/16)mv^2$
- The constant $3/16$ is dimensionless.
- Dimension of $m$: $[M]$.
- Dimension of $v^2$: $[LT^{-1}]^2 = [L^2 T^{-2}]$.
Dimension of RHS $= [M] [L^2 T^{-2}] = [ML^2 T^{-2}]$.
This matches $[ML^2 T^{-2}]$. Formula (d) is dimensionally consistent, but not necessarily correct.
(e) $K = (1/2)mv^2 + ma$
This formula involves adding two terms: $(1/2)mv^2$ and $ma$.
- Dimension of $(1/2)mv^2$ is $[ML^2 T^{-2}]$ (from part b).
- Dimension of $ma$ is $[MLT^{-2}]$ (from part c).
According to the principle of homogeneity, quantities with different dimensions cannot be added together. Since $[ML^2 T^{-2}] \neq [MLT^{-2}]$, this addition is dimensionally incorrect.
Formula (e) is ruled out.
Based on dimensional arguments, formulas (a), (c), and (e) are ruled out. Formulas (b) and (d) are dimensionally consistent with the dimensions of kinetic energy. Dimensional analysis cannot distinguish between (b) and (d) because they differ only by a dimensionless constant. The correct formula for kinetic energy is indeed (b).
Deducing Relation Among The Physical Quantities
Dimensional analysis can sometimes help in deducing a possible relationship between a physical quantity and other quantities it depends on. This method works best when the dependence is of a product type and involves a limited number of variables (typically up to three, in simple cases).
The procedure is to assume the physical quantity is proportional to a product of the other quantities raised to unknown powers, and then use dimensional homogeneity to solve for the exponents.
Example 2.17. Consider a simple pendulum, having a bob attached to a string, that oscillates under the action of the force of gravity. Suppose that the period of oscillation of the simple pendulum depends on its length (l), mass of the bob (m) and acceleration due to gravity (g). Derive the expression for its time period using method of dimensions.
Answer:
We are given that the period of oscillation ($T$) of a simple pendulum depends on the length of the pendulum ($l$), the mass of the bob ($m$), and the acceleration due to gravity ($g$).
We assume a relationship of the product form:
$$T \propto l^x m^y g^z$$Introducing a dimensionless constant of proportionality $k$, we can write:
$$T = k \, l^x m^y g^z$$Now, write the dimensions of each quantity:
- Dimension of Period $[T]$: $[T^1]$ or $[M^0 L^0 T^1]$
- Dimension of Length $[l]$: $[L^1]$ or $[M^0 L^1 T^0]$
- Dimension of Mass $[m]$: $[M^1]$ or $[M^1 L^0 T^0]$
- Dimension of acceleration due to gravity $[g]$: $[LT^{-2}]$ or $[M^0 L^1 T^{-2}]$
- The constant $k$ is dimensionless: $[M^0 L^0 T^0]$
Substitute these dimensions into the assumed equation:
$$[M^0 L^0 T^1] = [M^0 L^0 T^0] \times [M^0 L^1 T^0]^x \times [M^1 L^0 T^0]^y \times [M^0 L^1 T^{-2}]^z$$ $$[M^0 L^0 T^1] = [L^{x+z} M^{y} T^{-2z}]$$Applying the Principle of Homogeneity of Dimensions, the powers of corresponding base dimensions on both sides must be equal:
- For [M]: $0 = y$
- For [L]: $0 = x + z$
- For [T]: $1 = -2z$
Solve this system of linear equations for $x, y, z$:
- From the [M] equation, $y = 0$. This implies the period does not depend on the mass of the bob (for a simple pendulum, ignoring air resistance etc.).
- From the [T] equation, $1 = -2z$, so $z = -\frac{1}{2}$.
- From the [L] equation, $0 = x + z$. Substituting the value of $z$, $0 = x - \frac{1}{2}$, so $x = \frac{1}{2}$.
Now substitute the values of $x, y, z$ back into the assumed relationship:
$$T = k \, l^{1/2} m^0 g^{-1/2}$$ $$T = k \, l^{1/2} g^{-1/2}$$ $$T = k \sqrt{\frac{l}{g}}$$This is the dimensionally derived expression for the period of a simple pendulum. Dimensional analysis cannot determine the value of the dimensionless constant $k$. From experiments and more advanced theoretical analysis, it is known that $k = 2\pi$.
Thus, the full formula is $T = 2\pi \sqrt{\frac{l}{g}}$.
Limitations of Dimensional Analysis:
- Cannot determine dimensionless constants ($k$).
- Cannot determine the exact mathematical form of a relationship if it involves sums, differences, logarithmic, exponential, or trigonometric functions (as these don't follow the simple product rule assumed).
- Cannot distinguish between physical quantities that have the same dimensions (e.g., work and torque both have dimensions $[ML^2 T^{-2}]$).
- Can only deduce relations for quantities that depend on a limited number of other quantities in a simple product form.
Despite its limitations, dimensional analysis is a powerful and useful tool for checking consistency and gaining insight into the possible forms of physical relationships.
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.
Answer:
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.
Answer:
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 ?
Answer:
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 ?
Answer:
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 ?
Answer:
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 ?
Answer:
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.
Answer:
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$^2$
Answer:
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.
Answer:
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 ?
Answer:
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.
Answer:
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.
Answer:
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 ?
Answer:
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 ?
Answer:
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).
Answer:
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 ?
Answer:
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 ?
Answer:
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.
Answer:
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.
Answer:
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.
Answer:
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.
Answer:
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.
Answer:
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 ?
Answer:
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$^{–1}$).
Answer:
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 ?
Answer:
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?
Answer: