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Chapter 1 Units And Measurement
Introduction
At its core, physics is a quantitative science that relies heavily on measurement. Measurement involves comparing a physical quantity to a defined, internationally agreed-upon reference standard called a unit. The result of any measurement is expressed as a numerical value (or numerical measure) accompanied by its unit.
Although there are many physical quantities, they are often interrelated. This allows us to define a limited set of basic quantities, whose units are called fundamental or base units. The units for all other physical quantities, called derived units, can be expressed as combinations of these base units. A complete collection of both base units and derived units forms a system of units.
The International System Of Units
Historically, different countries used various systems of units like the CGS (centimetre, gram, second), FPS (foot, pound, second), and MKS (metre, kilogram, second) systems. To standardize measurements globally for scientific, technical, industrial, and commercial purposes, the Système Internationale d’ Unites (SI) was established and adopted internationally.
The SI system is based on seven fundamental base units:
| Base Quantity | Name of Unit | Symbol |
|---|---|---|
| Length | metre | m |
| Mass | kilogram | kg |
| Time | second | s |
| Electric Current | ampere | A |
| Thermodynamic Temperature | kelvin | K |
| Amount of Substance | mole | mol |
| Luminous Intensity | candela | cd |
The definitions of these base units are precise and based on fundamental physical constants or properties, ensuring reproducibility. For instance, the metre is defined based on the speed of light, the second on the frequency of caesium atom radiation, and the kilogram on Planck's constant.
In addition to the seven base units, SI includes two supplementary units for angles:
- Radian (rad): For plane angle (ratio of arc length to radius).
- Steradian (sr): For solid angle (ratio of intercepted area on a sphere to the square of the radius).
Both radian and steradian are dimensionless quantities.
Other physical quantities have derived units, which are combinations of the base units (e.g., velocity in m/s). Some derived units are given special names (e.g., Newton for force, Joule for energy, Watt for power).
The SI system is a decimal system, making conversions between multiples and sub-multiples convenient using standard prefixes (like kilo-, milli-, micro- etc.). Proper guidelines for using symbols for quantities and units are important for clear communication.
Significant Figures
Every physical measurement inherently involves some degree of uncertainty or error, primarily due to the limitations of the measuring instrument's precision (its least count) and the experimental conditions. The result of a measurement should reflect this precision.
The digits in a measured number that are known reliably, plus the very first digit that is uncertain, are called significant digits or significant figures. The number of significant figures indicates the precision of the measurement.
Changing the units of a measurement does not change the number of significant figures (e.g., 1.62 m has 3 significant figures, and so does 162 cm or 0.00162 km).
Rules for Determining the Number of Significant Figures:
- All non-zero digits are significant. (e.g., 285 cm has 3 significant figures).
- Zeros between two non-zero digits are significant, regardless of the decimal point. (e.g., 2005 g has 4 significant figures).
- Leading zeros (zeros to the left of the first non-zero digit) are NOT significant. They only indicate the position of the decimal point. (e.g., 0.0023 m has 2 significant figures).
- Trailing zeros (zeros at the end of a number) are significant ONLY if the number contains a decimal point.
- If there is no decimal point, trailing zeros are generally NOT significant (unless explicitly stated by scientific notation). (e.g., 12300 m might have 3, 4, or 5 significant figures depending on context).
- If there is a decimal point, trailing zeros ARE significant (they indicate precision). (e.g., 123.00 m has 5 significant figures, 0.06900 kg has 4 significant figures).
- To avoid ambiguity with trailing zeros, scientific notation is preferred ($a \times 10^b$, where $1 \le a < 10$). In scientific notation, all digits in the 'a' part are significant. (e.g., $1.23 \times 10^4 \text{ m}$ has 3 significant figures; $1.2300 \times 10^4 \text{ m}$ has 5 significant figures).
- Exact numbers (from counting or definitions, like '2' in $2\pi r$) have an infinite number of significant figures.
Rules For Arithmetic Operations With Significant Figures
When performing calculations with measured values, the result's precision must reflect the least precise input value. The final result should not have more significant figures (or decimal places) than justified by the original measurements.
- Multiplication or Division: The final result should have the same number of significant figures as the original number with the least number of significant figures.
Example 1. 5.74 g of a substance occupies 1.2 cm³. Express its density by keeping the significant figures in view.
Answer:
Mass = 5.74 g (3 significant figures)
Volume = 1.2 cm³ (2 significant figures)
Density = Mass / Volume = 5.74 g / 1.2 cm³ = 4.7833... g/cm³
The least precise measurement has 2 significant figures (1.2 cm³). Therefore, the density should be reported to 2 significant figures.
Density = 4.8 g cm⁻³.
- Addition or Subtraction: The final result should retain as many decimal places as are there in the number with the least number of decimal places.
Example: $436.32 \text{ g} + 227.2 \text{ g} + 0.301 \text{ g} = 663.821 \text{ g}$. The number $227.2 \text{ g}$ has the fewest decimal places (one). The result is rounded to one decimal place: 663.8 g.
Rounding Off The Uncertain Digits
When a calculation result has more digits than are significant, it must be rounded off. The rules for rounding are:
- If the digit to be dropped is greater than 5, increase the preceding digit by 1. (e.g., 2.746 rounded to 3 s.f. is 2.75).
- If the digit to be dropped is less than 5, leave the preceding digit unchanged. (e.g., 1.743 rounded to 3 s.f. is 1.74).
- 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 s.f. is 2.74).
- If the preceding digit is odd, increase it by 1. (e.g., 2.735 rounded to 3 s.f. is 2.74).
In multi-step calculations, it is advisable to retain one extra significant digit in intermediate results to minimize rounding errors, and round off to the final required significant figures only at the end.
Rules For Determining The Uncertainty In The Results Of Arithmetic Calculations
The uncertainty or error in experimental results propagates through calculations. Specific rules govern how uncertainties combine in arithmetic operations:
- In multiplication or division, the relative error of the result is the sum of the relative errors of the quantities being multiplied or divided. The number of significant figures in the result is limited by the number with the fewest significant figures.
- In addition or subtraction, the absolute error of the result is the sum of the absolute errors of the quantities being added or subtracted. The precision of the result is limited by the number with the smallest number of decimal places. Subtracting numbers can sometimes significantly reduce the number of significant figures.
- The relative error of a value expressed with $n$ significant figures also depends on the value itself (e.g., the relative error in 1.02 g is larger than in 9.89 g, even if both are measured to $\pm 0.01$ g).
Example 1. 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 length of the side of the cube is $l = 7.203 \text{ m}$. This value has 4 significant figures (all non-zero digits and the trailing zero after the decimal). The calculated area and volume should be reported to the same number of significant figures (4).
Total surface area of the cube = $6 \times l^2 = 6 \times (7.203 \text{ m})^2 = 6 \times 51.883209 \text{ m}^2 = 311.299254 \text{ m}^2$.
Rounding off to 4 significant figures: 311.3 m².
Volume of the cube = $l^3 = (7.203 \text{ m})^3 = 373.714754 \text{ m}^3$.
Rounding off to 4 significant figures: 373.7 m³.
The total surface area is 311.3 m², and the volume is 373.7 m³, reported to appropriate significant figures.
Dimensions Of Physical Quantities
The dimensions of a physical quantity describe its fundamental nature by expressing it in terms of the base quantities. The seven base quantities are considered the seven dimensions of the physical world and are denoted by square brackets: [L] for length, [M] for mass, [T] for time, [A] for electric current, [K] for thermodynamic temperature, [cd] for luminous intensity, and [mol] for amount of substance.
The dimensions of a derived physical quantity are the powers (exponents) to which the base quantity dimensions are raised to represent that quantity. For example:
- Volume: Product of three lengths. Dimensions: $[L] \times [L] \times [L] = [L]^3 = [L^3]$. (Zero dimension in M and T, three in L).
- Speed or Velocity: Length divided by time. Dimensions: $[L]/[T] = [LT^{-1}]$. (Zero dimension in M, one in L, minus one in T).
- Force: Mass times acceleration. Acceleration is length divided by time squared ($[LT^{-2}]$). Dimensions: $[M] \times [LT^{-2}] = [MLT^{-2}]$. (One dimension in M, one in L, minus two in T).
Dimensions focus on the type of physical quantity, not its magnitude. Different quantities of the same type (e.g., initial velocity, final velocity, speed) have the same dimensions.
Dimensional Formulae And Dimensional Equations
The expression showing how the base quantities are related to the dimensions of a physical quantity is its dimensional formula. It's written using the base dimensions with their appropriate powers. Example: Dimensional formula of volume is $[M^0 L^3 T^0]$, speed is $[M^0 L T^{-1}]$, force is $[M L T^{-2}]$, mass density is $[M L^{-3} T^0]$.
A dimensional equation is formed by equating a physical quantity (represented by its symbol in brackets) to its dimensional formula. Example: $[V] = [M^0 L^3 T^0]$, $[v] = [M^0 L T^{-1}]$, $[F] = [M L T^{-2}]$, $[\rho] = [M L^{-3} T^0]$. Dimensional equations are used to represent the dimensions explicitly.
Dimensional Analysis And Its Applications
Dimensional analysis is a tool based on the principle that physical quantities can only be added or subtracted if they have the same dimensions. This principle, called the principle of homogeneity of dimensions, also applies to equations: the dimensions of terms on both sides of a mathematical equation must be identical when simplified. Treating units/dimensions like algebraic symbols in multiplication and division allows for manipulation.
Checking The Dimensional Consistency Of Equations
The principle of homogeneity is used to check if an equation is dimensionally consistent. If the dimensions of all terms on one side of an equation match the dimensions of the terms on the other side, the equation is dimensionally correct. If they do not match, the equation is definitely wrong. However, a dimensionally correct equation is not guaranteed to be correct; it might be incorrect by a dimensionless constant or function.
Arguments of dimensionless mathematical functions (trigonometric, logarithmic, exponential) must be dimensionless themselves. Pure numbers (like angles, ratios of similar quantities) are dimensionless.
Example 3. Let us consider an equation $1/2 mv^2 = mgh$, 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:
We need to check if the dimensions on the Left Hand Side (LHS) are equal to the dimensions on the Right Hand Side (RHS).
LHS: $1/2 mv^2$. The factor 1/2 is a dimensionless constant. Mass $m$ has dimensions [M]. Velocity $v$ has dimensions [LT⁻¹]. So, $v^2$ has dimensions [LT⁻¹]² = [L²T⁻²].
Dimensions of LHS = $[M] \times [L^2T^{-2}] = [ML^2T^{-2}]$.
RHS: $mgh$. Mass $m$ has dimensions [M]. Acceleration due to gravity $g$ has dimensions [LT⁻²]. Height $h$ has dimensions [L].
Dimensions of RHS = $[M] \times [LT^{-2}] \times [L] = [ML^2T^{-2}]$.
Since the dimensions of LHS ($[ML^2T^{-2}]$) are the same as the dimensions of RHS ($[ML^2T^{-2}]$), the equation is dimensionally correct.
Example 4. The SI unit of energy is J = kg m² 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) $K = m² v³$ (b) $K = (1/2)mv²$ (c) $K = ma$ (d) $K = (3/16)mv²$ (e) $K = (1/2)mv² + ma$
Answer:
The dimensions of energy (K) are given by the units J = kg m² s⁻², which corresponds to dimensions $[ML^2T^{-2}]$. We need to check the dimensions of each proposed formula for K.
- (a) $K = m² v³$. Dimensions = $[M]^2 \times [LT^{-1}]^3 = [M^2L^3T^{-3}]$. These dimensions do not match $[ML^2T^{-2}]$. Rule out (a).
- (b) $K = (1/2)mv²$. The constant 1/2 is dimensionless. Dimensions = $[M] \times [LT^{-1}]^2 = [M] \times [L^2T^{-2}] = [ML^2T^{-2}]$. These dimensions match $[ML^2T^{-2}]$. This formula is dimensionally correct.
- (c) $K = ma$. Dimensions = $[M] \times [LT^{-2}] = [MLT^{-2}]$. These dimensions do not match $[ML^2T^{-2}]$. Rule out (c).
- (d) $K = (3/16)mv²$. The constant 3/16 is dimensionless. Dimensions = $[M] \times [LT^{-1}]^2 = [M] \times [L^2T^{-2}] = [ML^2T^{-2}]$. These dimensions match $[ML^2T^{-2}]$. This formula is dimensionally correct.
- (e) $K = (1/2)mv² + ma$. This involves the addition of two terms. The first term $(1/2)mv^2$ has dimensions $[ML^2T^{-2}]$ (as shown in b). The second term $ma$ has dimensions $[MLT^{-2}]$ (as shown in c). Quantities with different dimensions cannot be added. Therefore, the expression on the RHS has inconsistent dimensions. Rule out (e).
On the basis of dimensional arguments, formulae (a), (c), and (e) can be ruled out.
Deducing Relation Among The Physical Quantities
Dimensional analysis can sometimes be used to derive relationships between physical quantities, assuming a product form of dependence (e.g., $Q = k \cdot A^x B^y C^z$). By equating the dimensions on both sides of the equation, one can solve for the exponents (x, y, z). However, this method cannot determine the value of dimensionless constants (k) and is limited to cases where the dependence is a simple power product involving a limited number of variables.
Example 5. 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 ($T$) depends on length ($l$), mass ($m$), and acceleration due to gravity ($g$). We assume a product relationship:
$T = k \cdot l^x m^y g^z$
where $k$ is a dimensionless constant and $x, y, z$ are exponents to be determined.
Write the dimensions of each quantity:
- Period ($T$): $[T] = [M^0 L^0 T^1]$
- Length ($l$): $[L] = [M^0 L^1 T^0]$
- Mass ($m$): $[M] = [M^1 L^0 T^0]$
- Acceleration ($g$): $[LT^{-2}] = [M^0 L^1 T^{-2}]$
Substitute these dimensions into the assumed relationship:
$[M^0 L^0 T^1] = [L]^x [M]^y [LT^{-2}]^z$
$[M^0 L^0 T^1] = [L^x] [M^y] [L^z T^{-2z}]$
Combine the exponents for each base dimension on the right side:
$[M^0 L^0 T^1] = [M^y L^{x+z} T^{-2z}]$
Equate the exponents for each base dimension on both sides of the equation:
- For M: $0 = y \implies y = 0$.
- For L: $0 = x + z \implies x = -z$.
- For T: $1 = -2z \implies z = -1/2$.
Now substitute the value of $z$ into the equation for $x$:
$x = -z = -(-1/2) = +1/2$.
The exponents are $x = 1/2$, $y = 0$, $z = -1/2$.
Substitute these exponents back into the assumed relationship:
$T = k \cdot l^{1/2} m^0 g^{-1/2}$
$T = k \cdot l^{1/2} \cdot 1 \cdot \frac{1}{g^{1/2}}$
$T = k \sqrt{\frac{l}{g}}$
The method of dimensions derives the relationship $T = k \sqrt{l/g}$. The dimensionless constant $k$ cannot be determined by this method.
Exercises
Question 1.1. 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 1.2. 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 = $6.67 \times 10^{-11} N m^2 (kg)^{-2}$ = .... $(cm)^3 s^{-2} g^{-1}$.
Answer:
Question 1.3. 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 1.4. 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 1.5. 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 1.6. 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 1.7. 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 1.8. 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 1.9. 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 1.10. 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 1.11. 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 1.12. 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 1.13. A famous relation in physics relates ‘moving mass’ m to the ‘rest mass’ $m_o$ 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 1.14. 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 1.15. 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 1.16. 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 1.17. 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 = $2.0 \times 10^{30}$ kg, radius of the Sun = $7.0 \times 10^8$ m.
Answer: