1.1 Introduction to Physical Quantities
Physics is fundamentally a quantitative science. Every scientific inquiry, theoretical prediction, and technological advancement relies on our ability to measure physical properties accurately. Measurement involves comparing a physical quantity with a globally accepted standard. This comparison results in a two-part expression: a numerical value and a specific unit.
Physical quantities are broadly classified into two categories:
- Fundamental (Base) Quantities: Independent quantities that cannot be derived from one another (e.g., length, mass, time).
- Derived Quantities: Quantities obtained by mathematically combining base quantities (e.g., velocity, force, density).
1.2 The International System of Units (SI)
To ensure global consistency, the scientific community adopted the International System of Units (SI). The SI system ties units to universal physical constants. These include the speed of light (c) and the Planck constant (h). This method provides stability and reproducibility across the globe.
The SI system is built upon seven foundational base units:
- Metre (m): Length
- Kilogram (kg): Mass
- Second (s): Time
- Ampere (A): Electric current
- Kelvin (K): Thermodynamic temperature
- Mole (mol): Amount of substance
- Candela (cd): Luminous intensity
Additionally, the system defines two dimensionless units for geometric measurements. These units are the radian for plane angles and the steradian for solid angles.
1.3 Precision, Certainty, and Significant Figures
A measurement is only as valuable as its reliability. Significant figures show a measurement’s precision. They encompass all certain digits. They also include the first uncertain (estimated) digit. Including superfluous digits implies a false sense of accuracy, while providing too few underrepresents the data’s reliability.
The Role of Scientific Notation
Ambiguity often arises when dealing with trailing zeros. For example, converting 4.700 m to centimeters yields 470.0 cm. Both values contain four significant digits, proving that simply changing units does not alter the precision of the instrument used. To remove confusion, scientists rely on scientific notation (a × 10ᵇ). Expressing the earlier measurement as 4.700 × 10² cm unambiguously maintains the four significant figures.
Arithmetic Operations with Measured Values
When processing experimental data, the mathematical result cannot be more precise than the least precise measurement used in the calculation.
- Multiplication and Division: The final result must be reported with the same number of significant figures. This is based on the input that has the fewest significant figures. Example: Dividing a mass of 4.237 g (4 sig figs) by a volume of 2.51 cm³ (3 sig figs) yields a calculator output of 1.688047… g/cm³. The correct reported density is 1.69 g/cm³.
- Addition and Subtraction: The final result is limited by the input with the fewest decimal places. Example: Adding 436.32 g, 227.2 g, and 0.301 g. The least precise measurement (227.2 g) is known to only one decimal place. The sum is correctly reported as 663.8 g.
1.4 Dimensional Analysis
Dimensions provide a qualitative description of physical quantities, independent of their numerical magnitude or the specific unit system used. They are expressed as powers of base quantities and are conventionally enclosed in square brackets. The primary mechanical dimensions are mass [M], length [L], and time [T].
Using these building blocks, we can construct the dimensional formulae for derived quantities:
- Velocity (v): [L T⁻¹]
- Force (F): [M L T⁻²]
- Density (ρ): [M L⁻³]
The Principle of Dimensional Homogeneity
For any mathematical equation describing a physical reality to be valid, it must be dimensionally homogeneous. This principle dictates that every term separated by a plus, minus, or equals sign must have the exact same dimensions.
Consider the kinematic equation for displacement: x = x₀ + v₀t + ½at²
Analyzing the dimensions of each term:
- x and x₀ have the dimension [L]
- v₀t evaluates to [L T⁻¹][T] = [L]
- ½at² evaluates to [L T⁻²][T²] = [L] (Note: fractions like ½ are dimensionless)
Because every term reduces to the dimension of length [L], the equation is dimensionally consistent.
Deriving Physical Relationships
Dimensional analysis is a powerful theoretical tool used to deduce relationships between variables. If we assume a physical quantity depends on a set of variables, we can construct an equation. We do this by equating their dimensions.
Example: Determine the period (T) of a simple pendulum. Assume it depends on the length of the string (l). It also depends on the mass of the bob (m) and the acceleration due to gravity (g).
Let the relationship be: T = k lˣ mʸ gᶻ
Where k is a dimensionless constant. Substituting the dimensions for each variable yields: [T] = [L]ˣ [M]ʸ [L T⁻²]ᶻ = [L]ˣ⁺ᶻ [M]ʸ [T]⁻²ᶻ
By equating the exponents of identical base dimensions on both sides:
- For Time ([T]): 1 = -2z ⇒ z = -½
- For Mass ([M]): 0 = y ⇒ y = 0
- For Length ([L]): 0 = x + z ⇒ x = ½
Substituting these exponents back into our original assumption provides the physical relationship: T = k √(l/g)
(Through advanced mechanics or experiment, k is later determined to be 2π.)
1.5 Error Estimation and Propagation
All physical measurements have inherent uncertainty. Understanding how these errors propagate through calculations is vital for determining the reliability of the final result.
- In addition, or subtraction, the absolute errors of the individual measurements are combined.
- In multiplication or division, the relative (or percentage) errors of the individual measurements are combined.
For instance, if calculating the area of a rectangle with length 16.2 ± 0.1 cm and breadth 10.1 ± 0.1 cm, calculating the joint relative error proves that the total area is 163.6 ± 2.6 cm². Proper error handling ensures that scientific conclusions are neither overly confident nor excessively cautious.
Chapter Summary
This chapter establishes the framework for all quantitative physics. Standardization through the SI system ensures universal communication. The strict rules of significant figures and error propagation maintain the integrity of our data. Finally, dimensional analysis acts as a logic check for physical equations. It also serves as a theoretical tool for uncovering the fundamental laws governing the universe.
Practice Problems: Chapter 1 – Foundations of Measurement and Dimensional Analysis
Instructions: These problems are designed to test your understanding of significant figures, dimensional analysis, and error propagation. Show all your work for full credit.
Section A: Significant Figures and Arithmetic
Problem 1: Find the number of significant figures in the next measurements:
a) 0.0050 m
b) 2.050 x 10^3 kg
c) 400.0 cm
d) 0.03040 s
Problem 2: Perform the following calculations and report the answer to the correct number of significant figures:
a) 15.23 cm + 2.1 cm + 0.034 cm
b) 14.5 g – 2.34 g
c) (5.42 x 10^3 m) * (2.0 x 10^2 m)
d) 48.24 g / 12.1 cm^3
Section B: Dimensional Analysis
Problem 3: State the principle of dimensional homogeneity. Why is it useful, and what are its limitations?
Problem 4: Consider the equation v = u + at. In this equation, ‘v’ represents final velocity and ‘u’ is initial velocity. The term ‘a’ denotes acceleration, and ‘t’ indicates time. Verify if the equation is dimensionally consistent. Provide the dimensional formula for each term.
Problem 5: The escape velocity (v_e) from a planet may depend on the universal gravitational constant (G), the mass of the planet (M), and the radius of the planet (R).
Given the dimensional formula for G is [M^-1 L^3 T^-2], use dimensional analysis to derive the relationship between v_e, G, M, and R.
Section C: Error Estimation
Problem 6: The length of a rectangular block is measured as 10.5 +/- 0.2 cm, and its width is measured as 5.2 +/- 0.1 cm.
a) Calculate the perimeter of the block and its absolute error.
b) Calculate the area of the block and its relative error (expressed as a percentage).
Problem 7: The resistance R is given by the formula R = V / I. In this formula, V is voltage and I is current. In an experiment, V is measured as 100 +/- 5 Volts, and I is measured as 10 +/- 0.2 Amperes. Calculate the percentage error in the measurement of resistance R.
1.1 Fill in the blanks
Q (a): Volume of a cube of side 1 cm = ?
A: 1 cm³ = 1 × 10⁻⁶ m³
Q (b): Surface area of a cylinder (r = 2 cm, h = 10 cm) = ?
A: 2πrh + 2πr² = 48π cm² ≈ 4800 mm²
Q (c): Vehicle speed 18 km/h → distance in 1 s = ?
A: 18 km/h = 5 m/s → covers 5 m in 1 s
Q (d): Relative density of lead = 11.3 → density = ?
A: 11.3 g/cm³ = 11300 kg/m³
1.2 Unit conversions
Q (a): Convert 1 kg·m²/s² into g·cm²/s²
A: 1 × 10⁷ g·cm²/s²
Q (b): Convert 1 m into light years
A: 1.057 × 10⁻¹⁶ ly
Q (c): Convert 3.0 m/s² into km/h²
A: 38880 km/h²
Q (d): Express G in cm³/s²·g⁻¹
A: 6.67 × 10⁻⁸ cm³/s²·g⁻¹
1.3 Calorie in new units
Q: Show calorie magnitude in new units (mass = α kg, length = β m, time = γ s).
A: 1 calorie = 4.2 J = 4.2 kg·m²/s² → magnitude = 4.2 α⁻¹ β² γ⁻²
1.4 Reframe statements
Q: Why is “large” or “small” meaningless without comparison? Reframe:
A:
- Atoms are smaller than visible objects.
- Jet plane moves faster than cars.
- Jupiter’s mass is larger than Earth’s.
- Air in room has more molecules than a balloon.
- Proton is heavier than electron.
- Sound is slower than light.
1.5 New unit of length
Q: If speed of light = 1 unit, find Sun–Earth distance (light takes 8 min 20 s).
A: Time = 500 s → distance = 500 new units
1.6 Most precise device
Q: Which device is most precise: vernier, screw gauge, or optical instrument?
A: Optical instrument (precision ~ wavelength of light, ~10⁻⁷ m)
1.7 Hair thickness
Q: Student measures hair width = 3.5 mm at 100× magnification. Actual thickness = ?
A: 3.5 mm ÷ 100 = 0.035 mm = 35 micrometers
Here’s a full analysis of the Units and Measurement exercises (1.8–1.13) with each question written first, followed by its detailed answer in plain text:
1.8 Answer the following
(a) Q: You are given a thread and a metre scale. How will you estimate the diameter of the thread?
A: Wrap the thread tightly around the scale for several turns, measure the total length, then divide by the number of turns. This gives the diameter more accurately than measuring once.
(b) Q: A screw gauge has a pitch of 1.0 mm and 200 divisions on the circular scale. Can accuracy be increased arbitrarily by increasing divisions?
A: No. Accuracy is limited by practical factors like manufacturing defects, zero error, and least count. Beyond a point, more divisions don’t improve reliability.
(c) Q: Why is a set of 100 measurements of brass rod diameter more reliable than 5 measurements?
A: Larger sample size reduces random errors and gives a better average value, improving reliability.
1.9 Linear magnification
Q: A house photo occupies 1.75 cm² on a slide. On screen it is 1.55 m². Find linear magnification.
A: Magnification = √(area on screen ÷ area on slide) = √(1.55 m² ÷ 1.75 cm²). Convert units: 1.55 m² = 15,500 cm². Ratio = 15,500 ÷ 1.75 ≈ 8857. √8857 ≈ 94.1. Magnification ≈ 94.
1.10 Significant figures
Q: State number of significant figures:
(a) 0.007 m² → 1
(b) 2.64 × 10²⁴ kg → 3
(c) 0.2370 g/cm³ → 4
(d) 6.320 J → 4
(e) 6.032 N/m² → 4
(f) 0.0006032 m² → 4
1.11 Area and volume of sheet
Q: Length = 4.234 m, breadth = 1.005 m, thickness = 2.01 cm. Find area and volume with correct significant figures.
A:
- Area = 4.234 × 1.005 = 4.255 m² (4 significant figures).
- Thickness = 0.0201 m. Volume = 4.255 × 0.0201 = 0.0856 m³ (3 significant figures).
1.12 Mass of box and gold pieces
Q: Box mass = 2.30 kg. Gold pieces = 20.15 g and 20.17 g. Find total mass and difference.
A:
- Convert gold masses: 20.15 g = 0.02015 kg, 20.17 g = 0.02017 kg.
- Total mass = 2.30 + 0.02015 + 0.02017 = 2.340 kg (3 significant figures).
- Difference = 0.02 g (1 significant figure).
1.13 Relativistic mass relation
Q: A boy writes m = m₀ / √(1 – v²). Where should the missing c be?
A: Correct formula is m = m₀ / √(1 – v²/c²). The constant c (speed of light) must appear with v² to make the ratio dimensionless.
✅ Now you have questions + answers side by side, clearly explained and in non‑LaTeX format.
Here’s a clear analysis of the Units and Measurement exercises (1.14–1.17) with each question written first, followed by its detailed answer in plain text:
1.14 Atomic volume of hydrogen
Q: The unit of length convenient on the atomic scale is the angstrom (Å), where 1 Å = 10⁻¹⁰ m. The size of a hydrogen atom is about 0.5 Å. What is the total atomic volume in m³ of a mole of hydrogen atoms?
A:
- Radius of hydrogen atom ≈ 0.5 Å = 0.5 × 10⁻¹⁰ m.
- Volume of one atom ≈ (4/3)πr³ ≈ (4/3)π(0.5 × 10⁻¹⁰)³ ≈ 5.2 × 10⁻³¹ m³.
- Number of atoms in 1 mole = 6.022 × 10²³.
- Total volume ≈ 5.2 × 10⁻³¹ × 6.022 × 10²³ ≈ 3.1 × 10⁻⁷ m³.
1.15 Ratio of molar volume to atomic volume
Q: One mole of an ideal gas at STP occupies 22.4 L. What is the ratio of molar volume to atomic volume of a mole of hydrogen (size of molecule ≈ 1 Å)? Why is this ratio so large?
A:
- Radius of hydrogen molecule ≈ 1 Å = 1 × 10⁻¹⁰ m.
- Volume of one molecule ≈ (4/3)πr³ ≈ 4.2 × 10⁻³⁰ m³.
- Volume of 1 mole of molecules ≈ 4.2 × 10⁻³⁰ × 6.022 × 10²³ ≈ 2.5 × 10⁻⁶ m³.
- Molar volume at STP = 22.4 L = 2.24 × 10⁻² m³.
- Ratio = (2.24 × 10⁻²) ÷ (2.5 × 10⁻⁶) ≈ 9 × 10³.
- Reason: Gas molecules are extremely small compared to the empty space between them, so molar volume is much larger.
1.16 Observation from a moving train
Q: Why do nearby trees and houses seem to move rapidly opposite to the train’s motion, while distant objects like hills, Moon, or stars appear stationary?
A:
Nearby objects subtend a large angle at the eye and shift position quickly relative to you, so they appear to move fast. Distant objects subtend a very small angle and their relative shift is negligible, so they appear stationary.
1.17 Density of the Sun
Q: The Sun’s mass = 2.0 × 10³⁰ kg, radius = 7.0 × 10⁸ m. Estimate its density and compare with solids, liquids, or gases.
A:
- Volume of Sun = (4/3)πr³ = (4/3)π(7 × 10⁸)³ ≈ 1.4 × 10²⁷ m³.
- Density = mass ÷ volume = (2.0 × 10³⁰) ÷ (1.4 × 10²⁷) ≈ 1.4 × 10³ kg/m³.
- This is close to density of liquids (like water, ~1000 kg/m³), much higher than gases at STP.
- So the Sun’s average density lies in the range of liquids, even though it is plasma.
1.14 Atomic volume of hydrogen
Q: The unit of length convenient on the atomic scale is the angstrom (Å), where 1 Å = 10⁻¹⁰ m. The size of a hydrogen atom is about 0.5 Å. What is the total atomic volume in m³ of a mole of hydrogen atoms?
A:
- Radius of hydrogen atom ≈ 0.5 Å = 0.5 × 10⁻¹⁰ m.
- Volume of one atom ≈ (4/3)πr³ ≈ (4/3)π(0.5 × 10⁻¹⁰)³ ≈ 5.2 × 10⁻³¹ m³.
- Number of atoms in 1 mole = 6.022 × 10²³.
- Total volume ≈ 5.2 × 10⁻³¹ × 6.022 × 10²³ ≈ 3.1 × 10⁻⁷ m³.
1.15 Ratio of molar volume to atomic volume
Q: One mole of an ideal gas at STP occupies 22.4 L. What is the ratio of molar volume to atomic volume of a mole of hydrogen (size of molecule ≈ 1 Å)? Why is this ratio so large?
A:
- Radius of hydrogen molecule ≈ 1 Å = 1 × 10⁻¹⁰ m.
- Volume of one molecule ≈ (4/3)πr³ ≈ 4.2 × 10⁻³⁰ m³.
- Volume of 1 mole of molecules ≈ 4.2 × 10⁻³⁰ × 6.022 × 10²³ ≈ 2.5 × 10⁻⁶ m³.
- Molar volume at STP = 22.4 L = 2.24 × 10⁻² m³.
- Ratio = (2.24 × 10⁻²) ÷ (2.5 × 10⁻⁶) ≈ 9 × 10³.
- Reason: Gas molecules are extremely small compared to the empty space between them, so molar volume is much larger.
1.16 Observation from a moving train
Q: Why do nearby trees and houses seem to move rapidly in the opposite direction of the train’s movement? Distant objects, like hills, the Moon, or stars, appear stationary.
A:
Nearby objects subtend a large angle at the eye. They shift position quickly relative to you, so they appear to move fast. Distant objects subtend a very small angle and their relative shift is negligible, so they appear stationary.
1.17 Density of the Sun
Q: The Sun’s mass = 2.0 × 10³⁰ kg, radius = 7.0 × 10⁸ m. Estimate its density and compare with solids, liquids, or gases.
A:
- Volume of Sun = (4/3)πr³ = (4/3)π(7 × 10⁸)³ ≈ 1.4 × 10²⁷ m³.
- Density = mass ÷ volume = (2.0 × 10³⁰) ÷ (1.4 × 10²⁷) ≈ 1.4 × 10³ kg/m³.
- This is close to density of liquids (like water, ~1000 kg/m³), much higher than gases at STP.
- So the Sun’s average density lies in the range of liquids, even though it is plasma.
Leave a Reply