Chapter 1: Units and Measurements
Unit I – Physical World and Measurement
1.1 Introduction
Physics is the study of the fundamental laws of nature and their applications. Every observation in physics involves measurement. A measurement compares a quantity to a standard called a unit.
1.2 The International System of Units (SI)
The SI system (Système International d’Unités) is the modern metric system adopted internationally.
1.2.1 Fundamental (Base) Quantities and Units
| Physical Quantity | SI Unit | Symbol |
|---|---|---|
| Length | metre | m |
| Mass | kilogram | kg |
| Time | second | s |
| Electric current | ampere | A |
| Temperature | kelvin | K |
| Amount of substance | mole | mol |
| Luminous intensity | candela | cd |
1.2.2 Derived Units
Derived units are combinations of base units.
Examples:
- Speed: \(\text{m s}^{-1}\)
- Force: \(\text{N} = \text{kg m s}^{-2}\)
- Energy: \(\text{J} = \text{kg m}^2\text{s}^{-2}\)
- Pressure: \(\text{Pa} = \text{kg m}^{-1}\text{s}^{-2}\)
1.3 Significant Figures
Significant figures (sig figs) are the meaningful digits in a measured or calculated quantity.
Rules for Counting Significant Figures
- All non-zero digits are significant. (e.g., 2345 → 4 sig figs)
- Zeros between non-zero digits are significant. (e.g., 1024 → 4 sig figs)
- Leading zeros are not significant. (e.g., 0.0023 → 2 sig figs)
- Trailing zeros in a decimal number are significant. (e.g., 2.300 → 4 sig figs)
- Trailing zeros in an integer may or may not be significant.
Operations with Significant Figures
- Addition/Subtraction: Result has as many decimal places as the least precise measurement.
- Multiplication/Division: Result has as many sig figs as the least number of sig figs in the inputs.
Example: \(4.56 \times 1.4 = 6.384 \approx 6.4\) (2 sig figs)
1.4 Dimensions of Physical Quantities
The dimension of a physical quantity represents how it depends on the base quantities.
Using notation: M (mass), L (length), T (time), A (current), K (temperature):
| Quantity | Dimensional Formula |
|---|---|
| Velocity | \([M^0 L T^{-1}]\) |
| Acceleration | \([M^0 L T^{-2}]\) |
| Force | \([M L T^{-2}]\) |
| Work/Energy | \([M L^2 T^{-2}]\) |
| Power | \([M L^2 T^{-3}]\) |
| Pressure | \([M L^{-1} T^{-2}]\) |
| Impulse | \([M L T^{-1}]\) |
| Gravitational Constant G | \([M^{-1} L^3 T^{-2}]\) |
1.5 Dimensional Analysis
1.5.1 Checking Dimensional Consistency
A physical equation is dimensionally consistent if both sides have the same dimensions.
Example: Verify \(v = u + at\)
\[ [v] = [u] = [M^0 L T^{-1}] \] \[ [at] = [M^0 L T^{-2}][T] = [M^0 L T^{-1}] \checkmark \]
1.5.2 Deriving Relations Using Dimensions
The time period \(T\) of a simple pendulum may depend on length \(l\) and \(g\):
\[ T \propto l^a , g^b \]
Equating dimensions:
\[ [T] = [L]^a [L T^{-2}]^b = L^{a+b}, T^{-2b} \]
So \(-2b = 1 \Rightarrow b = -\dfrac{1}{2}\) and \(a + b = 0 \Rightarrow a = \dfrac{1}{2}\)
\[ \boxed{T = 2\pi\sqrt{\frac{l}{g}}} \]
1.5.3 Unit Conversion Using Dimensions
To convert \(1 \text{ J}\) into \(\text{erg}\) (CGS):
\[ \frac{n_2}{n_1} = \left(\frac{M_1}{M_2}\right)\left(\frac{L_1}{L_2}\right)^2\left(\frac{T_1}{T_2}\right)^{-2} \]
\[ = \left(\frac{\text{kg}}{\text{g}}\right)\left(\frac{\text{m}}{\text{cm}}\right)^2 = 10^3 \times 10^4 = 10^7 \]
\[ 1 \text{ J} = 10^7 \text{ erg} \]
1.6 Errors in Measurement
Every measurement has some uncertainty or error.
Types of Errors
- Systematic errors: Consistent bias (instrument error, personal error)
- Random errors: Irregular variations due to unpredictable factors
- Gross errors: Mistakes in reading instruments
Absolute, Relative, and Percentage Error
If the true value is \(a\) and measured value is \(a_m\):
\[ \text{Absolute error} = |\Delta a| = |a - a_m| \]
\[ \text{Relative error} = \frac{|\Delta a|}{a} \]
\[ \text{Percentage error} = \frac{|\Delta a|}{a} \times 100% \]
Error Propagation
For \(Z = A + B\): \(\Delta Z = \Delta A + \Delta B\)
For \(Z = A \times B\): \(\dfrac{\Delta Z}{Z} = \dfrac{\Delta A}{A} + \dfrac{\Delta B}{B}\)
For \(Z = \dfrac{A^p B^q}{C^r}\): \(\dfrac{\Delta Z}{Z} = p\dfrac{\Delta A}{A} + q\dfrac{\Delta B}{B} + r\dfrac{\Delta C}{C}\)
1.7 Unit Prefixes
| Prefix | Symbol | Power |
|---|---|---|
| tera | T | \(10^{12}\) |
| giga | G | \(10^{9}\) |
| mega | M | \(10^{6}\) |
| kilo | k | \(10^{3}\) |
| milli | m | \(10^{-3}\) |
| micro | μ | \(10^{-6}\) |
| nano | n | \(10^{-9}\) |
| pico | p | \(10^{-12}\) |
1.8 Diagram: Dimensional Analysis Flowchart
Key Formulas Summary
| Formula | Meaning |
|---|---|
| \(\dfrac{\Delta Z}{Z} = p\dfrac{\Delta A}{A} + q\dfrac{\Delta B}{B} + r\dfrac{\Delta C}{C}\) | Error in \(Z = \dfrac{A^p B^q}{C^r}\) |
| \(n_1 u_1 = n_2 u_2\) | Unit conversion |
| Dimensional formula of Force | \([MLT^{-2}]\) |
| Dimensional formula of Energy | \([ML^2T^{-2}]\) |
Practice Questions
Section A – Multiple Choice Questions (1 mark each)
Q1. The dimensional formula for Planck’s constant \(h\) is:
(a) \([ML^2T^{-1}]\) (b) \([ML^2T^{-2}]\) (c) \([MLT^{-1}]\) (d) \([ML^2T^{-3}]\)
Answer
(a) \([ML^2T^{-1}]\)
Since \(E = h\nu\(, we have \)h = \dfrac{E}{\nu}\).
\([h] = \dfrac{[ML^2T^{-2}]}{[T^{-1}]} = [ML^2T^{-1}]\)
Q2. The number of significant figures in \(0.006020\) is:
(a) 2 (b) 3 (c) 4 (d) 7
Answer
(c) 4 — digits 6, 0, 2, 0 are significant (leading zeros are not; trailing zero after decimal IS significant).
Q3. Which of the following pairs has the same dimensional formula?
(a) Force and Torque (b) Work and Power (c) Impulse and Momentum (d) Stress and Strain
Answer
(c) Impulse and Momentum — both have dimension \([MLT^{-1}]\).
Section B – Short Answer Questions (2 marks each)
Q4. The velocity of a body is given by \(v = At^2 + Bt + C\). Find the dimensions of \(A\), \(B\), and \(C\).
Answer
Since \(v\) has dimensions \([LT^{-1}]\):
- \([C] = [LT^{-1}]\)
- \([Bt] = [LT^{-1}]\) ⟹ \([B] = [LT^{-2}]\)
- \([At^2] = [LT^{-1}]\) ⟹ \([A] = [LT^{-3}]\)
Q5. A physical quantity \(P\) is given by \(P = \dfrac{a^3 b^2}{\sqrt{c},d}\). If the percentage errors in \(a, b, c, d\) are \(1%, 2%, 4%, 2%\) respectively, find the percentage error in \(P\).
Answer
\[ \frac{\Delta P}{P} \times 100 = 3\left(\frac{\Delta a}{a}\right) + 2\left(\frac{\Delta b}{b}\right) + \frac{1}{2}\left(\frac{\Delta c}{c}\right) + \frac{\Delta d}{d} \]
\[ = 3(1) + 2(2) + \frac{1}{2}(4) + 2 = 3 + 4 + 2 + 2 = \mathbf{11%} \]
Section C – Long Answer Questions (5 marks each)
Q6. State and explain the applications of dimensional analysis. Use it to derive the formula for the time period of a simple pendulum.
Answer
Applications of Dimensional Analysis:
- Checking consistency of physical equations.
- Deriving relations between physical quantities.
- Converting units from one system to another.
Derivation of time period \(T\) of simple pendulum:
Let \(T \propto m^a, l^b, g^c\).
Writing dimensions:
\[ [T] = [M]^a [L]^b [LT^{-2}]^c = M^a, L^{b+c}, T^{-2c} \]
Comparing exponents:
- Time: \(1 = -2c \Rightarrow c = -\dfrac{1}{2}\)
- Length: \(0 = b + c \Rightarrow b = \dfrac{1}{2}\)
- Mass: \(0 = a \Rightarrow a = 0\)
\[ \boxed{T = 2\pi\sqrt{\frac{l}{g}}} \]
(The constant \(2\pi\) cannot be determined by dimensional analysis alone.)
Section D – Competency-Based Questions (CBSE Pattern)
Q7. (Case Study) A student uses a vernier calliper to measure the diameter of a rod. The main scale reads \(2.3 \text{ cm}\) and the vernier scale coincidence is at the \(5\text{th}\) division. The least count of the vernier is \(0.01 \text{ cm}\).
(i) What is the diameter of the rod?
(ii) If the student repeats the measurement five times and gets readings: 2.35, 2.34, 2.36, 2.35, 2.35 cm — find the mean absolute error.
(iii) Express the result with appropriate significant figures and error.
Answer
(i) Diameter = Main scale + (VSC × LC) = \(2.3 + 5 \times 0.01 = \mathbf{2.35 \text{ cm}}\)
(ii) Mean = \(\dfrac{2.35 + 2.34 + 2.36 + 2.35 + 2.35}{5} = 2.35 \text{ cm}\)
Absolute errors: \(|0.00|, |0.01|, |0.01|, |0.00|, |0.00|\)
Mean absolute error = \(\dfrac{0 + 0.01 + 0.01 + 0 + 0}{5} = 0.004 \approx 0.01 \text{ cm}\)
(iii) Result: \(d = (2.35 \pm 0.01) \text{ cm}\)
Q8. Two resistances \(R_1 = (100 \pm 3),\Omega\) and \(R_2 = (200 \pm 4),\Omega\) are connected:
(i) In series — find the resultant resistance with error.
(ii) In parallel — find the resultant resistance with error.
Answer
(i) Series: \(R_s = R_1 + R_2 = 300,\Omega\), \(\Delta R_s = \Delta R_1 + \Delta R_2 = 7,\Omega\)
\(\boxed{R_s = (300 \pm 7),\Omega}\)
(ii) Parallel: \(R_p = \dfrac{R_1 R_2}{R_1 + R_2} = \dfrac{100 \times 200}{300} = 66.7,\Omega\)
\[ \frac{\Delta R_p}{R_p^2} = \frac{\Delta R_1}{R_1^2} + \frac{\Delta R_2}{R_2^2} \]
\[ \Delta R_p = R_p^2\left(\frac{3}{10000} + \frac{4}{40000}\right) = 66.7^2 \times (0.0003 + 0.0001) = 4449 \times 0.0004 \approx 1.8,\Omega \]
\(\boxed{R_p = (66.7 \pm 1.8),\Omega}\)