Table of Contents
Physical Quantities
Physical quantities are quantities that can be measured and expressed in numerical values with appropriate units. They are broadly classified into two categories:
Fundamental Quantities
Fundamental quantities are independent quantities that cannot be expressed in terms of other physical quantities. In the SI system, there are seven fundamental quantities:
| Physical Quantity | SI Unit | Symbol |
|---|---|---|
| Length | meter | m |
| Mass | kilogram | kg |
| Time | second | s |
| Electric Current | ampere | A |
| Temperature | kelvin | K |
| Amount of Substance | mole | mol |
| Luminous Intensity | candela | cd |
Derived Quantities
Derived quantities are those which can be expressed in terms of fundamental quantities. Examples:
- Area: \([L^2]\)
- Volume: \([L^3]\)
- Velocity: \([LT^{-1}]\)
- Acceleration: \([LT^{-2}]\)
- Force: \([MLT^{-2}]\)
- Energy: \([ML^2T^{-2}]\)
Systems of Units
SI System (International System of Units)
The SI system is the most widely used system globally and is based on seven fundamental units mentioned above. Some important SI derived units include:
- Force: Newton (N) = \(kg \cdot m \cdot s^{-2}\)
- Energy: Joule (J) = \(kg \cdot m^2 \cdot s^{-2}\)
- Power: Watt (W) = \(kg \cdot m^2 \cdot s^{-3}\)
- Pressure: Pascal (Pa) = \(kg \cdot m^{-1} \cdot s^{-2}\)
Important Prefixes
| Prefix | Symbol | Power of 10 |
|---|---|---|
| tera | T | \(10^{12}\) |
| giga | G | \(10^9\) |
| mega | M | \(10^6\) |
| kilo | k | \(10^3\) |
| centi | c | \(10^{-2}\) |
| milli | m | \(10^{-3}\) |
| micro | μ | \(10^{-6}\) |
| nano | n | \(10^{-9}\) |
| pico | p | \(10^{-12}\) |
Dimensional Analysis
Dimensions of a physical quantity are the powers to which the fundamental units must be raised to represent that quantity. The dimensional formula is expressed in terms of \([M]\), \([L]\), \([T]\), \([I]\), \([K]\), where M = Mass, L = Length, T = Time, I = Current, K = Temperature.
Dimensional Formulae of Common Physical Quantities
| Physical Quantity | Formula | Dimensional Formula |
|---|---|---|
| Velocity | \(v = \frac{s}{t}\) | \([M^0L^1T^{-1}]\) |
| Acceleration | \(a = \frac{v}{t}\) | \([M^0L^1T^{-2}]\) |
| Force | \(F = ma\) | \([M^1L^1T^{-2}]\) |
| Momentum | \(p = mv\) | \([M^1L^1T^{-1}]\) |
| Work/Energy | \(W = F \cdot s\) | \([M^1L^2T^{-2}]\) |
| Power | \(P = \frac{W}{t}\) | \([M^1L^2T^{-3}]\) |
| Pressure | \(P = \frac{F}{A}\) | \([M^1L^{-1}T^{-2}]\) |
| Density | \(\rho = \frac{m}{V}\) | \([M^1L^{-3}T^0]\) |
| Surface Tension | \(\gamma = \frac{F}{l}\) | \([M^1L^0T^{-2}]\) |
| Coefficient of Viscosity | \(\eta\) | \([M^1L^{-1}T^{-1}]\) |
| Planck’s Constant | \(h = \frac{E}{\nu}\) | \([M^1L^2T^{-1}]\) |
| Gravitational Constant | \(G = \frac{Fr^2}{m_1m_2}\) | \([M^{-1}L^3T^{-2}]\) |
Principle of Homogeneity of Dimensions
The principle states that the dimensions of all terms in a physical equation must be the same. This principle is used to check the correctness of physical equations.
Applications of Dimensional Analysis
a) To check the dimensional correctness of equations
b) To derive relations between physical quantities
c) To convert units from one system to another
Solved Example 1: Checking Dimensional Correctness
Problem: Check whether the equation \(v^2 = u^2 + 2as\) is dimensionally correct.
Solution:
LHS: Dimensions of \(v^2 = [LT^{-1}]^2 = [L^2T^{-2}]\)
RHS: Dimensions of \(u^2 = [LT^{-1}]^2 = [L^2T^{-2}]\)
Dimensions of \(2as = [LT^{-2}][L] = [L^2T^{-2}]\)
Note: The numerical constant ‘2’ has no dimensions.
Since dimensions of LHS = dimensions of each term on RHS = \([L^2T^{-2}]\)
Answer: The equation is dimensionally correct.
Solved Example 2: Deriving Relations Using Dimensions
Problem: The time period \(T\) of a simple pendulum depends on its length \(l\) and acceleration due to gravity \(g\). Derive the relation using dimensional analysis.
Solution:
Let \(T \propto l^a g^b\)
Therefore, \(T = k l^a g^b\) where \(k\) is a dimensionless constant.
Writing dimensions:
\([M^0L^0T^1] = [L]^a [LT^{-2}]^b\)
\([M^0L^0T^1] = [L^{a+b}T^{-2b}]\)
Comparing powers:
For M: \(0 = 0\) ✓
For L: \(0 = a + b\) … (1)
For T: \(1 = -2b\) … (2)
From equation (2): \(b = -\frac{1}{2}\)
From equation (1): \(a = -b = \frac{1}{2}\)
Therefore, \(T = k l^{1/2} g^{-1/2} = k\sqrt{\frac{l}{g}}\)
Answer: \(T = k\sqrt{\frac{l}{g}}\) (where \(k = 2\pi\) from exact derivation)
Solved Example 3: Unit Conversion
Problem: Convert the value of gravitational constant \(G = 6.67 \times 10^{-11}\) N·m²·kg⁻² to CGS system.
Solution:
Dimensional formula of \(G = [M^{-1}L^3T^{-2}]\)
Using the conversion formula:
\(n_2 = n_1 \left[\frac{M_1}{M_2}\right]^{-1} \left[\frac{L_1}{L_2}\right]^3 \left[\frac{T_1}{T_2}\right]^{-2}\)
Where:
- \(n_1 = 6.67 \times 10^{-11}\) (SI)
- \(M_1 = 1\) kg, \(M_2 = 1\) g = \(10^{-3}\) kg
- \(L_1 = 1\) m, \(L_2 = 1\) cm = \(10^{-2}\) m
- \(T_1 = 1\) s, \(T_2 = 1\) s
\(n_2 = 6.67 \times 10^{-11} \times \left[\frac{1}{10^{-3}}\right]^{-1} \times \left[\frac{1}{10^{-2}}\right]^3 \times [1]^{-2}\)
\(n_2 = 6.67 \times 10^{-11} \times 10^{-3} \times 10^{6}\)
\(n_2 = 6.67 \times 10^{-8}\) dyne·cm²·g⁻²
Answer: \(G = 6.67 \times 10^{-8}\) dyne·cm²·g⁻² in CGS system
Limitations of Dimensional Analysis
- Cannot determine dimensionless constants
- Cannot be used for equations involving trigonometric, exponential, or logarithmic functions
- Cannot distinguish between physical quantities having the same dimensions (e.g., work and torque)
- Fails for equations containing more than one term with different dimensional structures
Significant Figures
Significant figures are the meaningful digits in a measured or calculated quantity. They indicate the precision of measurement.
Rules for Significant Figures
- Non-zero digits: All non-zero digits are significant. Example: 253 has 3 significant figures.
- Zeros between non-zero digits: All zeros between non-zero digits are significant. Example: 2005 has 4 significant figures.
- Leading zeros: Leading zeros (zeros to the left of the first non-zero digit) are NOT significant. Example: 0.0025 has 2 significant figures.
- Trailing zeros: Trailing zeros in a number with a decimal point are significant. Example: 2.500 has 4 significant figures.
- Trailing zeros without decimal: Trailing zeros in a whole number without a decimal point are ambiguous. Example: 2500 may have 2, 3, or 4 significant figures (use scientific notation to clarify).
Operations with Significant Figures
Addition and Subtraction: The result should have the same number of decimal places as the measurement with the fewest decimal places.
Multiplication and Division: The result should have the same number of significant figures as the measurement with the fewest significant figures.
Solved Example 4: Significant Figures in Calculations
Problem: Calculate the area of a rectangular plate with length \(l = 12.5\) cm and width \(w = 8.32\) cm. Express the answer with the correct number of significant figures.
Solution:
Area = \(l \times w = 12.5 \times 8.32\)
Area = \(104.0\) cm²
Number of significant figures:
- \(l = 12.5\) has 3 significant figures
- \(w = 8.32\) has 3 significant figures
Since this is multiplication, the result should have 3 significant figures (minimum of the two).
Answer: Area = \(104\) cm² (3 significant figures)
Solved Example 5: Addition with Significant Figures
Problem: Add the following measurements: 25.4 cm + 3.26 cm + 0.182 cm
Solution:
25.4 cm (1 decimal place)
+ 3.26 cm (2 decimal places)
+ 0.182 cm (3 decimal places)
= 28.842 cm
The result should be rounded to 1 decimal place (the least precise measurement).
Answer: 28.8 cm
Errors in Measurement
No measurement is perfectly accurate. Every measurement has some uncertainty called error.
Types of Errors
a) Systematic Errors: These errors occur due to faulty equipment or experimental technique. They can be reduced by calibration. Types include:
- Instrumental errors
- Environmental errors
- Observational errors
- Personal errors
b) Random Errors: These occur irregularly and are variable in magnitude and sign. They can be reduced by taking multiple measurements and calculating the mean.
c) Gross Errors: These arise due to human carelessness or mistakes in reading instruments.
Absolute Error, Relative Error, and Percentage Error
If \(a_1, a_2, a_3, …, a_n\) are measured values of a quantity, then:
Mean value: \(a_{mean} = \frac{a_1 + a_2 + a_3 + … + a_n}{n}\)
Absolute error: \(\Delta a_i = |a_{mean} – a_i|\)
Mean absolute error: \(\Delta a_{mean} = \frac{\Delta a_1 + \Delta a_2 + … + \Delta a_n}{n}\)
Relative error: \(\frac{\Delta a_{mean}}{a_{mean}}\)
Percentage error: \(\frac{\Delta a_{mean}}{a_{mean}} \times 100\%\)
Combination of Errors
a) Error in Sum or Difference:
If \(Z = A + B\) or \(Z = A – B\), then:
\(\Delta Z = \Delta A + \Delta B\)
b) Error in Product:
If \(Z = A \times B\), then:
\(\frac{\Delta Z}{Z} = \frac{\Delta A}{A} + \frac{\Delta B}{B}\)
c) Error in Division:
If \(Z = \frac{A}{B}\), then:
\(\frac{\Delta Z}{Z} = \frac{\Delta A}{A} + \frac{\Delta B}{B}\)
d) Error in Power:
If \(Z = A^n\), then:
\(\frac{\Delta Z}{Z} = n\frac{\Delta A}{A}\)
e) General form:
If \(Z = \frac{A^p B^q}{C^r}\), then:
\(\frac{\Delta Z}{Z} = p\frac{\Delta A}{A} + q\frac{\Delta B}{B} + r\frac{\Delta C}{C}\)
Solved Example 6: Error Calculation
Problem: The period of oscillation of a simple pendulum is measured as 2.50 s, 2.48 s, 2.49 s, 2.51 s, and 2.52 s in five successive measurements. Calculate:
- Mean value of period
- Mean absolute error
- Relative error
- Percentage error
Solution:
a) Mean value:
\(T_{mean} = \frac{2.50 + 2.48 + 2.49 + 2.51 + 2.52}{5} = \frac{12.50}{5} = 2.50\) s
b) Absolute errors:
- \(\Delta T_1 = |2.50 – 2.50| = 0.00\) s
- \(\Delta T_2 = |2.50 – 2.48| = 0.02\) s
- \(\Delta T_3 = |2.50 – 2.49| = 0.01\) s
- \(\Delta T_4 = |2.50 – 2.51| = 0.01\) s
- \(\Delta T_5 = |2.50 – 2.52| = 0.02\) s
Mean absolute error:
\(\Delta T_{mean} = \frac{0.00 + 0.02 + 0.01 + 0.01 + 0.02}{5} = \frac{0.06}{5} = 0.012\) s
c) Relative error:
\(\frac{\Delta T_{mean}}{T_{mean}} = \frac{0.012}{2.50} = 0.0048\)
d) Percentage error:
\(= 0.0048 \times 100\% = 0.48\%\)
Final Answer: \(T = (2.50 \pm 0.01)\) s with a percentage error of 0.48%
Solved Example 7: Combination of Errors
Problem: The resistance \(R = \frac{V}{I}\) where \(V = (100 \pm 5)\) V and \(I = (10 \pm 0.2)\) A. Calculate the percentage error in \(R\).
Solution:
Given:
- \(V = 100\) V, \(\Delta V = 5\) V
- \(I = 10\) A, \(\Delta I = 0.2\) A
Since \(R = \frac{V}{I}\), the relative error is:
\(\frac{\Delta R}{R} = \frac{\Delta V}{V} + \frac{\Delta I}{I}\)
\(\frac{\Delta R}{R} = \frac{5}{100} + \frac{0.2}{10}\)
\(\frac{\Delta R}{R} = 0.05 + 0.02 = 0.07\)
Percentage error:
\(= 0.07 \times 100\% = 7\%\)
Also, \(R = \frac{100}{10} = 10\) Ω
\(\Delta R = 0.07 \times 10 = 0.7\) Ω
Answer: \(R = (10 \pm 0.7)\) Ω with a percentage error of 7%
Solved Example 8: Error in Power
Problem: The density of a sphere is measured using the formula \(\rho = \frac{M}{\frac{4}{3}\pi r^3}\). If the percentage error in measuring mass \(M\) is 2% and in measuring radius \(r\) is 1%, find the percentage error in density.
Solution:
The formula can be written as: \(\rho = \frac{3M}{4\pi r^3}\)
Or: \(\rho = M \cdot r^{-3}\) (ignoring constants)
Using the error formula for \(\rho = M^1 \cdot r^{-3}\):
\(\frac{\Delta \rho}{\rho} = 1 \cdot \frac{\Delta M}{M} + |-3| \cdot \frac{\Delta r}{r}\)
\(\frac{\Delta \rho}{\rho} = \frac{\Delta M}{M} + 3\frac{\Delta r}{r}\)
Given:
- Percentage error in \(M = 2\%\), so \(\frac{\Delta M}{M} = 0.02\)
- Percentage error in \(r = 1\%\), so \(\frac{\Delta r}{r} = 0.01\)
\(\frac{\Delta \rho}{\rho} = 0.02 + 3(0.01) = 0.02 + 0.03 = 0.05\)
Percentage error in density: \(= 0.05 \times 100\% = 5\%\)
Answer: The percentage error in density is 5%
Precision and Accuracy
Accuracy: How close a measured value is to the true value. It is related to systematic errors.
Precision: How close multiple measurements are to each other. It is related to random errors.
A measurement can be precise without being accurate, and vice versa. Ideally, we want measurements that are both accurate and precise.
Important Points for JEE
Least Count
The least count is the smallest measurement that can be made accurately with a given measuring instrument.
- Vernier Callipers: LC = 1 MSD – 1 VSD = \(\frac{\text{Value of 1 MSD}}{\text{Total number of VSD}}\)
- Screw Gauge: LC = \(\frac{\text{Pitch}}{\text{Total number of divisions on circular scale}}\)
Dimensional Constants vs Dimensionless Constants
Dimensional constants: Have dimensions (e.g., gravitational constant \(G\), Planck’s constant \(h\), Boltzmann constant \(k\))
Dimensionless constants: Have no dimensions (e.g., \(\pi\), \(e\), trigonometric ratios, pure numbers)
Solved Example 9: Vernier Callipers
Problem: In a vernier callipers, 1 cm on the main scale is divided into 20 equal parts. 10 divisions of vernier scale coincide with 9 divisions of the main scale. Find the least count.
Solution:
1 MSD (Main Scale Division) = \(\frac{1}{20}\) cm = 0.05 cm
10 VSD = 9 MSD
1 VSD = \(\frac{9}{10}\) MSD = \(\frac{9}{10} \times 0.05 = 0.045\) cm
Least Count = 1 MSD – 1 VSD
LC = \(0.05 – 0.045 = 0.005\) cm = 0.05 mm
Answer: Least count = 0.005 cm or 0.05 mm
Solved Example 10: Screw Gauge
Problem: A screw gauge has a pitch of 0.5 mm and 50 divisions on its circular scale. What is its least count?
Solution:
Least Count = \(\frac{\text{Pitch}}{\text{Number of divisions on circular scale}}\)
LC = \(\frac{0.5}{50} = 0.01\) mm
Answer: Least count = 0.01 mm
Important Formulas Summary
| Concept | Formula |
|---|---|
| Mean absolute error | \(\Delta a_{mean} = \frac{\sum \Delta a_i}{n}\) |
| Relative error | \(\frac{\Delta a}{a}\) |
| Percentage error | \(\frac{\Delta a}{a} \times 100\%\) |
| Error in sum/difference | \(\Delta Z = \Delta A + \Delta B\) |
| Error in product/quotient | \(\frac{\Delta Z}{Z} = \frac{\Delta A}{A} + \frac{\Delta B}{B}\) |
| Error in power | \(\frac{\Delta Z}{Z} = n\frac{\Delta A}{A}\) for \(Z = A^n\) |
| Vernier Callipers LC | 1 MSD – 1 VSD |
| Screw Gauge LC | \(\frac{\text{Pitch}}{\text{No. of circular divisions}}\) |
Practice Problems
Problem 1:
If the velocity \(v\) of a particle depends on time \(t\) according to the equation \(v = a + bt + \frac{c}{d+t}\), find the dimensions of \(a\), \(b\), \(c\), and \(d\).
Problem 2:
The length and breadth of a rectangle are \((5.7 \pm 0.1)\) cm and \((3.4 \pm 0.1)\) cm respectively. Calculate the area with error limits.
Problem 3:
In an experiment, the percentage errors in the measurements of mass and speed are 2% and 3% respectively. Calculate the maximum percentage error in the kinetic energy.
Problem 4:
Using dimensional analysis, show that the expression \(T = 2\pi\sqrt{\frac{l}{g}}\) for the time period of a simple pendulum is correct dimensionally.