AS Physics - SI Units Study Notes - New Syllabus-2025-2027
AS Physics – SI Units Study Notes
AS Physics – SI Units Study Notes at IITian Academy focus on specific topic and type of questions asked in actual exam. Study Notes focus on AS Physics Study Notes syllabus Candidates should be able to:
- recall the following SI base quantities and their units: mass (kg), length (m), time (s), current (A), temperature (K)
- express derived units as products or quotients of the SI base units and use the derived units for quantities listed in this syllabus as appropriate
- use SI base units to check the homogeneity of physical equations
- recall and use the following prefixes and their symbols to indicate decimal submultiples or multiples of both base and derived units: pico (p), nano (n), micro (μ), milli (m), centi (c), deci (d), kilo (k), mega (M), giga (G), tera (T)
SI Base Quantities and Their Units
The SI base quantities are the fundamental physical quantities on which all other derived quantities are defined. Each base quantity has a unique and independent SI unit that serves as the standard for measurement.
The Seven SI Base Quantities:
| Base Quantity | Symbol | SI Unit | Unit Symbol |
|---|---|---|---|
| Mass | \( \mathrm{m} \) | kilogram | \( \mathrm{kg} \) |
| Length | \( \mathrm{l} \) | metre | \( \mathrm{m} \) |
| Time | \( \mathrm{t} \) | second | \( \mathrm{s} \) |
| Electric current | \( \mathrm{I} \) | ampere | \( \mathrm{A} \) |
| Thermodynamic temperature | \( \mathrm{T} \) | kelvin | \( \mathrm{K} \) |
| Amount of substance | \( \mathrm{n} \) | mole | \( \mathrm{mol} \) |
| Luminous intensity | \( \mathrm{I_v} \) | candela | \( \mathrm{cd} \) |
All measurable physical quantities in physics can be expressed in terms of these seven base quantities. For example, force, energy, and pressure are all derived quantities formed from combinations of base units.
Example
Show that the SI unit of force (newton) can be expressed in terms of the base units.
▶️ Answer / Explanation
Step 1: By definition, \( \mathrm{F = ma} \).
\( \mathrm{[F] = [m][a]} \)
Step 2: Substitute base units for each quantity:
\( \mathrm{[m] = kg,\quad [a] = m/s^2} \)
Step 3: Therefore,
\( \mathrm{[F] = kg\,m/s^2} \)
Result: The newton (N) is a derived unit expressed in base units as \( \mathrm{kg\,m/s^2} \).
Derived Quantities and Their Units
Derived quantities are physical quantities that are defined in terms of the SI base quantities through mathematical relationships such as products or quotients.
Their corresponding derived units are combinations of the SI base units expressed as products or quotients (multiplication or division).
General Form:
\( \mathrm{Derived\ unit = (Base\ units)^{n_1}\times(Base\ units)^{n_2}\times\dots} \)
Each derived quantity can thus be expressed entirely in base units such as \( \mathrm{kg} \), \( \mathrm{m} \), \( \mathrm{s} \), \( \mathrm{A} \), \( \mathrm{K} \), \( \mathrm{mol} \), and \( \mathrm{cd} \).
All physical quantities — whether mechanical, electrical, or thermal — can be represented as combinations of base units. Derived units ensure consistency and dimensional coherence in all physical equations.
Common Derived Quantities and Their SI Units:
| Derived Quantity | Formula Definition | SI Unit Name | Derived Unit Symbol | In Base SI Units |
|---|---|---|---|---|
| Area | \( \mathrm{A = l \times b} \) | square metre | \( \mathrm{m^2} \) | \( \mathrm{m^2} \) |
| Volume | \( \mathrm{V = l \times b \times h} \) | cubic metre | \( \mathrm{m^3} \) | \( \mathrm{m^3} \) |
| Velocity | \( \mathrm{v = \dfrac{d}{t}} \) | metre per second | \( \mathrm{m/s} \) | \( \mathrm{m\,s^{-1}} \) |
| Acceleration | \( \mathrm{a = \dfrac{v}{t}} \) | metre per second squared | \( \mathrm{m/s^2} \) | \( \mathrm{m\,s^{-2}} \) |
| Force | \( \mathrm{F = m a} \) | newton | \( \mathrm{N} \) | \( \mathrm{kg\,m\,s^{-2}} \) |
| Pressure | \( \mathrm{p = \dfrac{F}{A}} \) | pascal | \( \mathrm{Pa} \) | \( \mathrm{kg\,m^{-1}\,s^{-2}} \) |
| Work / Energy | \( \mathrm{W = Fd} \) | joule | \( \mathrm{J} \) | \( \mathrm{kg\,m^2\,s^{-2}} \) |
| Power | \( \mathrm{P = \dfrac{W}{t}} \) | watt | \( \mathrm{W} \) | \( \mathrm{kg\,m^2\,s^{-3}} \) |
| Momentum | \( \mathrm{p = mv} \) | — | \( \mathrm{kg\,m/s} \) | \( \mathrm{kg\,m\,s^{-1}} \) |
| Charge | \( \mathrm{Q = I t} \) | coulomb | \( \mathrm{C} \) | \( \mathrm{A\,s} \) |
| Potential difference | \( \mathrm{V = \dfrac{W}{Q}} \) | volt | \( \mathrm{V} \) | \( \mathrm{kg\,m^2\,s^{-3}\,A^{-1}} \) |
| Resistance | \( \mathrm{R = \dfrac{V}{I}} \) | ohm | \( \mathrm{\Omega} \) | \( \mathrm{kg\,m^2\,s^{-3}\,A^{-2}} \) |
| Capacitance | \( \mathrm{C = \dfrac{Q}{V}} \) | farad | \( \mathrm{F} \) | \( \mathrm{A^2\,s^4\,kg^{-1}\,m^{-2}} \) |
| Frequency | \( \mathrm{f = \dfrac{1}{T}} \) | hertz | \( \mathrm{Hz} \) | \( \mathrm{s^{-1}} \) |
| Density | \( \mathrm{\rho = \dfrac{m}{V}} \) | — | \( \mathrm{kg/m^3} \) | \( \mathrm{kg\,m^{-3}} \) |
| Specific heat capacity | \( \mathrm{c = \dfrac{Q}{m\Delta T}} \) | — | \( \mathrm{J/(kg\,K)} \) | \( \mathrm{m^2\,s^{-2}\,K^{-1}} \) |
| Specific latent heat | \( \mathrm{L = \dfrac{Q}{m}} \) | — | \( \mathrm{J/kg} \) | \( \mathrm{m^2\,s^{-2}} \) |
| Magnetic flux density | \( \mathrm{B = \dfrac{F}{I l}} \) | tesla | \( \mathrm{T} \) | \( \mathrm{kg\,s^{-2}\,A^{-1}} \) |
Example
Express the unit of power (watt) in terms of the SI base units.
▶️ Answer / Explanation
Step 1: Start from the definition \( \mathrm{P = \dfrac{W}{t}} \).
\( \mathrm{[P] = [W][t]^{-1}} \)
Step 2: Substitute the unit of work \( \mathrm{W = Fd = kg\,m^2/s^2} \).
\( \mathrm{[P] = kg\,m^2\,s^{-3}} \)
Result: The watt (W) is a derived SI unit expressed in base units as \( \mathrm{1\,W = 1\,kg\,m^2\,s^{-3}} \).
Using SI Base Units to Check the Homogeneity of Physical Equations
The principle of homogeneity states that in any physically meaningful equation, all terms must have the same dimensions and hence the same SI base units. This ensures the equation is dimensionally consistent and can represent a real physical relationship.
The base units of every term on both sides of an equation must match. An equation that fails this test cannot be correct, no matter what numerical values are used.
SI Base Quantities and Their Dimensions:
| Physical Quantity | Symbol | SI Base Unit | Dimensional Formula |
|---|---|---|---|
| Mass | \( \mathrm{m} \) | \( \mathrm{kg} \) | \( \mathrm{[M]} \) |
| Length | \( \mathrm{l} \) | \( \mathrm{m} \) | \( \mathrm{[L]} \) |
| Time | \( \mathrm{t} \) | \( \mathrm{s} \) | \( \mathrm{[T]} \) |
| Electric current | \( \mathrm{I} \) | \( \mathrm{A} \) | \( \mathrm{[I]} \) |
| Temperature | \( \mathrm{T} \) | \( \mathrm{K} \) | \( \mathrm{[\Theta]} \) |
Steps to Check Homogeneity:
- Write down the given physical equation.
- Replace each quantity by its base SI unit or dimensional formula.
- Simplify both sides to express in terms of \( \mathrm{M, L, T, I, \Theta} \).
- If both sides are dimensionally the same, the equation is homogeneous.
Applications of Dimensional Homogeneity:
- To verify the correctness of a physical equation.
- To derive relationships between quantities (dimensional analysis).
- To convert quantities between different systems of units.
Limitations:
- Homogeneity cannot verify numerical constants (like ½, π, etc.).
- It cannot detect missing dimensionless factors.
- It only confirms consistency, not accuracy.
Example
Verify the dimensional homogeneity of the equation \( \mathrm{s = ut + \dfrac{1}{2} a t^2} \).
▶️ Answer / Explanation
Step 1: Write down the dimensions of each term.
| Quantity | Symbol | Dimensional Formula |
|---|---|---|
| Displacement | \( \mathrm{s} \) | \( \mathrm{[L]} \) |
| Velocity | \( \mathrm{u} \) | \( \mathrm{[L\,T^{-1}]} \) |
| Acceleration | \( \mathrm{a} \) | \( \mathrm{[L\,T^{-2}]} \) |
| Time | \( \mathrm{t} \) | \( \mathrm{[T]} \) |
Step 2: Check each term:
\( \mathrm{[u t] = [L\,T^{-1}][T] = [L]} \)
\( \mathrm{[a t^2] = [L\,T^{-2}][T^2] = [L]} \)
Step 3: Compare both sides.
LHS: \( \mathrm{[L]} \), RHS: \( \mathrm{[L]} \)
Result: The equation is dimensionally homogeneous — hence physically consistent.
Example
Check if \( \mathrm{v = u + a t^2} \) is homogeneous.
▶️ Answer / Explanation
Step 1: Write the dimensions of each term.
\( \mathrm{[v] = [L\,T^{-1}]} \)
\( \mathrm{[u] = [L\,T^{-1}]} \)
\( \mathrm{[a t^2] = [L\,T^{-2}][T^2] = [L]} \)
Step 2: Compare both sides.
LHS: \( \mathrm{[L\,T^{-1}]} \), RHS: \( \mathrm{[L]} \)
Result: The equation is not dimensionally homogeneous therefore, it is physically invalid.
SI Prefixes and Their Symbols
Prefixes are added to SI units to represent decimal multiples or submultiples of both base and derived units. They provide a convenient way to express very large or very small quantities without using cumbersome powers of ten.
Key Idea: Prefixes simplify the expression of physical quantities by scaling the unit appropriately — for example, \( \mathrm{1\,km = 10^3\,m} \), \( \mathrm{1\,ms = 10^{-3}\,s} \).
Common SI Prefixes Used in A Level Physics:
| Prefix Name | Symbol | Multiplying Factor | Equivalent Power of Ten | Example |
|---|---|---|---|---|
| tera | \( \mathrm{T} \) | \( \mathrm{1\,000\,000\,000\,000} \) | \( \mathrm{10^{12}} \) | \( \mathrm{1\,TW = 10^{12}\,W} \) |
| giga | \( \mathrm{G} \) | \( \mathrm{1\,000\,000\,000} \) | \( \mathrm{10^9} \) | \( \mathrm{1\,GHz = 10^9\,Hz} \) |
| mega | \( \mathrm{M} \) | \( \mathrm{1\,000\,000} \) | \( \mathrm{10^6} \) | \( \mathrm{1\,MJ = 10^6\,J} \) |
| kilo | \( \mathrm{k} \) | \( \mathrm{1\,000} \) | \( \mathrm{10^3} \) | \( \mathrm{1\,km = 10^3\,m} \) |
| deci | \( \mathrm{d} \) | \( \mathrm{0.1} \) | \( \mathrm{10^{-1}} \) | \( \mathrm{1\,dm = 10^{-1}\,m} \) |
| centi | \( \mathrm{c} \) | \( \mathrm{0.01} \) | \( \mathrm{10^{-2}} \) | \( \mathrm{1\,cm = 10^{-2}\,m} \) |
| milli | \( \mathrm{m} \) | \( \mathrm{0.001} \) | \( \mathrm{10^{-3}} \) | \( \mathrm{1\,ms = 10^{-3}\,s} \) |
| micro | \( \mathrm{\mu} \) | \( \mathrm{0.000001} \) | \( \mathrm{10^{-6}} \) | \( \mathrm{1\,\mu F = 10^{-6}\,F} \) |
| nano | \( \mathrm{n} \) | \( \mathrm{0.000000001} \) | \( \mathrm{10^{-9}} \) | \( \mathrm{1\,nm = 10^{-9}\,m} \) |
| pico | \( \mathrm{p} \) | \( \mathrm{0.000000000001} \) | \( \mathrm{10^{-12}} \) | \( \mathrm{1\,pF = 10^{-12}\,F} \) |
Tips for Using SI Prefixes:
- Prefixes multiply the base unit never change the physical dimension of the quantity.
- Always write the prefix immediately before the unit symbol (e.g., km, not k m).
- Avoid mixing prefixes in a single expression — use one consistent prefix for clarity.
Example
Express \( \mathrm{2.5\,km} \) in metres and \( \mathrm{4.0\,\mu s} \) in seconds.
▶️ Answer / Explanation
Step 1: Convert kilometres to metres.
\( \mathrm{1\,km = 10^3\,m} \Rightarrow 2.5\,km = 2.5\times10^3\,m} \)
Step 2: Convert microseconds to seconds.
\( \mathrm{1\,\mu s = 10^{-6}\,s} \Rightarrow 4.0\,\mu s = 4.0\times10^{-6}\,s} \)
Result: \( \mathrm{2.5\,km = 2.5\times10^3\,m} \), \( \mathrm{4.0\,\mu s = 4.0\times10^{-6}\,s} \).