Units And Measurement | Class 11 Physics Notes
By ConceptScroll Team · Published on 17 July 2026 · 4 min read
Units And Measurement – this guide gives you a concise, exam-ready overview of Units And Measurement from Class 11 Physics, written by ConceptScroll editors and reviewed against the latest NCERT textbook.
1.5 Dimensional formulae and dimensional equations
The dimensional formula of a physical quantity expresses how the quantity relates to the base quantities through powers of their dimensions. For example, volume has dimensional formula [M⁰ L³ T⁰], indicating it depends on length cubed and is independent of mass and time. Dimensional equations equate a physical quantity to its dimensional formula, representing the dimensions on both sides. For instance, the dimensional equation of force is [F] = [M][L][T⁻²]. Dimensional formulae can be derived from the physical relations between quantities. They are essential tools for verifying the dimensional consistency of equations, ensuring that both sides have the same dimensions. Appendix 9 in the NCERT textbook lists dimensional formulae for various physical quantities. Understanding dimensional formulae helps in checking equations, deriving relations among quantities, and converting units correctly.
📊 Diagram: No specific diagram in this section.
🧪 Activity: No specific activity in this section.
🔗 Connection: Introduces dimensional analysis and its applications in the next section.
Frequently asked questions
1.1 Fill in the blanks (a) The volume of a cube of side $1\mathrm{cm}$ is equal to $\dots \mathrm{m}^3$ (b) The surface area of a solid cylinder of radius 2.0 cm and height 10.0 cm is equal to $\dots (\mathrm{mm})^2$ (c) A vehicle moving with a speed of $18\mathrm{km}\mathrm{h}^{-1}$ covers...m in 1 s (d) The relative density of lead is 11.3. Its density is ...g cm$^{-3}$ or ...kg m$^{-3}$.
(a) Volume of cube = side^3 = (1 cm)^3 = (1 × 10^{-2} m)^3 = 1 × 10^{-6} m^3
(b) Surface area of cylinder = 2\pi r (r + h) = 2 × 3.1416 × 2 cm × (2 cm + 10 cm) = 2 × 3.1416 × 2 × 12 = 150.8 cm^2 Convert to mm^2: 1 cm^2 = 100 mm^2 So, 150.8 cm^2 = 150.8 × 100 = 15080 mm^2
(c) Speed = 18 km/h = 18 × (1000 m / 3600 s) = 5 m/s Distance covered in 1 s = 5 m
(d) Relative density (RD) = density of lead / density of water Density of water = 1 g/cm^3 So, density of lead = RD × density of water = 11.3
1.2 Fill in the blanks by suitable conversion of units (a) $1\mathrm{kg}\mathrm{m}^2\mathrm{s}^{-2} = \dots \mathrm{g}\mathrm{cm}^2\mathrm{s}^{-2}$ (b) $1\mathrm{m} = \dots \mathrm{ly}$ (c) $3.0\mathrm{ms}^{-2} = \dots \mathrm{km}\mathrm{h}^{-2}$ (d) $G = 6.67 \times 10^{-11} \mathrm{~N} \mathrm{~m}^{2} (\mathrm{kg})^{-2} = \dots (\mathrm{cm})^{3} \mathrm{s}^{-2} \mathrm{g}^{-1}$.
(a) 1 kg m^2 s^{-2} = 1 × (10^3 g) × (10^2 cm)^2 × s^{-2} = 1 × 10^7 g cm^2 s^{-2}
(b) 1 m = 1 / (9.46 × 10^{15}) ly ≈ 1.057 × 10^{-16} ly
(c) 3.0 m s^{-2} = 3.0 × (3600)^2 km h^{-2} = 3.0 × 12960000 = 3.888 × 10^{7} km h^{-2}
(d) G = 6.67 × 10^{-11} N m^2 kg^{-2} 1 N = 1 kg m s^{-2} Convert units: 1 m = 100 cm, 1 kg = 1000 g So, G = 6.67 × 10^{-11} × (100)^3 cm^3 s^{-2} (1000)^{-1} g^{-1} = 6.67 × 10^{-8} cm^3 s^{-2} g^{-1}
1.3 A calorie is a unit of heat (energy in transit) and it equals about $4.2 \, \text{J}$ where $1 \, \text{J} = 1 \, \text{kg} \, \text{m}^2 \, \text{s}^{-2}$. Suppose we employ a system of units in which the unit of mass equals $\alpha$ kg, the unit of length equals $\beta \, \text{m}$, the unit of time is $\gamma \, \text{s}$. Show that a calorie has a magnitude $4.2 \, \alpha^{-1} \beta^{-2} \gamma^2$ in terms of the new units.
Given: 1 calorie = 4.2 J 1 J = 1 kg m^2 s^{-2} In new units: Mass unit = α kg Length unit = β m Time unit = γ s
Express 1 J in new units: 1 J = 1 kg m^2 s^{-2} = (1/α) mass units × (1/β)^2 length units^2 × (1/γ)^(-2) time units^{-2} = (1/α) × (1/β^2) × (γ^2) in new units
Therefore, 1 calorie = 4.2 J = 4.2 × α^{-1} × β^{-2} × γ^{2} in new units.
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.
The statement means that 'large' or 'small' are relative terms and only meaningful when compared to a reference or standard.
Reframed statements: (a) Atoms are very small compared to everyday objects. (b) A jet plane moves with great speed compared to a car. (c) The mass of Jupiter is very large compared to Earth. (d) The air inside this room contains a large number of molecules compared to a vacuum. (e) A proton is much more massive than an electron (mass of proton is about 1836 times that of
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