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PhysicsClass 11Units And Measurement

Units And Measurement | Class 11 Physics Notes

By ConceptScroll Team · Published on 17 July 2026 · 5 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.

1.3 Significant figures

Measurements inherently involve errors and uncertainties. To convey the precision of a measurement, results are reported with significant figures, which include all reliably known digits plus the first uncertain digit. The number of significant figures reflects the precision of the measuring instrument and the measurement process. The position of the decimal point does not affect the count of significant figures. Rules for determining significant figures include: all non-zero digits are significant; zeros between non-zero digits are significant; leading zeros in numbers less than one are not significant; trailing zeros in numbers without a decimal point are not significant; trailing zeros in numbers with a decimal point are significant. To avoid ambiguity, measurements are best reported in scientific notation, where all digits in the base number are significant. Arithmetic operations with significant figures follow specific rules: in multiplication/division, the result should have the same number of significant figures as the least precise factor; in addition/subtraction, the result should have the same number of decimal places as the least precise term. Rounding off rules apply to uncertain digits, with conventions for digits equal to 5. Exact numbers, such as constants or counting numbers, have infinite significant figures. Proper use of significant figures ensures that reported results reflect the true precision of measurements and avoid misleading over-precision.

📊 Diagram: No specific diagram in this section.

🧪 Activity: No specific activity in this section.

🔗 Connection: Leads to rules for arithmetic operations with significant figures and uncertainty estimation in measurements.

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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