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.
1.3.2 Rounding off the Uncertain Digits
In computations involving approximate numbers, results often contain more digits than justified by the precision of the input data. Rounding off is the process of reducing digits to reflect the true precision. The general rule is: if the digit to be dropped is greater than 5, the preceding digit is increased by 1; if less than 5, the preceding digit remains unchanged. When the digit to be dropped is exactly 5, the preceding digit is rounded to make it even (round half to even rule). For example, 2.746 rounded to three significant figures becomes 2.75, 1.743 becomes 1.74, 2.745 becomes 2.74 (preceding digit even), and 2.735 becomes 2.74 (preceding digit odd). In multi-step calculations, it is advisable to retain one extra digit in intermediate steps and round off only at the end to minimize cumulative rounding errors. Constants like π or exact numbers in formulae have effectively infinite significant figures and can be used with appropriate rounding as needed.
📊 Diagram: No specific diagram in this section.
🧪 Activity: No specific activity in this section.
🔗 Connection: Leads to examples illustrating significant figures and uncertainty 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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