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Measurement – Foundation of Science - International System of Units (SI)-advanced

Grade 9CBSE

Review the key concepts, formulae, and examples before starting your quiz.

🔑Concepts

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The International System of Units (SI) is a decimal system of weights and measures based on seven base units: meter (mm) for length, kilogram (kgkg) for mass, second (ss) for time, ampere (AA) for electric current, kelvin (KK) for thermodynamic temperature, mole (molmol) for amount of substance, and candela (cdcd) for luminous intensity.

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Derived units are formed by powers, products, or quotients of the base units. For example, the unit of force, the Newton (NN), is derived as 1 N=1 kg⋅m⋅s−21\text{ N} = 1\text{ kg} \cdot \text{m} \cdot \text{s}^{-2}.

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Standard Prefixes are used to express very large or small magnitudes. Common prefixes include Nano (10−910^{-9}), Micro (10−610^{-6}), Milli (10−310^{-3}), Centi (10−210^{-2}), Kilo (10310^{3}), Mega (10610^{6}), and Giga (10910^{9}).

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Scientific Notation is used to represent numbers in the form a×10na \times 10^n, where 1≤a<101 \le a < 10 and nn is an integer. This is essential for maintaining precision in measurements.

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The Principle of Homogeneity states that the units of all terms in a physical equation must be the same. For example, in v=u+atv = u + at, all three terms (vv, uu, and atat) must have the unit m⋅s−1\text{m} \cdot \text{s}^{-1}.

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Significant figures in a measurement consist of all the digits known with certainty plus one digit that is uncertain. Rules for rounding and arithmetic operations must be followed to maintain the accuracy of the measured data.

📐Formulae

Density (ρ)=Mass (m)Volume (V)\text{Density } (\rho) = \frac{\text{Mass } (m)}{\text{Volume } (V)}

Force (F)=mass (m)×acceleration (a)=kg⋅m⋅s−2=Newton (N)\text{Force } (F) = \text{mass } (m) \times \text{acceleration } (a) = \text{kg} \cdot \text{m} \cdot \text{s}^{-2} = \text{Newton } (N) wheel

Pressure (P)=Force (F)Area (A)=N⋅m−2=Pascal (Pa)\text{Pressure } (P) = \frac{\text{Force } (F)}{\text{Area } (A)} = \text{N} \cdot \text{m}^{-2} = \text{Pascal } (Pa) wheel

Work (W)=Force (F)×Displacement (s)=kg⋅m2⋅s−2=Joule (J)\text{Work } (W) = \text{Force } (F) \times \text{Displacement } (s) = \text{kg} \cdot \text{m}^2 \cdot \text{s}^{-2} = \text{Joule } (J) wheel

Power (P)=Work (W)Time (t)=J⋅s−1=Watt (W)\text{Power } (P) = \frac{\text{Work } (W)}{\text{Time } (t)} = \text{J} \cdot \text{s}^{-1} = \text{Watt } (W) wheel

💡Examples

Problem 1:

The density of mercury is 13.6 g/cm313.6 \text{ g/cm}^3. Convert this value into the SI unit (kg/m3\text{kg/m}^3).

Solution:

Given density ρ=13.6 g/cm3\rho = 13.6 \text{ g/cm}^3. We know: 1 g=10−3 kg1 \text{ g} = 10^{-3} \text{ kg} 1 cm=10−2 m  ⟹  1 cm3=(10−2 m)3=10−6 m31 \text{ cm} = 10^{-2} \text{ m} \implies 1 \text{ cm}^3 = (10^{-2} \text{ m})^3 = 10^{-6} \text{ m}^3 Substituting these into the expression: ρ=13.6×10−3 kg10−6 m3\rho = 13.6 \times \frac{10^{-3} \text{ kg}}{10^{-6} \text{ m}^3} ρ=13.6×103 kg/m3\rho = 13.6 \times 10^{3} \text{ kg/m}^3 ρ=13600 kg/m3\rho = 13600 \text{ kg/m}^3

Explanation:

To convert from CGS to SI, we substitute the equivalent base unit values for mass (grams to kilograms) and volume (cubic centimeters to cubic meters).

Problem 2:

Check the dimensional consistency of the equation s=ut+12at2s = ut + \frac{1}{2}at^2 using SI units.

Solution:

LHS (Left Hand Side): ss (displacement) is measured in meters (mm). RHS (Right Hand Side) consists of two terms: Term 1: ut=(m⋅s−1)×(s)=mut = (\text{m} \cdot \text{s}^{-1}) \times (\text{s}) = m Term 2: 12at2=(m⋅s−2)×(s2)=m\frac{1}{2}at^2 = (\text{m} \cdot \text{s}^{-2}) \times (\text{s}^2) = m (Note: The constant 12\frac{1}{2} is dimensionless). Since both terms on the RHS and the term on the LHS have the unit meter (mm), the equation is dimensionally consistent.

Explanation:

According to the principle of homogeneity, for an equation to be correct, every term separated by a plus or minus sign must have the same fundamental units.