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Measurement – Foundation of Science - Different Systems of Units-advanced

Grade 9CBSE

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

🔑Concepts

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A physical quantity QQ is expressed as the product of its numerical value nn and its unit uu, represented by the relation Q=nuQ = n u.

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Measurement is the process of comparing an unknown physical quantity with a known fixed standard unit of the same nature.

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Fundamental Quantities are independent of other quantities (e.g., Mass, Length, Time). There are 7 base SI units and 2 supplementary units (Radian for plane angle and Steradian for solid angle).

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Derived Quantities are those expressed in terms of fundamental quantities (e.g., Velocity, Force, Pressure).

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The SI (Système International d'Unités) is the modern version of the MKS (Metre-Kilogram-Second) system and is globally accepted.

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Systems of Units: CGS (Centimetre, Gram, Second), FPS (Foot, Pound, Second), and MKS (Metre, Kilogram, Second).

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Dimensions of a physical quantity represent the powers to which the fundamental units are raised to represent that quantity, usually written as [MaLbTc][M^a L^b T^c].

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Relation between unit size and numerical value: Since n1u1=n2u2n_1 u_1 = n_2 u_2, if the size of the unit increases, the numerical value decreases (n∝1un \propto \frac{1}{u}).

📐Formulae

Q=n×uQ = n \times u

n1u1=n2u2n_1 u_1 = n_2 u_2

Density (ρ)=MassVolume=ML3=[ML−3T0]\text{Density } (\rho) = \frac{\text{Mass}}{\text{Volume}} = \frac{M}{L^3} = [M L^{-3} T^0]

Force (F)=Mass×Acceleration=[M]×[LT−2]=[MLT−2]\text{Force } (F) = \text{Mass} \times \text{Acceleration} = [M] \times [L T^{-2}] = [M L T^{-2}]

Pressure (P)=ForceArea=[MLT−2][L2]=[ML−1T−2]\text{Pressure } (P) = \frac{\text{Force}}{\text{Area}} = \frac{[M L T^{-2}]}{[L^2]} = [M L^{-1} T^{-2}]

1 Newton (N)=105 dynes1 \text{ Newton (N)} = 10^5 \text{ dynes}

1 Joule (J)=107 ergs1 \text{ Joule (J)} = 10^7 \text{ ergs}

💡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/m3kg/m^3).

Solution:

Given n1=13.6n_1 = 13.6, u1=g/cm3u_1 = \text{g/cm}^3. In SI, u2=kg/m3u_2 = \text{kg/m}^3. We know 1 g=10−3 kg1 \text{ g} = 10^{-3} \text{ kg} and 1 cm=10−2 m1 \text{ cm} = 10^{-2} \text{ m}. So, 1 cm3=(10−2 m)3=10−6 m31 \text{ cm}^3 = (10^{-2} \text{ m})^3 = 10^{-6} \text{ m}^3. Using n1u1=n2u2n_1 u_1 = n_2 u_2: 13.6×1 g1 cm3=n2×1 kg1 m313.6 \times \frac{1 \text{ g}}{1 \text{ cm}^3} = n_2 \times \frac{1 \text{ kg}}{1 \text{ m}^3} n2=13.6×10−3 kg10−6 m3n_2 = 13.6 \times \frac{10^{-3} \text{ kg}}{10^{-6} \text{ m}^3} n2=13.6×103=13600n_2 = 13.6 \times 10^3 = 13600 Therefore, the density in SI units is 13600 kg/m313600 \text{ kg/m}^3.

Explanation:

To convert units, we substitute the equivalent value of the CGS units in terms of SI units and simplify the numerical factor.

Problem 2:

Find the SI unit and dimensional formula for the Universal Gravitational Constant GG using the formula F=Gm1m2r2F = G \frac{m_1 m_2}{r^2}.

Solution:

Rearranging for GG: G=F×r2m1×m2G = \frac{F \times r^2}{m_1 \times m_2} In terms of units: Unit of G=Newton×metre2kg×kg=N⋅m2⋅kg−2\text{Unit of } G = \frac{\text{Newton} \times \text{metre}^2}{\text{kg} \times \text{kg}} = N \cdot m^2 \cdot kg^{-2} In terms of dimensions: [G]=[MLT−2][L2][M][M]=[ML3T−2][M2]=[M−1L3T−2][G] = \frac{[M L T^{-2}] [L^2]}{[M] [M]} = \frac{[M L^3 T^{-2}]}{[M^2]} = [M^{-1} L^3 T^{-2}]

Explanation:

The unit is derived by isolating the constant in the physical equation and substituting the units of the other variables. Dimensions are found by substituting the base dimensions of force, distance, and mass.