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Measurement – Foundation of Science - Conversion of Units between different Systems-advanced

Grade 9CBSE

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

🔑Concepts

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The fundamental principle of unit conversion is based on the fact that the physical quantity remains the same regardless of the system of units used: Q=n1u1=n2u2Q = n_1 u_1 = n_2 u_2, where nn is the numerical value and uu is the unit.

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The SI system (International System of Units) is the standard modern form of the metric system, based on seven base units including meter (mm), kilogram (kgkg), and second (ss).

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The CGS system (Centimeter-Gram-Second) is an older metric system where length is in cmcm, mass in gg, and time in ss.

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Derived units like Force (NewtonNewton) and Energy (JouleJoule) can be expressed in terms of base units. For example, 1 N=1 kg⋅m⋅s−21 \text{ N} = 1 \text{ kg} \cdot \text{m} \cdot \text{s}^{-2}.

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Prefixes are used to express very large or small quantities using powers of 10, such as Micro (10−610^{-6}), Nano (10−910^{-9}), Mega (10610^{6}), and Giga (10910^{9}).

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Conversion factors are ratios used to express the same quantity in different units, such as 1 m=100 cm1 \text{ m} = 100 \text{ cm} or 1 kg=1000 g1 \text{ kg} = 1000 \text{ g}.

📐Formulae

n1u1=n2u2n_1 u_1 = n_2 u_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}

1 g/cm3=1000 kg/m31 \text{ g/cm}^3 = 1000 \text{ kg/m}^3

1 km/h=518 m/s1 \text{ km/h} = \frac{5}{18} \text{ m/s}

1 Pascal (Pa)=1 N/m2=10 dynes/cm21 \text{ Pascal (Pa)} = 1 \text{ N/m}^2 = 10 \text{ dynes/cm}^2

💡Examples

Problem 1:

Convert the value of the Universal Gravitational Constant G=6.67×10−11 N m2 kg−2G = 6.67 \times 10^{-11} \text{ N m}^2 \text{ kg}^{-2} into the CGS system (dynes⋅cm2⋅g−2dynes \cdot cm^2 \cdot g^{-2}).

Solution:

We know the conversion factors:

  1. 1 N=105 dyne1 \text{ N} = 10^5 \text{ dyne}
  2. 1 m=102 cm  ⟹  1 m2=104 cm21 \text{ m} = 10^2 \text{ cm} \implies 1 \text{ m}^2 = 10^4 \text{ cm}^2
  3. 1 kg=103 g  ⟹  1 kg−2=(103 g)−2=10−6 g−21 \text{ kg} = 10^3 \text{ g} \implies 1 \text{ kg}^{-2} = (10^3 \text{ g})^{-2} = 10^{-6} \text{ g}^{-2}

Now, substitute these into the value of GG: G=6.67×10−11×(105)×(104)×(10−6)G = 6.67 \times 10^{-11} \times (10^5) \times (10^4) \times (10^{-6}) G=6.67×10−11+5+4−6G = 6.67 \times 10^{-11 + 5 + 4 - 6} G=6.67×10−8 dyne cm2 g−2G = 6.67 \times 10^{-8} \text{ dyne cm}^2 \text{ g}^{-2}

Explanation:

To convert a complex derived unit, convert each individual component (Newton, meter, kilogram) to its CGS equivalent and multiply the powers of 10.

Problem 2:

The density of mercury is 13.6 g/cm313.6 \text{ g/cm}^3. Convert this into SI units (kg/m3kg/m^3).

Solution:

In the CGS system, 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 values: ρ=13.6×10−3 kg10−6 m3\rho = 13.6 \times \frac{10^{-3} \text{ kg}}{10^{-6} \text{ m}^3} ρ=13.6×10−3×106 kg/m3\rho = 13.6 \times 10^{-3} \times 10^6 \text{ kg/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:

Conversion of density requires dividing the mass conversion factor by the volume conversion factor. Note that 1 g/cm31 \text{ g/cm}^3 is always equal to 1000 kg/m31000 \text{ kg/m}^3.

Problem 3:

A car is moving at a speed of 72 km/h72 \text{ km/h}. Express this speed in m/sm/s.

Solution:

To convert from km/hkm/h to m/sm/s, we use the conversion factor 518\frac{5}{18}: Speed=72×518 m/sSpeed = 72 \times \frac{5}{18} \text{ m/s} Speed=4×5 m/sSpeed = 4 \times 5 \text{ m/s} Speed=20 m/sSpeed = 20 \text{ m/s}

Explanation:

Since 1 km=1000 m1 \text{ km} = 1000 \text{ m} and 1 hour=3600 s1 \text{ hour} = 3600 \text{ s}, the ratio is 1000/36001000/3600, which simplifies to 5/185/18.