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Physics - Physical Quantities and Measurement (Density and Relative Density)

Grade 8ICSE

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

🔑Concepts

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Density is defined as the mass of a substance per unit volume. It is represented by the Greek letter ρ\rho (rho).

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The S.I. unit of density is kg m−3kg \, m^{-3}, while the C.G.S. unit is g cm−3g \, cm^{-3}.

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Conversion factor between units: 1 g cm−3=1000 kg m−31 \, g \, cm^{-3} = 1000 \, kg \, m^{-3}.

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Relative Density (R.D.R.D.), also known as Specific Gravity, is the ratio of the density of a substance to the density of water at 4∘C4^{\circ}C.

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Since R.D.R.D. is a ratio of two similar quantities, it has no units.

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Water has its maximum density at 4∘C4^{\circ}C, which is 1 g cm−31 \, g \, cm^{-3} or 1000 kg m−31000 \, kg \, m^{-3}.

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A body floats in a liquid if its density is less than or equal to the density of the liquid (rhobody≤ρliquid\\rho_{body} \le \rho_{liquid}). It sinks if its density is greater than that of the liquid (hobody>ρliquid ho_{body} > \rho_{liquid}).

📐Formulae

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

Relative Density (R.D.)=Density of substanceDensity of water at 4∘C\text{Relative Density } (R.D.) = \frac{\text{Density of substance}}{\text{Density of water at } 4^{\circ}C}

Relative Density (R.D.)=Mass of substanceMass of an equal volume of water at 4∘C\text{Relative Density } (R.D.) = \frac{\text{Mass of substance}}{\text{Mass of an equal volume of water at } 4^{\circ}C}

Density of substance in S.I. units=R.D.×1000 kg m−3\text{Density of substance in S.I. units} = R.D. \times 1000 \, kg \, m^{-3}

💡Examples

Problem 1:

An iron cylinder has a mass of 632 g632 \, g and a volume of 80 cm380 \, cm^3. Calculate its density in kg m−3kg \, m^{-3} and find its Relative Density.

Solution:

  1. Density in g cm−3g \, cm^{-3}: ρ=MV=632 g80 cm3=7.9 g cm−3\rho = \frac{M}{V} = \frac{632 \, g}{80 \, cm^3} = 7.9 \, g \, cm^{-3}.
  2. Density in kg m−3kg \, m^{-3}: ρ=7.9×1000=7900 kg m−3\rho = 7.9 \times 1000 = 7900 \, kg \, m^{-3}.
  3. R.D.=Density in g cm−31 g cm−3=7.91=7.9R.D. = \frac{\text{Density in } g \, cm^{-3}}{1 \, g \, cm^{-3}} = \frac{7.9}{1} = 7.9.

Explanation:

First, we use the density formula in C.G.S. units. Then, we convert it to S.I. units by multiplying by 10001000. Since the density of water is 1 g cm−31 \, g \, cm^{-3}, the numerical value of R.D.R.D. is equal to the density in C.G.S. units.

Problem 2:

The Relative Density of silver is 10.810.8. Find the mass of a silver block of volume 20 cm320 \, cm^3.

Solution:

  1. Density of silver (ho ho) = R.D.×Density of water=10.8×1 g cm−3=10.8 g cm−3R.D. \times \text{Density of water} = 10.8 \times 1 \, g \, cm^{-3} = 10.8 \, g \, cm^{-3}.
  2. Mass (MM) = ρ×V=10.8 g cm−3×20 cm3=216 g\rho \times V = 10.8 \, g \, cm^{-3} \times 20 \, cm^3 = 216 \, g.

Explanation:

By knowing the R.D.R.D., we determine the density of the substance in C.G.S. units. Using the rearranged density formula M=ρ×VM = \rho \times V, we calculate the mass.

Physical Quantities and Measurement (Density and Relative Density) Revision - Class 8 Science ICSE