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ICSE • Class 7 • Science • Ch 1
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Physical Quantities and Measurement

In ICSE Class 7 Science (Physics), "Physical Quantities and Measurement" provides an authoritative, experimentally grounded master study guide analyzing the measurement of area, volume, mass, and density of regular and irregular solids, as well as liquids. This comprehensive chapter explores Fundamental vs Derived Physical Quantities (Measurement of Area of regular shapes [rectangles, squares, triangles, circles] and irregular lamina using graph grid method; Measurement of Volume of regular solids [$V = l \times b \times h, V = \frac{4}{3}\pi r^3, V = \pi r^2 h$] and irregular solids using the liquid displacement method with a graduated cylinder and eureka/overflow can; Measurement of volume of liquids using graduated cylinders and burettes), Concept of Density (Mass per unit volume: $d = \frac{M}{V}$; SI unit $\text{kg/m}^3$ and CGS unit $\text{g/cm}^3$; Fundamental conversion relationship: $1\text{ g/cm}^3 = 1,000\text{ kg/m}^3$; Density of water at $4^\circ\text{C} = 1\text{ g/cm}^3 = 1,000\text{ kg/m}^3$), Relative Density / Specific Gravity (Ratio of density of a substance to density of pure water at $4^\circ\text{C}$: $\text{RD} = \frac{\text{Density of substance}}{\text{Density of water at } 4^\circ\text{C}} = \frac{\text{Mass of any volume of substance}}{\text{Mass of equal volume of water at } 4^\circ\text{C}}$; Dimensionless and unitless nature of RD; Measurement of liquid density using a Constant Volume Specific Gravity Bottle / Pycnometer), Concept of Speed ($S = \frac{D}{T}$, SI unit $\text{m/s}$, conversion to $\text{km/h}$ by multiplying by $\frac{18}{5}$), and Experimental Precision (Parallax error elimination, zero error corrections) aligned with the 2026–27 CISCE ICSE curriculum.

How Did a Single Wooden Density Bottle Expose the Most Dangerous Counterfeit Gold Scam in the Mediterranean World?

Imagine stepping into the royal treasury of King Hiero in ancient Sicily. On the marble altar rests a gleaming golden crown, matching the exact weight of solid gold entrusted to the artisan. Yet King Hiero was tormented by suspicion: silver is far cheaper than gold, but if the goldsmith melted pure gold with lightweight silver, how could anyone prove fraud without melting the sacred crown into molten slag? The genius philosopher Archimedes discovered the secret while lowering his body into a brimming bathhouse pool: every solid object, regardless of its shape, displaces a volume of water strictly identical to its own solid volume! Because gold is almost twice as dense as silver ($19.3\text{ g/cm}^3$ vs $10.5\text{ g/cm}^3$), a one-kilogram lump of pure gold occupies a remarkably tiny volume, whereas a one-kilogram alloy of gold and silver occupies a significantly larger volume and spills more water! By measuring Volume through liquid displacement, Archimedes mathematically proved the crown was counterfeit! Why does an iron nail sink in water while a massive steel ship floats? What is the exact mathematical relationship between density in $\text{g/cm}^3$ and $\text{kg/m}^3$? Let's master physical quantities and measurement.

Why This Chapter Matters

Measurement is the foundation of all physical sciences, aerospace engineering, pharmacology dosage formulation, and industrial manufacturing. Without precise measurement of volume, density, and relative density, cargo ships would capsize under unbalanced loading, hydraulic brakes would fail, and chemical syntheses would produce dangerous impurities.

Before You Begin (Prerequisites)

  • Fundamental SI units of length (meter, m) and mass (kilogram, kg).
  • Basic arithmetic: Division of decimals and cross-multiplication.
  • Basic geometric volume formulas for cubes and cuboids.

What You Will Learn (Core Objectives)

  • Differentiate between fundamental and derived physical quantities.
  • Measure the area of irregular lamina using graph paper counting techniques.
  • Determine the volume of irregular solids using measuring cylinders and eureka overflow cans.
  • Calculate density using $d = \frac{M}{V}$ and convert between CGS ($\text{g/cm}^3$) and SI ($\text{kg/m}^3$) units.
  • Define Relative Density (RD) and explain why it is a dimensionless pure number.
  • Determine the relative density of liquids using a specific gravity bottle (density bottle).

Chapter Roadmap & Progression

1 1. Measurement of Area & Volume: Re...
2 2. Density: Physical Significance &...
3 3. Relative Density (Specific Gravi...
4 4. Concept of Speed & Laboratory Pr...

Complete Concept Guide (100% Curriculum Coverage)

1. Measurement of Area & Volume: Regular & Irregular Solids

Understand
A. Measurement of Surface Area:
  • Regular Shapes: Square ($A = s^2$), Rectangle ($A = l \times b$), Triangle ($A = \frac{1}{2} b h$), Circle ($A = \pi r^2$).
  • Irregular Lamina (Graph Paper Method):
    1. Place the irregular leaf or lamina on a $1\text{ cm}^2$ grid paper and trace its boundary with a sharp pencil.
    2. Count the number of fully enclosed squares ($N_f$).
    3. Count the number of squares that are half or more than half enclosed ($N_h$). Ignore squares less than half enclosed.
    4. $$\mathbf{\text{Approximate Area} = (N_f + N_h) \text{ cm}^2}$$
B. Measurement of Volume:
  • SI Unit: Cubic meter ($\text{m}^3$). Practical units: Cubic centimeter ($\text{cm}^3$ or $\text{cc}$), Liter ($\text{L}$), Milliliter ($\text{mL}$). $$1\text{ m}^3 = 10^6\text{ cm}^3; \quad 1\text{ L} = 1,000\text{ mL} = 1,000\text{ cm}^3; \quad 1\text{ mL} = 1\text{ cm}^3$$
  • Irregular Solid (Liquid Displacement Method):
    1. Fill a graduated cylinder with water to an initial level $V_1$.
    2. Tie the irregular insoluble solid (e.g., a stone) with a thin thread and gently submerge it completely without splashing.
    3. Read the new water level $V_2$ at the bottom of the meniscus.
    4. $$\mathbf{\text{Volume of Solid } V = V_2 - V_1}$$
  • Eureka Can / Overflow Can Method: When the solid is too large for a measuring cylinder, fill a eureka can until water overflows through the spout. Place an empty measuring cylinder under the spout, gently lower the solid, and the overflow volume directly equals the solid volume!

2. Density: Physical Significance & Unit Conversions

Density
A. Definition & Formula:

The Density ($d$ or $\rho$) of a substance is defined as its mass per unit volume:

$$\mathbf{d = \frac{\text{Mass } (M)}{\text{Volume } (V)}}$$
  • SI Unit: Kilogram per cubic meter ($\text{kg/m}^3$ or $\text{kg m}^{-3}$).
  • CGS Unit: Gram per cubic centimeter ($\text{g/cm}^3$ or $\text{g cm}^{-3}$).
B. Master Unit Conversion Proof:
$$1\text{ g/cm}^3 = \frac{1\text{ g}}{1\text{ cm}^3} = \frac{10^{-3}\text{ kg}}{(10^{-2}\text{ m})^3} = \frac{10^{-3}\text{ kg}}{10^{-6}\text{ m}^3} = 10^3\text{ kg/m}^3 = \mathbf{1,000\text{ kg/m}^3}$$
  • To convert from $\text{g/cm}^3$ to $\text{kg/m}^3$: Multiply by $1,000$.
  • To convert from $\text{kg/m}^3$ to $\text{g/cm}^3$: Divide by $1,000$.
  • Example: Density of pure water at $4^\circ\text{C} = 1\text{ g/cm}^3 = \mathbf{1,000\text{ kg/m}^3}$.
  • Example: Density of liquid mercury $= 13.6\text{ g/cm}^3 = \mathbf{13,600\text{ kg/m}^3}$.

3. Relative Density (Specific Gravity) & The Density Bottle

Relative Density
A. Definition of Relative Density (RD):

The Relative Density of a substance is the ratio of the density of the substance to the density of pure water at $4^\circ\text{C}$:

$$\mathbf{\text{RD} = \frac{\text{Density of substance}}{\text{Density of water at } 4^\circ\text{C}} = \frac{\text{Mass of unit volume of substance}}{\text{Mass of equal volume of water at } 4^\circ\text{C}}}$$
  • No Units: Because Relative Density is the ratio of two identical physical quantities (densities), it is a pure number having NO units.
  • Numerical Equivalence: In the CGS system, since the density of water is $1\text{ g/cm}^3$, the numerical value of relative density is identical to its density in $\text{g/cm}^3$! (e.g., if RD of copper is $8.9$, its density is $8.9\text{ g/cm}^3 = 8,900\text{ kg/m}^3$).
B. Measurement of Relative Density using a Density Bottle:

A Specific Gravity Bottle (Pycnometer) is a specially blown glass flask with a ground glass stopper having a fine capillary bore to ensure a strictly constant volume $V$ of liquid.

  1. Weigh the clean, dry empty bottle with stopper: $M_1$.
  2. Fill the bottle completely with the experimental liquid, insert the stopper (excess liquid escapes through the capillary), wipe dry, and weigh: $M_2$. $$\text{Mass of liquid} = M_2 - M_1$$
  3. Empty, rinse, and fill completely with pure water, insert stopper, wipe dry, and weigh: $M_3$. $$\text{Mass of equal volume of water} = M_3 - M_1$$
  4. $$\mathbf{\text{Relative Density of Liquid (RD)} = \frac{M_2 - M_1}{M_3 - M_1}}$$

4. Concept of Speed & Laboratory Precautions

Speed & Precautions
A. Speed ($S$):

Speed is the rate of change of distance with respect to time: $\text{Speed} = \frac{\text{Distance}}{\text{Time}} = \frac{D}{T}$.

  • SI Unit: meters per second ($\text{m/s}$). Commercial Unit: kilometers per hour ($\text{km/h}$).
  • Conversion: $1\text{ km/h} = \frac{1000\text{ m}}{3600\text{ s}} = \mathbf{\frac{5}{18}\text{ m/s}}$, and $1\text{ m/s} = \mathbf{\frac{18}{5}\text{ km/h} = 3.6\text{ km/h}}$.
B. Experimental Precautions:
  • Meniscus Reading: For transparent wetting liquids (water), read the lowest point of the concave meniscus. For non-wetting convex liquids (mercury), read the highest point of the convex meniscus.
  • Parallax Error: The observer's eye must be held exactly perpendicular and level with the liquid meniscus scale line.

Key Formulas, Reactions & Definitions

Density Equation
$$d = \frac{M}{V} \quad \left[\text{SI: kg/m}^3, \text{ CGS: g/cm}^3\right]$$
1 g/cm^3 = 1000 kg/m^3.
Relative Density (RD)
$$\text{RD} = \frac{d_{\text{substance}}}{d_{\text{water at } 4^\circ\text{C}}} = \frac{M_2 - M_1}{M_3 - M_1}$$
Dimensionless pure number with zero units.

Density, Displacement Method & Specific Gravity Bottle

Physics: Density, Displacement Method & Relative Density DENSITY & CONVERSIONS d = Mass / Volume • SI Unit: kg/m3 • CGS Unit: g/cm3 1 g/cm3 = 1,000 kg/m3 Water at 4°C = 1,000 kg/m3 Mercury: 13,600 kg/m3 Wood: ~700 kg/m3 (Floats) VOLUME DISPLACEMENT V1 = 50 mL V2 = 75 mL Vsolid = V2 - V1 = 25 cm3 RELATIVE DENSITY (RD) RD = dsubstance / dwater • Pure number: NO UNITS! Density Bottle Formula: RD = (M2 - M1) / (M3 - M1) M1 = Empty bottle M2 = Bottle + Liquid M3 = Bottle + Water • Capillary stopper ensures constant V DISPLACEMENT: V = V2 - V1 • 1 g/cm3 = 1,000 kg/m3 • RD HAS ZERO UNITS

Chapter Summary & 10 Key Takeaways

Takeaway 1
Area of irregular lamina is estimated by counting fully and half-or-more enclosed grid squares on graph paper.
Takeaway 2
Volume is the space occupied by a body; SI unit is m3, and 1 mL = 1 cm3 = 1 cc.
Takeaway 3
Volume of an insoluble irregular solid is measured by liquid displacement: V = V2 - V1.
Takeaway 4
Density is mass per unit volume: d = M / V.
Takeaway 5
SI unit of density is kg/m3; CGS unit is g/cm3.
Takeaway 6
Crucial conversion: 1 g/cm3 = 1,000 kg/m3.
Takeaway 7
Density of pure water at 4°C is 1 g/cm3 or 1,000 kg/m3.
Takeaway 8
Relative Density (RD) is the ratio of substance density to density of water at 4°C.
Takeaway 9
Relative Density is a pure ratio having NO physical dimensions or units.
Takeaway 10
A Specific Gravity Bottle determines RD of liquids via RD = (M2 - M1) / (M3 - M1).

Check Your Understanding (Diagnostic Practice Questions)

Diagnostic questions testing core conceptual clarity. Answers are hidden initially — solve each problem first, then click to reveal the step-by-step verified solution.

1
A piece of iron of mass $395\text{ g}$ has a volume of $50\text{ cm}^3$. Calculate its density in: (a) $\text{g/cm}^3$, (b) $\text{kg/m}^3$.
Reveal Answer & Explanation
Answer:

Given: Mass $M = 395\text{ g}$, Volume $V = 50\text{ cm}^3$.
• (a) In $\text{g/cm}^3$:

$$d = \frac{M}{V} = \frac{395}{50} = \mathbf{7.9\text{ g/cm}^3}$$


• (b) In $\text{kg/m}^3$:
Multiply by $1,000$:

$$d = 7.9 \times 1,000 = \mathbf{7,900\text{ kg/m}^3}$$

.


$d = 395 / 50 = 7.9\text{ g/cm}^3$. In SI units, multiply by $1,000$ to get $7,900\text{ kg/m}^3$.
2
Explain why Relative Density has no units, whereas Density has units.
Reveal Answer & Explanation
Answer:

• Density is the physical ratio of two different physical quantities—Mass divided by Volume ($d = \frac{M}{V}$). Therefore, it has derived dimensional units ($\text{kg/m}^3$ or $\text{g/cm}^3$).
• Relative Density (RD) is defined as the ratio of two identical physical quantities (the density of a substance divided by the density of water):

$$\text{RD} = \frac{\text{Density of substance}}{\text{Density of water}} = \frac{\text{kg/m}^3}{\text{kg/m}^3} = 1$$


The units cancel out completely, making Relative Density a pure, dimensionless number with NO units.


RD is a ratio of two identical quantities (density / density), so the units cancel out completely.
3
An empty density bottle weighs $25.2\text{ g}$. When filled with water it weighs $50.2\text{ g}$, and when filled with an unknown oil it weighs $45.2\text{ g}$. Calculate: (a) Volume of the bottle, (b) Relative density of the oil, (c) Density of the oil in SI units.
Reveal Answer & Explanation
Answer:

Given: $M_1 = 25.2\text{ g}$, $M_3 = 50.2\text{ g}$ (with water), $M_2 = 45.2\text{ g}$ (with oil).
• (a) Volume of the bottle:
Mass of water $= M_3 - M_1 = 50.2 - 25.2 = 25.0\text{ g}$.
Since density of water is $1\text{ g/cm}^3$, Volume $= \frac{\text{Mass}}{\text{Density}} = \frac{25.0}{1} = \mathbf{25\text{ cm}^3}$.
• (b) Relative density of oil:

$$\text{Mass of oil} = M_2 - M_1 = 45.2 - 25.2 = 20.0\text{ g}$$


$$\text{RD} = \frac{M_2 - M_1}{M_3 - M_1} = \frac{20.0}{25.0} = \mathbf{0.8}$$


• (c) Density of oil in SI units:

$$\text{Density} = \text{RD} \times 1,000\text{ kg/m}^3 = 0.8 \times 1,000 = \mathbf{800\text{ kg/m}^3}$$

.


Mass of water is $25\text{ g}$ (vol is $25\text{ cm}^3$). Mass of oil is $20\text{ g}$. $\text{RD} = 20/25 = 0.8$. Density is $800\text{ kg/m}^3$.
4
Describe how the volume of an irregular piece of stone can be determined using a graduated measuring cylinder.
Reveal Answer & Explanation
Answer:
  1. Pour clean water into a graduated cylinder until it is approximately half-full, recording the initial volume $V_1$ by viewing the bottom of the concave meniscus at eye level.
    2. Tie the irregular stone securely with a piece of thin, non-absorbent thread.
    3. Gently lower the stone into the cylinder until it is completely submerged under water without touching the sides or splashing.
    4. Note the raised water level $V_2$.
    5. The volume of the irregular stone is given by the displacement formula: $V = V_2 - V_1$ (expressed in $\text{cm}^3$ or $\text{mL}$).

Record initial level $V_1$, submerge tied stone, record final level $V_2$. Volume of stone $= V_2 - V_1$.
5
Convert a speed of $90\text{ km/h}$ into $\text{m/s}$, and a speed of $15\text{ m/s}$ into $\text{km/h}$.
Reveal Answer & Explanation
Answer:

• (a) Convert $90\text{ km/h}$ to $\text{m/s}$:
Multiply by $\frac{5}{18}$:

$$\text{Speed} = 90 \times \frac{5}{18} = 5 \times 5 = \mathbf{25\text{ m/s}}$$


• (b) Convert $15\text{ m/s}$ to $\text{km/h}$:
Multiply by $\frac{18}{5}$:

$$\text{Speed} = 15 \times \frac{18}{5} = 3 \times 18 = \mathbf{54\text{ km/h}}$$

.


Multiply $\text{km/h}$ by $5/18$ to get $\text{m/s}$; multiply $\text{m/s}$ by $18/5$ to get $\text{km/h}$.
6
A solid cube of side $4\text{ cm}$ has a mass of $512\text{ g}$. Find its density. Will it float or sink in water? Explain.
Reveal Answer & Explanation
Answer:

Step 1: Calculate volume of the cube:

$$V = s^3 = 4 \times 4 \times 4 = 64\text{ cm}^3$$


Step 2: Calculate density:

$$d = \frac{M}{V} = \frac{512\text{ g}}{64\text{ cm}^3} = \mathbf{8\text{ g/cm}^3}$$


Step 3: Flotation condition:
The density of water is $1\text{ g/cm}^3$. Since the density of the cube ($8\text{ g/cm}^3$) is greater than the density of water, the cube will SINK in water.


Volume $= 64\text{ cm}^3$. Density $= 512/64 = 8\text{ g/cm}^3$. Since $8 > 1$, it will sink.
7
What is the function of the fine capillary bore in the glass stopper of a specific gravity bottle?
Reveal Answer & Explanation
Answer:

The fine capillary bore running through the glass stopper allows excess liquid and trapped air bubbles to escape when the stopper is inserted into the full bottle. This ensures that the bottle always encloses the exact, strictly constant volume of liquid without any variability due to overfilling or trapped air.


It allows excess liquid and air bubbles to escape, maintaining an identical, constant volume every time.
8
State two precautions to be observed while measuring the volume of a liquid in a graduated measuring cylinder.
Reveal Answer & Explanation
Answer:
  1. Meniscus Alignment: For water and clear liquids, read the graduation line corresponding to the bottom of the concave meniscus. For mercury, read the top of the convex meniscus.
    2. Avoid Parallax Error: Keep the eye strictly at the same horizontal level as the liquid surface to avoid parallax error caused by looking from above or below.

Read the bottom of concave meniscus and keep eyes strictly level to avoid parallax error.
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