Physical Quantities and Measurement ICSE Class 8: Density, Floating and Sinking

What Physical Quantities and Measurement is about

Weightage: CISCE publishes no question paper, marking scheme or chapter weightage for Class VIII Physics: your school sets and marks the exam. Physical Quantities and Measurement sits in the Theme 2: Physical Quantities and Measurement theme of the CISCE Curriculum for Upper Primary Classes (VI-VIII).Source: CISCE Curriculum for Upper Primary Classes (VI-VIII), RDCD 2016

Key concepts

Density and its units

Density is the mass of a unit volume of a substance:
$\rho =\frac{m}{V}$.

A block of iron and a block of wood of the same size have the same volume, but the iron has much more mass, so iron is denser.

Units:

• SI unit: kg/m³.
• Practical unit: g/cm³.
• To change g/cm³ to kg/m³, multiply by 1000; to change kg/m³ to g/cm³, divide by 1000.

Density of a substance does not depend on how much of it you take: a small and a large piece of the same pure metal have the same density.

Worked example

A metal block has a mass of 240 g and a volume of 30 cm³. Find its density in g/cm³ and in kg/m³.

Try it first, then show the solutionSolution

1. $\rho =\frac{240\text{g}}{30{\text{cm}}^3}=8{\text{g/cm}}^3$.
2. In SI units: $8\times 1000=8000{\text{kg/m}}^3$.

Answer: $8{\text{g/cm}}^3=8000{\text{kg/m}}^3$

Volume of an irregular solid with a measuring cylinder

An irregular solid (a stone, a key) has no simple shape, so we find its volume by displacement of water.

Method:
1. Pour water into a measuring cylinder and note the reading $V_1$ (read the bottom of the meniscus at eye level).
2. Tie the solid with a thin thread and lower it gently until it is fully under water. Note the new reading $V_2$.
3. Volume of the solid $=V_2-V_1$.

Then find its mass with a balance and use $\rho =\frac{m}{V}$. The solid must sink and must not dissolve in or absorb water.

Worked example

The water level in a measuring cylinder rises from 50 mL to 68 mL when a stone of mass 45 g is lowered in. Find the density of the stone.

Try it first, then show the solutionSolution

1. Volume $=68-50=18$ mL $=18{\text{cm}}^3$.
2. $\rho =\frac{45}{18}=2.5{\text{g/cm}}^3$ ($2500{\text{kg/m}}^3$).

Answer: $2.5{\text{g/cm}}^3$

Using a eureka can

A eureka can (overflow can) has a spout near the top. It is used when the solid is too large for a measuring cylinder.

Method:
1. Fill the can with water until water flows out of the spout; wait until it stops dripping.
2. Place an empty measuring cylinder under the spout.
3. Gently lower the solid (on a thread) into the can. The water it pushes out flows into the cylinder.
4. Volume of water collected $=$ volume of the solid.

Draw-and-label checklist: eureka can with spout; water level at the spout; solid hanging by a thread; measuring cylinder collecting the overflow.

Worked example

A stone of mass 65 g is lowered into a full eureka can, and 25 cm³ of water is collected. Find the density.

Try it first, then show the solutionSolution

1. Volume of stone $=25{\text{cm}}^3$.
2. $\rho =\frac{65}{25}=2.6{\text{g/cm}}^3$.

Answer: $2.6{\text{g/cm}}^3$

Density of a liquid

To find the density of a liquid:
1. Find the mass of an empty beaker (or bottle), $m_1$.
2. Pour in a measured volume $V$ of the liquid (using a measuring cylinder) and find the new mass $m_2$.
3. Mass of liquid $=m_2-m_1$; density $=\frac{m_2-m_1}{V}$.

A density bottle that holds a fixed volume can also be used: fill it with water to find its volume (since water is 1 g/cm³), then fill it with the liquid.

Worked example

An empty beaker has a mass of 120 g. With 50 mL of a liquid in it, the mass is 160 g. Find the density of the liquid.

Try it first, then show the solutionSolution

1. Mass of liquid $=160-120=40$ g.
2. $\rho =\frac{40}{50}=0.8{\text{g/cm}}^3$.

Answer: $0.8{\text{g/cm}}^3$

Floating and sinking

Compare the density of the object with the density of the liquid:

• object less dense than the liquid → it floats;
• object denser than the liquid → it sinks;
• equal densities → it stays wherever it is placed inside the liquid.

Examples in water (about 1 g/cm³): wood and ice float; iron and stone sink. A liquid that is less dense floats on a denser one (oil on water).

A hollow object such as a steel ship floats because its average density (steel plus the air inside) is less than that of water.

Worked example

A block has a volume of 80 cm³ and a mass of 72 g. Will it float in water ($1{\text{g/cm}}^3$)? Will it float in an oil of density $0.8{\text{g/cm}}^3$?

Try it first, then show the solutionSolution

1. Density of block $=\frac{72}{80}=0.9{\text{g/cm}}^3$.
2. $0.9<1$: it floats in water.
3. $0.9>0.8$: it sinks in the oil.

Answer: Floats in water; sinks in the oil

Density in the three states

In general, for the same substance:
$\text{density of solid}>\text{density of liquid}\gg \text{density of gas}$.

In a solid the particles are packed most closely; in a gas they are far apart, so a gas has a very small density (air is about $1.3{\text{kg/m}}^3$, roughly 800 times less dense than water).

Water is an exception: ice (about $0.92{\text{g/cm}}^3$) is less dense than liquid water, which is why ice floats and ponds freeze from the top.

Worked example

A hall is 6 m × 5 m × 3 m. Taking the density of air as $1.3{\text{kg/m}}^3$, find the mass of air in it.

Try it first, then show the solutionSolution

1. Volume $=6\times 5\times 3=90{\text{m}}^3$.
2. Mass $=\rho \times V=1.3\times 90=117$ kg.

Answer: 117 kg

Physics in real life: Physical Quantities and Measurement

Each concept, met in an everyday situation. Read the situation, try the problem, then check the solution. The one-line hook is what to remember.

Topics in this chapter

• Density and Density Numericals
• Measuring the Density of Solids and Liquids; Floating and Sinking

Physical Quantities and Measurement formula sheet

Density

$\rho =\frac{m}{V}$

Also m = ρV and V = m/ρ.

Unit conversion

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

Multiply by 1000 to get SI units.

Displacement

$V=V_2-V_1$

Measuring cylinder readings.

Liquid density

$\rho =\frac{m_2-m_1}{V}$

m₁ empty, m₂ with liquid.

Volume units

$1\text{mL}=1{\text{cm}}^3$

1 L = 1000 cm³.

Physical Quantities and Measurement: diagrams to practise

Practise each one on paper: draw it from the checklist, then tick off every label.

How the school exam asks Physical Quantities and Measurement

The question patterns to expect, and the method that works for each.

Tricks to remember Physical Quantities and Measurement

Short memory aids. Say each one aloud once, then use it on a question the same day.

• Mass over volume

  'D-M-V': Density = Mass ÷ Volume. Cover the one you want in the triangle.

  Use it when: Rearranging the formula.

• × 1000 to SI

  g/cm³ → kg/m³: multiply by 1000 (water 1 → 1000).

  Use it when: Unit conversions.

• Less dense floats

  Lighter-for-its-size goes up, heavier-for-its-size goes down.

  Use it when: Floating and sinking.

• Rise = volume

  The rise in water level (or the overflow) equals the volume of the sunk object.

  Use it when: Displacement methods.

Physical Quantities and Measurement: exam-level solved questions

Ordered from exam level to HOTS. Cover the solution, write your own, then compare step by step.

Case-study practice (Section B style)

Read the situation, then answer every part. In a school exam set on the board pattern this is a 8-mark question split into 2 + 3 + 3 marks.

Where the marks are

How a typical answer is split into step marks. Show every step: a right final answer alone rarely earns full marks.

Where students lose marks in Physical Quantities and Measurement

The slips examiners see most often, and what to write instead.

• ✗Mistake: Writing density without units

  The syllabus asks for results in proper units with proper symbols.

  Wrong working

  Density $=2.7$

  Right working

  Density $=2.7{\text{g/cm}}^3$

  ✓Do this: Write the unit in every answer line.

  Usually costs: 1 mark.

• ✗Mistake: Wrong conversion to kg/m³

  1 g/cm³ = 1000 kg/m³, not 100 or 1/1000.

  Wrong working

  $2.7{\text{g/cm}}^3=0.0027{\text{kg/m}}^3$

  Right working

  $2700{\text{kg/m}}^3$

  ✓Do this: Remember water: 1 g/cm³ = 1000 kg/m³.

  Usually costs: 1 mark.

• ✗Mistake: Using the total water reading as the volume

  The volume of the solid is the rise in level, not the final reading.

  Wrong working

  $V=68{\text{cm}}^3$

  Right working

  $V=68-50=18{\text{cm}}^3$

  ✓Do this: Always subtract the initial reading.

  Usually costs: The answer.

• ✗Mistake: Not filling the eureka can to the spout first

  Some displaced water then stays in the can, so the measured volume is too small.

  ✓Do this: Fill until water drips from the spout and wait for it to stop.

  Usually costs: 1 mark in a description.

• ✗Mistake: Saying heavy things sink

  Floating depends on density, not on mass. A heavy ship floats; a small stone sinks.

  ✓Do this: Compare densities, not weights.

  Usually costs: 1 to 2 marks.

Practice questions on Physical Quantities and Measurement

Write each answer on paper first. Open the hint only if you are stuck, and the answer only when you have finished.

1. Q1 · 1 mark · MCQExam level

   Which of these floats on water?
   (a) iron nail
   (b) ice cube
   (c) stone
   (d) glass marble

   Hint

   Less dense than water.

   Answer

   (b)

2. Q2 · 1 mark · Very short answerExam level

   Mercury has a density of $13.6{\text{g/cm}}^3$. Express this in kg/m³.

   Hint

   × 1000.

   Answer

   $13600{\text{kg/m}}^3$.

3. Q3 · 2 marks · Short answerExam level

   A rectangular block measures 2 cm × 3 cm × 5 cm and has a mass of 81 g. Find its density.

   Hint

   Volume = l × b × h.

   Answer

   $V=30{\text{cm}}^3$; $\rho =2.7{\text{g/cm}}^3$.

4. Q4 · 2 marks · Short answerHarder exam level

   Find the mass of 2 litres of milk if its density is $1.03{\text{g/cm}}^3$.

   Hint

   2 L = 2000 cm³.

   Answer

   $m=1.03\times 2000=2060$ g $=2.06$ kg.

5. Q5 · 1 mark · Assertion-reasonHarder exam level

   Assertion (A): Ice floats on water.
   Reason (R): Ice is less dense than water.
   (a) (A) is true and (R) is false.
   (b) (A) is false and (R) is true.
   (c) Both (A) and (R) are true, and (R) is the correct explanation for (A).
   (d) Both (A) and (R) are true, but (R) is not the correct explanation for (A).

   Hint

   Floating depends on density.

   Answer

   (c)

6. Q6 · 2 marks · Short answerHarder exam level

   A metal has a density of $8{\text{g/cm}}^3$. Find the volume of 400 g of it.

   Hint

   V = m/ρ.

   Answer

   $V=\frac{400}{8}=50{\text{cm}}^3$.

7. Q7 · 3 marks · Short answerHarder exam level

   The water level in a measuring cylinder rises from 40.0 mL to 52.5 mL when a pebble of mass 31.5 g is lowered in. Find the density of the pebble.

   Hint

   Volume = rise.

   Answer

   $V=12.5{\text{cm}}^3$; $\rho =\frac{31.5}{12.5}=2.52{\text{g/cm}}^3$.

8. Q8 · 2 marks · Short answerHarder exam level

   Arrange in order of increasing density, for the same substance: its liquid, its gas, its solid (for a typical substance). Give the reason.

   Hint

   Spacing of particles.

   Answer

   Gas < liquid < solid, because the particles are farthest apart in a gas and closest in a solid.

9. Q9 · 3 marks · Short answerHarder exam level

   Liquid A ($1.2{\text{g/cm}}^3$) and liquid B ($0.9{\text{g/cm}}^3$) do not mix. (i) Which forms the top layer? (ii) Where will a bead of density $1.0{\text{g/cm}}^3$ settle?

   Hint

   Compare densities.

   Answer

   (i) B, the less dense liquid. (ii) It sinks through B ($1.0>0.9$) but floats on A ($1.0<1.2$), so it rests at the boundary between the layers.

10. Q10 · 2 marks · Short answerHarder exam level

    Find the mass of 25 cm³ of mercury (density $13.6{\text{g/cm}}^3$).

    Hint

    m = ρV.

    Answer

    $13.6\times 25=340$ g.

11. Q11 · 3 marks · Short answerHOTS / competency

    Two cubes have edges of 3 cm. Cube P has a mass of 270 g and cube Q 21.6 g. Find their densities and state which floats on water.

    Hint

    Each volume is 27 cm³.

    Answer

    P: $\frac{270}{27}=10{\text{g/cm}}^3$ (sinks). Q: $\frac{21.6}{27}=0.8{\text{g/cm}}^3$ (floats).

12. Q12 · 3 marks · Short answerHOTS / competency

    A steel ship has a mass of 2000 tonnes, and the volume enclosed by its hull is 5000 m³. Find its average density in kg/m³ and explain why it floats, although steel is about 8 times denser than water. (1 tonne = 1000 kg)

    Hint

    Average density = total mass ÷ total volume.

    Answer

    $\frac{2000000\text{kg}}{5000{\text{m}}^3}=400{\text{kg/m}}^3$, less than water's $1000{\text{kg/m}}^3$. The hull holds a lot of air, so the ship's average density is low and it floats.

Test your recall

Answer in your head or on paper first, then open the card. Pulling an answer out of memory makes it stick far better than re-reading.

1. 1.Define density.Show answer

   Mass per unit volume.

2. 2.SI unit of density?Show answer

   kg/m³.

3. 3.Convert 1 g/cm³ to kg/m³.Show answer

   1000 kg/m³.

4. 4.What does a eureka can measure?Show answer

   The volume of an irregular solid (by the water it displaces).

5. 5.When does an object float in a liquid?Show answer

   When it is less dense than the liquid.

6. 6.Why does ice float on water?Show answer

   Ice is less dense than water.

7. 7.1 mL equals how many cm³?Show answer

   1 cm³.

8. 8.Which state usually has the lowest density?Show answer

   Gas.

Quick revision checklist

The night before the exam: can you do each of these without looking?

• ☐Did you write the unit (g/cm³ or kg/m³)?
• ☐Did you multiply by 1000 to convert to kg/m³?
• ☐Did you subtract the initial reading for volume?
• ☐Did you read the bottom of the meniscus at eye level?
• ☐Did you compare densities, not masses, for floating?
• ☐Do you remember that ice is less dense than water?

Physical Quantities and Measurement: frequently asked questions