Physical Quantities and Measurement
1. Physical Quantities
A physical quantity is a quantity that can be MEASURED.
| Type | Definition | Examples |
|---|---|---|
| Scalar | Has ONLY magnitude | Mass, length, time, speed, density |
| Vector | Has BOTH magnitude and direction | Force, velocity, displacement, acceleration |
SI Base Units:
| Quantity | Unit | Symbol |
|---|---|---|
| Length | metre | m |
| Mass | kilogram | kg |
| Time | second | s |
| Temperature | kelvin | K |
| Electric current | ampere | A |
| Amount of substance | mole | mol |
| Luminous intensity | candela | cd |
2. Measurement of Length
Standard Units
1 km = 1000 m 1 m = 100 cm = 1000 mm 1 cm = 10 mm
Measuring Instruments
| Instrument | Use | Least Count | Accuracy |
|---|---|---|---|
| Metre scale | Length up to 1 m | 0.1 cm or 1 mm | ± 1 mm |
| Vernier callipers | Internal/external dimensions | 0.01 cm or 0.1 mm | ± 0.01 cm |
| Screw gauge | Very small thickness/diameter | 0.001 cm or 0.01 mm | ± 0.001 cm |
3. Vernier Callipers
Principle
The vernier scale has n divisions that cover (n — 1) main scale divisions.
Least Count (LC) = Value of 1 MSD / Number of VSD LC = 1 mm / 10 = 0.1 mm = 0.01 cm (for standard vernier)
Reading
Total reading = Main scale reading + (Vernier coincidence × LC)
Worked Example: In a vernier callipers, 10 VSD = 9 MSD (1 MSD = 1 mm). The main scale reads 3.2 cm and the 6th vernier division coincides with a main scale division. Find the measurement.
LC = 1 mm / 10 = 0.1 mm = 0.01 cm Reading = 3.2 cm + 6 × 0.01 cm = 3.2 + 0.06 = 3.26 cm
4. Screw Gauge (Micrometer)
Principle
When the screw is rotated through ONE complete rotation, it moves by a fixed distance called the PITCH.
Least Count = Pitch / Number of divisions on circular scale
Standard LC = 0.5 mm / 100 = 0.005 mm = 0.0005 cm
Reading
Total reading = Main scale reading + (Circular scale reading × LC)
Worked Example: A screw gauge has a pitch of 1 mm and 100 circular divisions. The main scale reads 2 mm and the circular scale reads 45. Find the thickness.
LC = 1/100 = 0.01 mm Reading = 2 + 45 × 0.01 = 2.45 mm
Zero Error: If the screw gauge reads NON-ZERO when fully closed, this must be CORRECTED.
5. Measurement of Area and Volume
Area
Regular shapes: Formula-based (l×b, πr², etc.) Irregular shapes: GRAPH PAPER method (count squares)
Volume
Regular solids: Formula-based (l³, l×b×h, πr²h) Irregular solids: DISPLACEMENT method using a measuring cylinder Volume of object = Final water level — Initial water level
Volume of liquids: Measuring cylinder or graduated beaker
6. Measurement of Mass
| Instrument | Use |
|---|---|
| Beam balance | Laboratory measurement of mass |
| Electronic balance | Precise digital measurement |
| Spring balance | Measures WEIGHT (force), not mass |
7. Density
Density = Mass / Volume
ρ = m/V
Units: kg/m³ (SI) or g/cm³ (CGS)
1 g/cm³ = 1000 kg/m³
Worked Example: A metal block has mass 500 g and volume 100 cm³. Find its density in g/cm³ and kg/m³.
Density = 500/100 = 5 g/cm³ = 5 × 1000 = 5000 kg/m³
Relative Density (R.D.) = Density of substance / Density of water
R.D. has NO units (it is a ratio).
Common Mistakes and Fixes
| Mistake | Fix |
|---|---|
| 'Confusing mass and weight' | Mass is CONSTANT. Weight = mg, varies with gravity |
| 'Not accounting for zero error' | ALWAYS check zero error BEFORE measuring. Subtract error from reading |
| 'Using m³ for density of small objects' | Use g/cm³ for small objects. 1 g/cm³ = 1000 kg/m³ |
| 'Forgetting units in density calculations' | Density requires BOTH mass and volume units. Write units in every step |
ICSE Exam Focus (5–7 marks)
- 2-mark questions: SI units, conversion of units
- 3-mark questions: Vernier callipers or screw gauge reading
- 4-mark questions: Density calculations with relative density
- 5-mark questions: Volume of irregular solids using displacement method
Self-Test
Q1. What is the SI unit of density? A1. kg/m³ (kilogram per cubic metre).
Q2. The least count of a vernier callipers is 0.01 cm. The main scale reads 2.5 cm and the 8th vernier division coincides. Find the measurement. A2. Reading = 2.5 + 8 × 0.01 = 2.58 cm.
Q3. An irregular stone has mass 60 g. When placed in a measuring cylinder with 50 mL water, the water rises to 75 mL. Find the density. A3. Volume = 75 — 50 = 25 mL = 25 cm³. Density = 60/25 = 2.4 g/cm³.
Q4. What is the difference between mass and weight? A4. Mass is the AMOUNT of matter (constant). Weight is the FORCE due to gravity (= mg). Weight varies with location, mass does not.
Q5. The density of mercury is 13.6 g/cm³. Find its relative density. A5. R.D. = 13.6/1 = 13.6 (density of water = 1 g/cm³).
Q6. A screw gauge has a pitch of 0.5 mm and 100 divisions. What is its least count? A6. LC = 0.5/100 = 0.005 mm.
