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Scientific Enquiry - Observation and Measurement

Grade 7IGCSE

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

🔑Concepts

Scientific observations can be qualitative (describing qualities using the five senses) or quantitative (using numerical values and units, such as 25.0 cm25.0 \text{ cm}).

Measurements must always include a magnitude and a unit. The SI (International System of Units) includes the meter (mm) for length, kilogram (kgkg) for mass, and second (ss) for time.

Accuracy refers to how close a measurement is to the true value, while Precision refers to how close a series of measurements are to each other.

Parallax error is a common mistake in measurement caused by viewing a scale at an angle. To avoid this, always read at eye level and perpendicular to the scale.

When measuring liquid in a graduated cylinder, measurements should be taken from the bottom of the meniscus (the curve of the liquid surface).

Independent variables are the factors we change, while Dependent variables are the factors we measure (e.g., measuring the change in temperature ΔT\Delta T over time tt).

Derived units are units calculated from SI base units, such as area (m2m^2), volume (m3m^3), and density (kg/m3kg/m^3).

📐Formulae

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

Volume of a rectangular prism=l×w×h\text{Volume of a rectangular prism} = l \times w \times h

Average (Mean)=recorded valuesn\text{Average (Mean)} = \frac{\sum \text{recorded values}}{n}

Area of a rectangle=l×w\text{Area of a rectangle} = l \times w

💡Examples

Problem 1:

A student measures the mass of a rock as 150 g150\text{ g} and its volume as 30 cm330\text{ cm}^3. Calculate the density of the rock.

Solution:

ρ=150 g30 cm3=5 g/cm3\rho = \frac{150\text{ g}}{30\text{ cm}^3} = 5\text{ g/cm}^3

Explanation:

To find the density, divide the mass (mm) by the volume (VV). The resulting unit is g/cm3\text{g/cm}^3 because the mass was in grams and the volume was in cubic centimeters.

Problem 2:

A beaker contains 50 ml50\text{ ml} of water. After placing a marble inside, the water level rises to 62 ml62\text{ ml}. What is the volume of the marble?

Solution:

Vmarble=VfinalVinitial=62 ml50 ml=12 mlV_{marble} = V_{final} - V_{initial} = 62\text{ ml} - 50\text{ ml} = 12\text{ ml}

Explanation:

This is the displacement method used to measure the volume of irregular objects. Since 1 ml=1 cm31\text{ ml} = 1\text{ cm}^3, the volume is 12 cm312\text{ cm}^3.

Problem 3:

Convert a length of 2.5 meters2.5\text{ meters} into centimeterscentimeters.

Solution:

2.5 m×100=250 cm2.5\text{ m} \times 100 = 250\text{ cm}

Explanation:

There are 100 cm100\text{ cm} in 1 m1\text{ m}, so we multiply the value in meters by 10210^2 to get the value in centimeters.

Observation and Measurement - Revision Notes & Key Formulas | IGCSE Grade 7 Science