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Physics - Physical Quantities and Measurement Techniques

Grade 9IGCSE

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

🔑Concepts

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Physical quantities consist of a numerical magnitude and a unit. For example, in 10 kg10 \text{ kg}, 1010 is the magnitude and kg\text{kg} is the unit.

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The SI base units required for IGCSE are: Length (meter, m\text{m}), Mass (kilogram, kg\text{kg}), Time (second, s\text{s}), Temperature (kelvin, K\text{K}), and Electric Current (ampere, A\text{A}).

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Scalar quantities have magnitude only (e.g., distance, speed, mass, time, energy). Vector quantities have both magnitude and direction (e.g., displacement, velocity, acceleration, force, weight).

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Precision of instruments: A ruler measures to the nearest 1 mm1 \text{ mm} (0.1 cm0.1 \text{ cm}), Vernier calipers to 0.1 mm0.1 \text{ mm} (0.01 cm0.01 \text{ cm}), and a Micrometer screw gauge to 0.01 mm0.01 \text{ mm} (0.001 cm0.001 \text{ cm}).

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To measure the volume of an irregular solid, use the displacement method with a measuring cylinder. The volume is the difference between the final and initial water levels: V=Vfinal−VinitialV = V_{final} - V_{initial}.

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Density is a measure of mass per unit volume. An object will float in a fluid if its density is less than the density of the fluid.

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To find the period TT of a pendulum accurately, measure the time for 2020 oscillations and divide the total time by 2020 to minimize human reaction time errors.

📐Formulae

ρ=mV\rho = \frac{m}{V}

Vcuboid=l×w×hV_{cuboid} = l \times w \times h

Vcylinder=πr2hV_{cylinder} = \pi r^2 h

T=Total TimeNumber of OscillationsT = \frac{\text{Total Time}}{\text{Number of Oscillations}}

Average Speed=Total DistanceTotal Time\text{Average Speed} = \frac{\text{Total Distance}}{\text{Total Time}}

💡Examples

Problem 1:

A metal cylinder has a mass of 135 g135 \text{ g} and a volume of 15 cm315 \text{ cm}^3. Calculate the density of the metal in g/cm3\text{g/cm}^3.

Solution:

ρ=mV=135 g15 cm3=9 g/cm3\rho = \frac{m}{V} = \frac{135 \text{ g}}{15 \text{ cm}^3} = 9 \text{ g/cm}^3

Explanation:

Substitute the mass m=135m = 135 and volume V=15V = 15 into the density formula ρ=mV\rho = \frac{m}{V} to find the result.

Problem 2:

A student measures the time for 4040 oscillations of a pendulum as 32.0 s32.0 \text{ s}. Calculate the period of the pendulum.

Solution:

T=32.0 s40=0.8 sT = \frac{32.0 \text{ s}}{40} = 0.8 \text{ s}

Explanation:

The period TT is the time for one single oscillation. Divide the total time by the total number of swings.

Problem 3:

A rectangular block has dimensions 2.0 cm2.0 \text{ cm} by 5.0 cm5.0 \text{ cm} by 10.0 cm10.0 \text{ cm}. If the block has a mass of 800 g800 \text{ g}, find its density.

Solution:

V=2.0×5.0×10.0=100 cm3V = 2.0 \times 5.0 \times 10.0 = 100 \text{ cm}^3 ρ=800 g100 cm3=8.0 g/cm3\rho = \frac{800 \text{ g}}{100 \text{ cm}^3} = 8.0 \text{ g/cm}^3

Explanation:

First, calculate the volume of the block using V=l×w×hV = l \times w \times h. Then, use the density formula.

Problem 4:

A measuring cylinder contains 50 cm350 \text{ cm}^3 of water. After a stone of mass 120 g120 \text{ g} is lowered into it, the water level rises to 80 cm380 \text{ cm}^3. Find the density of the stone.

Solution:

Vstone=80 cm3−50 cm3=30 cm3V_{stone} = 80 \text{ cm}^3 - 50 \text{ cm}^3 = 30 \text{ cm}^3 ρ=120 g30 cm3=4 g/cm3\rho = \frac{120 \text{ g}}{30 \text{ cm}^3} = 4 \text{ g/cm}^3

Explanation:

Calculate the volume of the stone by the displacement of water (80−5080 - 50). Then divide the mass by this volume.