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Measurements and Uncertainties - Propagation of Uncertainties

Grade 11IBPhysics

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

🔑Concepts

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Absolute Uncertainty: The actual range of values a measurement could take, denoted as Δx\Delta x. It has the same units as the measurement.

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Fractional (Relative) Uncertainty: The ratio of the absolute uncertainty to the measured value, calculated as Δxx\frac{\Delta x}{x}. This value is dimensionless.

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Percentage Uncertainty: The fractional uncertainty expressed as a percentage: Δxx×100%\frac{\Delta x}{x} \times 100\%.

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Addition and Subtraction Rule: When quantities are added or subtracted, the absolute uncertainties are added together.

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Multiplication and Division Rule: When quantities are multiplied or divided, the fractional (or percentage) uncertainties are added together.

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Power Rule: When a quantity is raised to a power nn, the fractional uncertainty of that quantity is multiplied by ∣n∣|n|.

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Rounding Rules: Calculated uncertainties are typically expressed to one or two significant figures, and the final result must be rounded to the same decimal place as the absolute uncertainty.

📐Formulae

If y=a±b, then Δy=Δa+Δb\text{If } y = a \pm b, \text{ then } \Delta y = \Delta a + \Delta b

If y=a⋅bc, then Δyy=Δaa+Δbb+Δcc\text{If } y = \frac{a \cdot b}{c}, \text{ then } \frac{\Delta y}{y} = \frac{\Delta a}{a} + \frac{\Delta b}{b} + \frac{\Delta c}{c}

If y=an, then Δyy=∣n∣Δaa\text{If } y = a^n, \text{ then } \frac{\Delta y}{y} = |n| \frac{\Delta a}{a}

Percentage Uncertainty=Δxx×100%\text{Percentage Uncertainty} = \frac{\Delta x}{x} \times 100\%

💡Examples

Problem 1:

A student measures the initial temperature of a liquid as T1=(20.0±0.5)∘CT_1 = (20.0 \pm 0.5) ^\circ C and the final temperature as T2=(55.0±0.5)∘CT_2 = (55.0 \pm 0.5) ^\circ C. Calculate the change in temperature ΔT\Delta T and its absolute uncertainty.

Solution:

ΔT=T2−T1=55.0−20.0=35.0∘C\Delta T = T_2 - T_1 = 55.0 - 20.0 = 35.0 ^\circ C. According to the addition/subtraction rule: Δ(ΔT)=ΔT1+ΔT2=0.5+0.5=1.0∘C\Delta (\Delta T) = \Delta T_1 + \Delta T_2 = 0.5 + 0.5 = 1.0 ^\circ C. Final answer: (35.0±1.0)∘C(35.0 \pm 1.0) ^\circ C.

Explanation:

Even though we are subtracting the temperatures, the uncertainties represent a range of possible error, so they must be added to find the maximum possible uncertainty in the result.

Problem 2:

The radius of a sphere is measured to be r=(5.0±0.2) cmr = (5.0 \pm 0.2) \text{ cm}. Calculate the volume VV and its percentage uncertainty. (Use V=43πr3V = \frac{4}{3}\pi r^3)

Solution:

V=43π(5.0)3≈523.6 cm3V = \frac{4}{3} \pi (5.0)^3 \approx 523.6 \text{ cm}^3. Using the power rule for r3r^3: ΔVV=3×Δrr=3×0.25.0=3×0.04=0.12\frac{\Delta V}{V} = 3 \times \frac{\Delta r}{r} = 3 \times \frac{0.2}{5.0} = 3 \times 0.04 = 0.12. Percentage uncertainty: 0.12×100%=12%0.12 \times 100\% = 12\%. Absolute uncertainty ΔV=0.12×523.6≈62.8 cm3\Delta V = 0.12 \times 523.6 \approx 62.8 \text{ cm}^3. Final answer: (520±60) cm3(520 \pm 60) \text{ cm}^3 (rounded to appropriate precision).

Explanation:

Since the volume depends on the cube of the radius, the fractional uncertainty of the radius is tripled to find the fractional uncertainty of the volume.

Problem 3:

An object travels a distance s=(100±2) ms = (100 \pm 2) \text{ m} in a time t=(20.0±0.5) st = (20.0 \pm 0.5) \text{ s}. Calculate the average speed vv and its absolute uncertainty.

Solution:

v=st=10020.0=5.0 m s−1v = \frac{s}{t} = \frac{100}{20.0} = 5.0 \text{ m s}^{-1}. Using the division rule: Δvv=Δss+Δtt=2100+0.520.0=0.02+0.025=0.045\frac{\Delta v}{v} = \frac{\Delta s}{s} + \frac{\Delta t}{t} = \frac{2}{100} + \frac{0.5}{20.0} = 0.02 + 0.025 = 0.045. Δv=0.045×5.0=0.225 m s−1\Delta v = 0.045 \times 5.0 = 0.225 \text{ m s}^{-1}. Final answer: (5.0±0.2) m s−1(5.0 \pm 0.2) \text{ m s}^{-1}.

Explanation:

For division, we sum the fractional uncertainties. The resulting absolute uncertainty is then rounded to one significant figure, and the mean value is adjusted to match that decimal place.