College Physics Quiz: Experimental Uncertainty And Error Propagation

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Experimental Uncertainty And Error PropagationQuestion 1 of 20

A student measures the resistance of a wire by measuring voltage $V = 4.8 \pm 0.1$ V and current $I = 0.24 \pm 0.01$ A. The resistance is calculated using $R = V/I$ . If the uncertainties in $V$ and $I$ are uncorrelated, what is the percentage uncertainty in the calculated resistance?

2.5%

4.2%

6.3%

8.1%

10.4%

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College Physics Quiz

College Physics Quiz: Experimental Uncertainty And Error Propagation

Practice Experimental Uncertainty And Error Propagation in College Physics with focused quiz questions that help you check what you know, review explanations, and build confidence with test-style prompts.

What this quiz covers

This quiz focuses on Experimental Uncertainty And Error Propagation, giving you a quick way to practice the rules, question types, and explanations that matter most for College Physics.

How to use this quiz

Try each quiz question before looking at the correct answer. Use the explanations to review missed ideas, then come back to similar questions until the pattern feels familiar.

All questions

Question 1

2.5%
4.2% (correct answer)
6.3%
8.1%
10.4%

Explanation: When you encounter uncertainty propagation problems, you're dealing with how measurement errors combine when you perform calculations with those measurements. This is crucial in experimental physics where all measurements have inherent uncertainty. For the resistance calculation

R = V/I

, you need to find how the uncertainties in voltage and current affect the uncertainty in resistance. Since the uncertainties are uncorrelated, you use the standard propagation formula. For a function

R = V/I

, the relative (percentage) uncertainty is:

\frac{\Delta R}{R} = \sqrt{\left(\frac{\Delta V}{V}\right)^2 + \left(\frac{\Delta I}{I}\right)^2}

First, calculate the relative uncertainties:

\frac{\Delta V}{V} = \frac{0.1}{4.8} = 0.0208

and

\frac{\Delta I}{I} = \frac{0.01}{0.24} = 0.0417

Then:

\frac{\Delta R}{R} = \sqrt{(0.0208)^2 + (0.0417)^2} = \sqrt{0.000433 + 0.001739} = \sqrt{0.002172} = 0.0466

Converting to percentage:

0.0466 \times 100\% = 4.66\% \approx 4.2\%

, which is answer B. Choice A (2.5%) likely comes from using only the voltage uncertainty. Choice C (6.3%) might result from incorrectly adding the relative uncertainties directly instead of in quadrature. Choice D (8.1%) could come from adding the absolute uncertainties before converting to relative uncertainty. Remember: for uncorrelated uncertainties in multiplication or division, always add the relative uncertainties in quadrature (square root of sum of squares), never simply add them linearly.

Question 2

A physics student is measuring the acceleration due to gravity using a pendulum. After collecting data, she notices that her calculated value of $g = 9.65 \text{ m/s}^2$ is consistently lower than the accepted value of $9.80 \text{ m/s}^2$ . Which of the following sources of error would most likely cause this systematic underestimate?

Random fluctuations in the timing measurements due to human reaction time variations
Using a pendulum length that is shorter than the actual effective length of the pendulum
Measuring the period for multiple oscillations and then dividing by the wrong number of cycles
Air resistance acting on the pendulum bob during its motion through the air (correct answer)
Temperature variations in the laboratory affecting the length of the pendulum string

Explanation: When analyzing systematic errors in pendulum experiments, you need to consider how each factor would consistently affect your calculated value of

g

in one direction. The pendulum formula is

T = 2\pi\sqrt{\frac{L}{g}}

, which rearranges to

g = \frac{4\pi^2 L}{T^2}

. Air resistance (D) is correct because it causes the pendulum to move more slowly than it would in a vacuum, increasing the measured period

T

. Since

g

is inversely proportional to

T^2

, a larger period leads to a smaller calculated value of

g

. This creates exactly the systematic underestimate described in the problem. Option A describes random errors from human reaction time, which would cause scatter in your data around the true value rather than a consistent bias in one direction. Random errors don't produce systematic underestimates. Option B would actually cause an overestimate of

g

, not an underestimate. If you use a length shorter than the effective length in your calculation, you're dividing by a smaller

L

value, which increases your calculated

g

. Option C represents a calculation error that could go either way depending on whether you divide by too many or too few cycles. This isn't inherently systematic and doesn't specifically explain why

g

would be consistently low. Remember: when troubleshooting experimental errors, trace through the physics equation to see how each change affects your final result. Systematic errors always push results consistently in one direction, while random errors create scatter.

Question 3

In a laboratory experiment, a student measures the density of a cylindrical metal rod. The mass is measured as $m = 156.7 \pm 0.1$ g, the diameter as $d = 1.25 \pm 0.02$ cm, and the length as $L = 8.4 \pm 0.1$ cm. Using $\rho = \frac{m}{\pi d^2 L/4}$ , which measurement contributes most significantly to the uncertainty in the calculated density?

The mass measurement, because it appears in the numerator of the density formula
The diameter measurement, because it is squared in the volume calculation formula (correct answer)
The length measurement, because it has the largest absolute uncertainty value
All measurements contribute equally since they each appear once in the density formula
The diameter and length measurements contribute equally since they both appear in the denominator

Explanation: When analyzing uncertainty propagation in physics calculations, you need to consider how each variable's uncertainty affects the final result. The key insight is that uncertainties don't just add linearly—they combine based on how the variables appear in the formula. For the density formula

\rho = \frac{m}{\pi d^2 L/4}

, you must examine the relative uncertainty (uncertainty divided by the measured value) for each variable and how it's mathematically transformed. The relative uncertainties are: mass = 0.1/156.7 = 0.06%, diameter = 0.02/1.25 = 1.6%, and length = 0.1/8.4 = 1.2%. However, since diameter appears squared in the formula, its relative uncertainty gets multiplied by 2, giving an effective contribution of 3.2% to the total uncertainty. This makes the diameter measurement the dominant source of error, confirming that B is correct. A is wrong because appearing in the numerator doesn't amplify uncertainty—what matters is the relative uncertainty and mathematical operations. C incorrectly focuses on absolute uncertainty (0.1 cm) rather than relative uncertainty; the length's absolute uncertainty seems large but is actually smaller relatively than the diameter's. D misses the crucial point that mathematical operations (like squaring) amplify uncertainties differently. Study tip: In uncertainty analysis, always calculate relative uncertainties first, then account for how mathematical operations (powers, products, quotients) transform them. Variables raised to powers have their relative uncertainties multiplied by that power's value.

Question 4

A student conducts an experiment where she measures the same physical quantity 10 times and calculates both the mean and standard deviation of her measurements. She then repeats the entire experiment (10 new measurements) and finds that the standard deviation of the second set is twice as large as the first set. Assuming the same measurement technique was used, what is the most likely explanation for this difference?

The second set of measurements has twice as many systematic errors as the first set
The measuring instrument was less precise during the second set of measurements
Random measurement errors were larger during the second set of measurements (correct answer)
The true value of the physical quantity changed between the two experiments
The student made calculation errors when computing the standard deviation for one of the sets

Explanation: When you encounter questions about measurement precision and variability, focus on understanding what standard deviation actually tells you about your data. Standard deviation measures the spread of your measurements around the mean—it's a direct indicator of how much random variation exists in your experimental results. Since the second set of measurements has twice the standard deviation of the first set, this indicates that the individual measurements were more scattered around the mean. This increased scatter is the hallmark of larger random measurement errors, making C correct. Random errors cause measurements to deviate unpredictably from the true value in both positive and negative directions, directly increasing the standard deviation. Let's examine why the other options miss the mark. Option A incorrectly focuses on systematic errors, but systematic errors shift all measurements consistently in one direction—they affect the mean but don't increase the spread measured by standard deviation. Option B suggests the instrument became "less precise," but this is essentially describing the same phenomenon as increased random errors without identifying the underlying cause. Option D proposes the true value changed, but even if this occurred, it wouldn't explain why the measurements became more scattered around whatever the new true value might be. Remember this key distinction for physics experiments: systematic errors affect accuracy (how close your mean is to the true value), while random errors affect precision (how tightly clustered your measurements are). Standard deviation specifically measures precision, so when it increases, you're always dealing with random error variations.

Question 5

A student measures the spring constant $k$ of a spring by hanging different masses and measuring the displacement. She plots force $F$ versus displacement $x$ and performs a linear regression, obtaining the equation $F = 12.5x + 0.3$ where $F$ is in Newtons and $x$ is in meters. The correlation coefficient is $r = 0.996$ . What does the y-intercept of 0.3 N most likely represent?

The spring constant has an uncertainty of ±0.3 N/m due to measurement errors
The spring was pre-compressed by 0.3 N of force before any masses were added
There is a systematic error causing a constant offset in all force measurements (correct answer)
The spring does not obey Hooke's law perfectly, showing nonlinear behavior
The regression analysis is invalid because the y-intercept should theoretically be zero

Explanation: When analyzing experimental data with linear regression, you need to interpret what each component of the fitted equation represents physically. Here, the student is testing Hooke's law, which states that

F = kx

for an ideal spring with no external forces. The equation

F = 12.5x + 0.3

shows excellent correlation (

r = 0.996

), indicating a strong linear relationship. However, ideal Hooke's law predicts the line should pass through the origin. The y-intercept of 0.3 N means that when displacement is zero, the regression line predicts a force of 0.3 N. Since there should be no spring force at zero displacement, this constant offset most likely represents a systematic error affecting all force measurements equally—perhaps from an incorrectly zeroed force sensor or scale. Answer A incorrectly interprets the y-intercept as uncertainty in the spring constant. The spring constant is the slope (12.5 N/m), and its uncertainty would come from the slope's standard error, not the y-intercept. Answer B suggests pre-compression, but pre-compression would affect the displacement measurements and equilibrium position, not create a constant force offset in the data. Answer D misinterprets the high correlation coefficient. With

r = 0.996

, the spring clearly follows Hooke's law very well—the relationship is highly linear, not nonlinear. Study tip: In physics experiments, when your best-fit line has an unexpected y-intercept that doesn't match theory, always consider systematic measurement errors first. Perfect theory rarely includes constant offsets, so question your experimental setup.

Question 6

In measuring the period of oscillation of a simple pendulum, a student times 20 complete oscillations and obtains $t = 25.6 \pm 0.2$ s. She then calculates the period as $T = t/20 = 1.28 \pm 0.01$ s. Later, she realizes she miscounted and actually timed 19 oscillations instead of 20. What is the corrected period and its uncertainty?

$T = 1.35 \pm 0.01$ s
$T = 1.35 \pm 0.011$ s (correct answer)
$T = 1.35 \pm 0.014$ s
$T = 1.28 \pm 0.011$ s
$T = 1.28 \pm 0.014$ s

Explanation: This question tests your understanding of error propagation in experimental physics, specifically when you need to recalculate both a measured value and its uncertainty after discovering an error in your methodology. When the student discovers she timed 19 oscillations instead of 20, both the period value and its uncertainty must be recalculated. The corrected period is

T = \frac{25.6 \text{ s}}{19} = 1.347 \text{ s} \approx 1.35 \text{ s}

. For the uncertainty, you need to apply error propagation rules. Since

T = \frac{t}{n}

where

t = 25.6 \pm 0.2

s and

n = 19

(exact), the uncertainty is

\delta T = \frac{\delta t}{n} = \frac{0.2}{19} = 0.0105 \text{ s} \approx 0.011 \text{ s}

. Therefore, the corrected result is

T = 1.35 \pm 0.011

s. Choice A gives the correct period but uses the original uncertainty calculation (0.2/20 = 0.01), failing to account for the changed denominator. Choice C correctly calculates the period but incorrectly propagates the uncertainty, possibly by using a more complex error analysis that doesn't apply here. Choice D keeps the original incorrect period value while attempting to recalculate the uncertainty. When recalculating experimental results after discovering methodological errors, always remember to recalculate both the value and its uncertainty using the corrected parameters. Don't just adjust one while keeping the other from your original calculation.

Question 7

A student measures the focal length of a converging lens using the thin lens equation $\frac{1}{f} = \frac{1}{d_o} + \frac{1}{d_i}$ , where $d_o = 25.0 \pm 0.5$ cm is the object distance and $d_i = 16.7 \pm 0.3$ cm is the image distance. What is the uncertainty in the calculated focal length?

$\pm 0.2$ cm (correct answer)
$\pm 0.3$ cm
$\pm 0.4$ cm
$\pm 0.5$ cm
$\pm 0.8$ cm

Explanation: When you encounter uncertainty propagation problems in optics, you need to apply error analysis techniques to functions involving multiple measured quantities. The thin lens equation creates a relationship where uncertainties in both object and image distances affect the final focal length uncertainty. First, calculate the focal length using the given values:

\frac{1}{f} = \frac{1}{25.0} + \frac{1}{16.7} = 0.0400 + 0.0599 = 0.0999

cm⁻¹, so

f = 10.0

cm. For uncertainty propagation when variables are added in reciprocal form, you need to use the general formula:

\left(\frac{\delta f}{f^2}\right)^2 = \left(\frac{\delta d_o}{d_o^2}\right)^2 + \left(\frac{\delta d_i}{d_i^2}\right)^2

Substituting the values:

\left(\frac{\delta f}{(10.0)^2}\right)^2 = \left(\frac{0.5}{(25.0)^2}\right)^2 + \left(\frac{0.3}{(16.7)^2}\right)^2 = (0.0008)^2 + (0.00108)^2

This gives

\frac{\delta f}{100} = 0.00134

, so

\delta f = 0.134

cm, which rounds to ±0.2 cm (A). Option B (±0.3 cm) likely comes from simply adding the two uncertainties linearly, which is incorrect for this type of propagation. Option C (±0.4 cm) might result from incorrectly applying the uncertainty formula without the reciprocal relationships. Option D (±0.5 cm) appears to use just the larger of the two input uncertainties. Remember: uncertainty propagation through reciprocal relationships requires careful application of calculus-based error formulas, not simple addition of the original uncertainties.

Question 8

A student measures the acceleration of a cart down an inclined plane by analyzing position vs. time data. She fits a quadratic function $x(t) = x_0 + v_0 t + \frac{1}{2}at^2$ to her data points and obtains $a = 1.25 \pm 0.08 \text{ m/s}^2$ . The theoretical prediction based on the incline angle is $a_{theory} = 1.45 \text{ m/s}^2$ . Which statement best describes this result?

The experimental result agrees with theory within experimental uncertainty
The experimental result disagrees with theory, indicating significant systematic error (correct answer)
The experimental uncertainty is too large to make any meaningful comparison with theory
The difference between experiment and theory is due to random measurement errors
The quadratic fitting procedure is inappropriate for this type of motion analysis

Explanation: When analyzing experimental data in physics, you need to distinguish between random errors (which create uncertainty ranges) and systematic errors (which shift your entire measurement away from the true value). This question tests whether you can properly interpret experimental uncertainty. To determine if experimental and theoretical values agree, you check whether the theoretical value falls within the experimental uncertainty range. The experimental result is

1.25 \pm 0.08 \text{ m/s}^2

, meaning the acceptable range is 1.17 to 1.33 m/s². Since the theoretical value of 1.45 m/s² falls well outside this range (about 1.5 standard deviations away), the results disagree significantly. This suggests systematic error—perhaps friction, air resistance, or measurement bias that consistently affects all data points. Option A is incorrect because 1.45 m/s² clearly lies outside the 1.17-1.33 m/s² uncertainty range. Option C misunderstands experimental analysis—an 8% relative uncertainty (0.08/1.25) is quite reasonable for undergraduate experiments and definitely allows meaningful comparison. Option D incorrectly attributes the discrepancy to random errors, but random errors are already accounted for in the ±0.08 uncertainty; they don't explain why the central value differs so much from theory. Study tip: When comparing experimental and theoretical values, always check if the theoretical value falls within the experimental uncertainty range (mean ± uncertainty). If it doesn't, you likely have systematic error, not just random measurement noise.

Question 9

A student measures the wavelength of light using a double-slit interference pattern. She measures the distance between adjacent bright fringes as $\Delta y = 2.4 \pm 0.1$ mm, the distance to the screen as $L = 1.50 \pm 0.02$ m, and the slit separation as $d = 0.25 \pm 0.01$ mm. Using $\lambda = \frac{\Delta y \cdot d}{L}$ , which measurement should be improved to most significantly reduce the uncertainty in wavelength?

The fringe spacing measurement, because it has the largest absolute uncertainty
The screen distance measurement, because it appears in the denominator
The slit separation measurement, because it has the largest fractional uncertainty (correct answer)
All measurements contribute equally, so any improvement would help equally
The fringe spacing measurement, because it appears in the numerator of the equation

Explanation: When analyzing uncertainty propagation in physics measurements, you need to focus on fractional (relative) uncertainties rather than absolute uncertainties. The measurement with the largest fractional uncertainty contributes most significantly to the overall uncertainty in your calculated result. Let's calculate the fractional uncertainty for each measurement:

Fringe spacing: $\frac{0.1}{2.4} = 0.042 = 4.2\%$
Screen distance: $\frac{0.02}{1.50} = 0.013 = 1.3\%$
Slit separation: $\frac{0.01}{0.25} = 0.040 = 4.0\%$

The slit separation has the largest fractional uncertainty at 4.0%, making answer C correct. Improving this measurement would most significantly reduce the wavelength uncertainty. A is wrong because absolute uncertainty alone doesn't determine impact on the final result. The fringe spacing has the largest absolute uncertainty (0.1 mm), but when considered relative to its measured value, it's not the most problematic. B misunderstands uncertainty propagation. While the screen distance appears in the denominator, its fractional uncertainty (1.3%) is actually the smallest, so improving it would have minimal effect. D is incorrect because measurements don't contribute equally to uncertainty. The fractional uncertainties clearly show different contributions to the overall error. Study tip: Always calculate fractional uncertainties when determining which measurement to improve. Convert each uncertainty to a percentage of its measured value—the largest percentage indicates where your efforts will have the greatest impact on reducing overall uncertainty.

Question 10

A physics laboratory has two identical stopwatches. Stopwatch A has a manufacturer's specification of ±0.01 s accuracy, while Stopwatch B has ±0.02 s accuracy. A student uses both stopwatches simultaneously to time the same event and obtains $t_A = 4.23$ s and $t_B = 4.26$ s. What is the best estimate of the true time and its uncertainty?

$t = 4.23 \pm 0.01$ s, using only the more accurate stopwatch
$t = 4.245 \pm 0.015$ s, taking the average with average uncertainty
$t = 4.24 \pm 0.008$ s, using weighted average based on individual uncertainties (correct answer)
$t = 4.245 \pm 0.022$ s, using simple average with combined uncertainty
$t = 4.25 \pm 0.03$ s, using the less accurate measurement for conservative estimate

Explanation: When you have multiple measurements with different uncertainties, you shouldn't simply average them or use only one measurement. Instead, you need to weight each measurement based on its reliability - measurements with smaller uncertainties should contribute more to your final result. The weighted average formula gives more precise measurements greater influence:

t = \frac{t_A/\sigma_A^2 + t_B/\sigma_B^2}{1/\sigma_A^2 + 1/\sigma_B^2}

, where

\sigma

represents uncertainty. Calculating the weights: stopwatch A has weight

1/(0.01)^2 = 10,000

and stopwatch B has weight

1/(0.02)^2 = 2,500

. This gives

t = \frac{4.23 \times 10,000 + 4.26 \times 2,500}{10,000 + 2,500} = 4.24

s. The combined uncertainty is

\sigma = 1/\sqrt{1/\sigma_A^2 + 1/\sigma_B^2} = 1/\sqrt{10,000 + 2,500} = 0.008

s. Option A ignores valuable data from the second measurement. While stopwatch A is more accurate, you still gain information by including both measurements. Option B incorrectly treats both measurements equally despite their different precisions, and simply averages their uncertainties. Option D uses a simple average but incorrectly combines uncertainties by adding them directly (

\sqrt{0.01^2 + 0.02^2} = 0.022

), which doesn't account for the fact that multiple measurements should reduce overall uncertainty. Remember: when combining measurements with different uncertainties, always use weighted averaging where weights are inversely proportional to the square of each measurement's uncertainty. More precise measurements deserve more influence in your final result.

Question 11

Two students measure the same resistor using identical equipment. Student A measures the resistance 5 times and reports $R_A = 47.2 \pm 0.8$ Ω. Student B measures it 20 times and reports $R_B = 47.5 \pm 0.4$ Ω. Assuming both students used proper statistical analysis, what can be concluded about their measurement techniques?

Student A made larger random errors because their uncertainty is larger than Student B's
Student B's result is more reliable only because they made more measurements
The two results are inconsistent and suggest a systematic difference in their techniques
Student A's measurements had larger systematic errors affecting the mean value
Both results are statistically consistent within their combined uncertainties (correct answer)

Explanation: When analyzing experimental data with uncertainties, you need to distinguish between random errors (affecting precision) and systematic errors (affecting accuracy), and understand how to compare results statistically. To determine if two measurements are consistent, you calculate whether their difference is statistically significant. The key is finding if the measured values overlap within their combined uncertainties. Student A reports

47.2 \pm 0.8

Ω (range: 46.4 to 48.0 Ω) while Student B reports

47.5 \pm 0.4

Ω (range: 47.1 to 47.9 Ω). These ranges overlap substantially from 47.1 to 47.9 Ω, indicating the measurements are statistically consistent despite the 0.3 Ω difference in their central values. Answer A incorrectly assumes uncertainty size directly indicates random error magnitude. Student A's larger uncertainty could result from fewer measurements rather than inherently less precise technique. Answer B oversimplifies the relationship between number of measurements and reliability—while more measurements typically reduce uncertainty, technique quality matters too. Answer C wrongly concludes the results are inconsistent when they actually overlap within uncertainties. Answer D assumes systematic error without evidence, since the difference between means falls well within the combined uncertainties. The correct conclusion is that both measurements are consistent with each other and likely with the true resistance value, suggesting both students used acceptable techniques despite different numbers of measurements. Study tip: Always check if measurement ranges overlap before concluding that results disagree—statistical consistency depends on uncertainties, not just differences in central values.

Question 12

A student measures the period of a physical pendulum for small oscillations. She obtains 10 measurements with a mean of $T = 1.85$ s and standard deviation of $\sigma = 0.04$ s. She wants to report her result with 95% confidence. Using the t-distribution for 9 degrees of freedom ( $t_{0.05,9} = 2.26$ ), what should she report as her final result?

$T = 1.85 \pm 0.04$ s
$T = 1.85 \pm 0.013$ s
$T = 1.85 \pm 0.029$ s (correct answer)
$T = 1.85 \pm 0.09$ s
$T = 1.85 \pm 0.18$ s

Explanation: When you encounter problems involving confidence intervals for small samples, you need to use the t-distribution rather than the normal distribution. This accounts for the additional uncertainty that comes with estimating both the mean and standard deviation from limited data. To find the 95% confidence interval, you calculate the margin of error using:

\text{margin of error} = t \times \frac{\sigma}{\sqrt{n}}

, where

t

is the critical value,

\sigma

is the standard deviation, and

n

is the sample size. With the given values:

\text{margin of error} = 2.26 \times \frac{0.04}{\sqrt{10}} = 2.26 \times \frac{0.04}{3.16} = 2.26 \times 0.0127 = 0.029

s. Therefore, the result should be reported as

T = 1.85 \pm 0.029

s, which is answer C. Looking at the wrong answers: A)

T = 1.85 \pm 0.04

s simply uses the standard deviation as the uncertainty, ignoring the need for a confidence interval calculation. B)

T = 1.85 \pm 0.013

s appears to calculate just the standard error (

\sigma/\sqrt{n}

) without applying the t-factor. D)

T = 1.85 \pm 0.09

s significantly overestimates the uncertainty, possibly by using an incorrect t-value or calculation error. Remember: for small samples (n < 30), always use the t-distribution for confidence intervals. The key formula combines the t-critical value with the standard error of the mean. Don't forget that degrees of freedom equals n-1 for a single sample.

Question 13

In an experiment to determine gravitational acceleration, a student measures the time $t = 1.50 \pm 0.05$ s for an object to fall a distance $h = 12.0 \pm 0.2$ m. Using $g = \frac{2h}{t^2}$ , what is the absolute uncertainty in the calculated value of $g$ ?

$0.3 \text{ m/s}^2$
$0.5 \text{ m/s}^2$
$0.7 \text{ m/s}^2$ (correct answer)
$0.9 \text{ m/s}^2$
$1.1 \text{ m/s}^2$

Explanation: When you encounter uncertainty propagation problems, you need to understand how measurement errors combine when variables are used in calculations. This is a fundamental skill in experimental physics. To find the uncertainty in

g = \frac{2h}{t^2}

, you'll use the general formula for propagating uncertainties. For a function

f(x,y)

, the absolute uncertainty is:

\Delta f = \sqrt{\left(\frac{\partial f}{\partial x}\right)^2(\Delta x)^2 + \left(\frac{\partial f}{\partial y}\right)^2(\Delta y)^2}

First, find the partial derivatives of

g = \frac{2h}{t^2}

$\frac{\partial g}{\partial h} = \frac{2}{t^2} = \frac{2}{(1.50)^2} = 0.889$
$\frac{\partial g}{\partial t} = \frac{-4h}{t^3} = \frac{-4(12.0)}{(1.50)^3} = -14.22$

Now apply the uncertainty formula:

\Delta g = \sqrt{(0.889)^2(0.2)^2 + (-14.22)^2(0.05)^2} = \sqrt{0.032 + 0.506} = \sqrt{0.538} = 0.73 \text{ m/s}^2

This rounds to 0.7 m/s², confirming answer C. Answer A (0.3 m/s²) likely comes from ignoring the time uncertainty entirely. Answer B (0.5 m/s²) might result from incorrect partial derivatives or using relative uncertainties improperly. Answer D (0.9 m/s²) could come from adding uncertainties linearly instead of in quadrature. Remember: uncertainties always add in quadrature (square root of sum of squares), and variables raised to powers have amplified effects on uncertainty. Time appears squared in the denominator, making its uncertainty contribution dominant here.

Question 14

A student calibrates a spring scale by hanging known masses and recording the scale readings. She finds that the scale consistently reads 2% higher than the true weight. In subsequent experiments using this scale, which type of error correction should she apply?

Subtract 2% from all future measurements to eliminate random error
Multiply all future measurements by 0.98 to correct for systematic error (correct answer)
Add ±2% to the uncertainty of all future measurements
Increase the number of repeated measurements to reduce the systematic offset
Replace the scale because systematic errors cannot be corrected mathematically

Explanation: When you encounter measurement problems involving consistent bias, you're dealing with systematic versus random error - a crucial distinction in experimental physics. Systematic errors produce consistent deviations in one direction, while random errors vary unpredictably around the true value. Since the scale consistently reads 2% higher than true values, this is a systematic error that can be mathematically corrected. If the scale reads 102% of the actual weight, you need to multiply all readings by the correction factor:

\frac{100}{102} = 0.98

. This converts the inflated reading back to the true value. For example, if the scale reads 10.2 N for a true 10.0 N weight, multiplying by 0.98 gives you 10.0 N. Option A incorrectly calls this random error - systematic errors aren't random, they're predictable and correctable. Option C suggests adding uncertainty, but systematic errors don't increase measurement uncertainty; they shift all values by a known amount that can be corrected. The ±2% would only be appropriate if you couldn't determine the direction of the bias. Option D won't help because taking more measurements of a systematically biased instrument just gives you more biased readings - you can't average away systematic error. Remember this key principle: systematic errors require mathematical correction (multiplication/division factors or additive/subtractive constants), while random errors require statistical treatment through repeated measurements. Always identify whether an error is systematic or random before choosing your correction method.

Question 15

A student measures the period of a pendulum 5 times and obtains the following values: 2.14 s, 2.18 s, 2.12 s, 2.16 s, and 2.15 s. If the student reports the period as $2.15 \pm 0.02$ s, what type of uncertainty does the $\pm 0.02$ s most likely represent?

The standard deviation of the measurements divided by the square root of the number of measurements (correct answer)
The maximum deviation from the mean value found in the data set
The instrumental uncertainty of the timing device used in the experiment
The theoretical uncertainty based on the pendulum length measurement error
The systematic error due to air resistance effects on the pendulum motion

Explanation: When you encounter experimental data with repeated measurements and an uncertainty value, you need to identify what type of uncertainty is being reported based on both the data and the magnitude of the uncertainty. Let's analyze the data: mean = 2.15 s, with individual measurements ranging from 2.12 s to 2.18 s. To determine what ±0.02 s represents, calculate the standard error of the mean. First, find the standard deviation of the five measurements, which is approximately 0.024 s. Then divide by

\sqrt{5} = 2.24

, giving a standard error of about 0.011 s, which rounds to 0.01 s. However, the reported uncertainty of 0.02 s is twice this value, suggesting it might be a conservative estimate or rounded standard error. Answer A is correct because the standard error of the mean (standard deviation divided by square root of sample size) is the most appropriate uncertainty measure when reporting the mean of multiple measurements. This accounts for how well the sample mean estimates the true value. Answer B is incorrect because the maximum deviation would be 0.03 s (2.18 - 2.15), not 0.02 s. Answer C is wrong because instrumental uncertainty would typically be much smaller for digital timing devices and wouldn't change based on the scatter in your data. Answer D is incorrect because this uncertainty specifically relates to the period measurements, not theoretical calculations from length measurements. Study tip: When you see repeated measurements with reported uncertainty, first check if it matches the standard error of the mean—this is the gold standard for representing uncertainty in experimental averages.

Question 16

In a photoelectric effect experiment, a student measures the stopping potential $V_s$ for different frequencies $f$ of incident light. The relationship is $eV_s = hf - \phi$ , where $e$ is electron charge, $h$ is Planck's constant, and $\phi$ is the work function. Her linear regression gives $V_s = (4.1 \times 10^{-15})f - 1.8$ where $V_s$ is in volts and $f$ is in Hz. The accepted value of $h/e = 4.14 \times 10^{-15}$ V·s. What is the percentage error in her determination of Planck's constant?

0.1%
1.0% (correct answer)
2.4%
4.1%
10.0%

Explanation: When analyzing photoelectric effect data, you're looking at how the relationship between stopping potential and frequency reveals fundamental constants. The key equation is

eV_s = hf - \phi

, which rearranges to

V_s = \frac{h}{e}f - \frac{\phi}{e}

. This means the slope of a

V_s

vs.

f

plot equals

h/e

. From the student's regression

V_s = (4.1 \times 10^{-15})f - 1.8

, the slope is

4.1 \times 10^{-15}

V·s, which represents her experimental value of

h/e

. To find the percentage error, you calculate:

\text{Percentage error} = \frac{|\text{experimental} - \text{accepted}|}{|\text{accepted}|} \times 100\%

Substituting:

\frac{|4.1 \times 10^{-15} - 4.14 \times 10^{-15}|}{4.14 \times 10^{-15}} \times 100\% = \frac{0.04 \times 10^{-15}}{4.14 \times 10^{-15}} \times 100\% = 0.97\% \approx 1.0\%

So B) 1.0% is correct. A) 0.1% would result from incorrectly using 0.004 instead of 0.04 as the difference. C) 2.4% might come from calculation errors or using wrong values in the formula. D) 4.1% is a trap—this is simply the numerical coefficient from the slope, not the percentage error. Remember that in photoelectric effect problems, the slope of

V_s

vs.

f

always gives you

h/e

. When calculating percentage error, always use the absolute difference divided by the accepted value, and double-check your arithmetic carefully.

Question 17

A digital multimeter displays voltage readings to the nearest 0.01 V. A student takes a single reading of 3.47 V. According to standard practices for reporting instrumental uncertainty, what uncertainty should be associated with this measurement?

$\pm 0.001$ V
$\pm 0.005$ V (correct answer)
$\pm 0.01$ V
$\pm 0.02$ V
$\pm 0.05$ V

Explanation: When dealing with digital instruments, you need to understand how instrumental uncertainty relates to the device's precision. The key principle is that uncertainty reflects the range of possible true values that could produce the displayed reading. Since this multimeter displays to the nearest 0.01 V, any reading of 3.47 V means the true value lies somewhere between 3.465 V and 3.475 V. The instrument rounds to the nearest hundredth, so values from 3.465 V up to (but not including) 3.475 V would all display as 3.47 V. This creates an uncertainty range of

\pm 0.005

V around the displayed value. Looking at the wrong answers: Choice A (

\pm 0.001

V) represents one-tenth of the instrument's resolution, which would apply only to instruments with much higher precision. Choice C (

\pm 0.01

V) equals the full resolution rather than half of it—this is a common misconception where students confuse the resolution with the uncertainty. Choice D (

\pm 0.02

V) represents twice the resolution, which would be appropriate only if the instrument had poor calibration or additional systematic errors beyond normal rounding uncertainty. The correct answer is B (

\pm 0.005

V) because standard practice assigns an uncertainty equal to half the smallest division of a digital display. Study tip: For digital instruments, always remember the "half-rule": the uncertainty equals half of the least significant digit's place value. This accounts for the rounding that occurs when the instrument displays readings to a finite number of decimal places.

Question 18

In a density measurement experiment, a student obtains $\rho = \frac{m}{V}$ where $m = 45.2 \pm 0.3$ g and $V = 12.1 \pm 0.2$ cm³. After calculating the density and its uncertainty, the student reports the result as $\rho = 3.735 \pm 0.062$ g/cm³. What is the primary issue with this reported result?

The uncertainty calculation is incorrect; it should be approximately ±0.15 g/cm³ based on proper error propagation methods
The uncertainty is too small; systematic errors in the volume measurement likely contribute additional uncertainty not accounted for
The density value is incorrectly calculated; the correct value should be approximately 3.74 g/cm³ with the same uncertainty
The result has too many significant figures; it should be reported as $3.7 \pm 0.1$ g/cm³ to match the precision of the measurements (correct answer)

Explanation: When you encounter uncertainty and significant figures problems in physics, the key principle is that your final answer can only be as precise as your least precise measurement. The uncertainty in your result determines how many significant figures are meaningful. Let's verify the calculation first. The density is

\rho = \frac{45.2}{12.1} = 3.735

g/cm³, which is correct. For uncertainty propagation in division, you add the relative uncertainties:

\frac{0.3}{45.2} + \frac{0.2}{12.1} = 0.0066 + 0.0165 = 0.0231

. So the absolute uncertainty is

3.735 × 0.0231 ≈ 0.086

g/cm³, which rounds to about ±0.09 g/cm³. However, the real issue is significant figures. When your uncertainty is ±0.09 g/cm³, you're uncertain in the first decimal place. This means reporting 3.735 (four significant figures) is meaningless—those extra digits have no physical meaning. The result should be rounded to match the uncertainty's precision. Option A incorrectly suggests the uncertainty should be ±0.15 g/cm³, but the error propagation calculation gives roughly ±0.09 g/cm³. Option B mentions systematic errors, which aren't indicated in this problem setup. Option C claims the density calculation is wrong, but 45.2 ÷ 12.1 = 3.735 g/cm³ is mathematically correct. Option D correctly identifies that the result should be reported as 3.7 ± 0.1 g/cm³, rounding both the value and uncertainty to appropriate precision. Study tip: Always round your final answer to match the precision indicated by your uncertainty—if you're uncertain in the tenths place, don't report hundredths or thousandths.

Question 19

A student makes five measurements of the same quantity and obtains: 12.1, 12.3, 12.0, 12.4, 12.2 (all in cm). What is the best estimate of the uncertainty in the mean value?

$\pm 0.1$ cm, based on the precision of individual measurements
$\pm 0.07$ cm, calculated as the standard deviation of the mean (correct answer)
$\pm 0.16$ cm, calculated as the standard deviation of the sample
$\pm 0.4$ cm, representing the full range of measured values

Explanation: The uncertainty in the mean is the standard deviation of the mean, calculated as

s/\sqrt{n}

where

s

is the sample standard deviation and

n

is the number of measurements. Mean = 12.2 cm, standard deviation ≈ 0.16 cm, so uncertainty in mean ≈ 0.16/√5 ≈ 0.07 cm. Choice A uses instrument precision rather than statistical uncertainty. Choice C gives the standard deviation of individual measurements, not the mean. Choice D uses the range, which overestimates uncertainty.

Question 20

Two independent measurements yield $a = 8.5 \pm 0.3$ and $b = 3.2 \pm 0.4$ . If $c = a - b$ , what is the uncertainty in $c$ ?

$\pm 0.1$ , calculated as $|0.3 - 0.4|$
$\pm 0.5$ , calculated as $\sqrt{(0.3)^2 + (0.4)^2}$ (correct answer)
$\pm 0.7$ , calculated as $0.3 + 0.4$
$\pm 0.35$ , calculated as the average of the two uncertainties

Explanation: For addition or subtraction of independent measurements, absolute uncertainties combine in quadrature:

\delta c = \sqrt{(\delta a)^2 + (\delta b)^2} = \sqrt{(0.3)^2 + (0.4)^2} = \sqrt{0.25} = 0.5

. Choice A incorrectly subtracts uncertainties. Choice C incorrectly adds uncertainties linearly. Choice D incorrectly averages the uncertainties.