201. A set of readings has mean value \(5.00\,\text{s}\). One reading is \(5.08\,\text{s}\). The absolute deviation of this reading from the mean is
ⓐ. \(10.08\,\text{s}\)
ⓑ. \(5.08\,\text{s}\)
ⓒ. \(0.08\,\text{s}\)
ⓓ. \(-0.08\,\text{s}\)
Correct Answer: \(0.08\,\text{s}\)
Explanation: \( \textbf{Mean value:} \) \(a_{\text{mean}}=5.00\,\text{s}\).
\( \textbf{Reading:} \) \(a=5.08\,\text{s}\).
\( \textbf{Deviation from mean:} \)
\[
a-a_{\text{mean}}=5.08\,\text{s}-5.00\,\text{s}
\]
\[
=0.08\,\text{s}
\]
\( \textbf{Absolute deviation:} \)
\[
|a-a_{\text{mean}}|=0.08\,\text{s}
\]
The absolute value is used because deviations above and below the mean should not cancel when estimating spread.
\( \textbf{Final answer:} \) The absolute deviation is \(0.08\,\text{s}\).
202. A measurement table gives the following readings of a length.
| Reading | Value |
|---|
| P | \(2.4\,\text{cm}\) |
| Q | \(2.5\,\text{cm}\) |
| R | \(2.6\,\text{cm}\) |
The mean absolute error is
ⓐ. \(2.500\,\text{cm}\)
ⓑ. \(0.100\,\text{cm}\)
ⓒ. \(0.067\,\text{cm}\)
ⓓ. \(0.000\,\text{cm}\)
Correct Answer: \(0.067\,\text{cm}\)
Explanation: \( \textbf{Readings:} \) \(2.4\,\text{cm}\), \(2.5\,\text{cm}\), and \(2.6\,\text{cm}\).
\( \textbf{Mean value:} \)
\[
a_{\text{mean}}=\frac{2.4+2.5+2.6}{3}\,\text{cm}
\]
\[
a_{\text{mean}}=\frac{7.5}{3}\,\text{cm}=2.5\,\text{cm}
\]
\( \textbf{Absolute deviations:} \)
\[
|2.4-2.5|=0.1\,\text{cm}
\]
\[
|2.5-2.5|=0\,\text{cm}
\]
\[
|2.6-2.5|=0.1\,\text{cm}
\]
\( \textbf{Mean absolute error:} \)
\[
\Delta a_{\text{mean}}=\frac{0.1+0+0.1}{3}\,\text{cm}
\]
\[
\Delta a_{\text{mean}}=0.067\,\text{cm}\ \text{approximately}
\]
The middle reading has zero deviation, but the spread of the other readings still gives a non-zero mean absolute error.
\( \textbf{Final answer:} \) The mean absolute error is approximately \(0.067\,\text{cm}\).
203. A measurement is reported as \(2.50\pm0.07\,\text{cm}\). The interval suggested by this report is approximately
ⓐ. \(2.50\,\text{cm}\) to \(2.57\,\text{cm}\) only
ⓑ. \(2.57\,\text{cm}\) to \(2.64\,\text{cm}\)
ⓒ. \(0.07\,\text{cm}\) to \(2.50\,\text{cm}\)
ⓓ. \(2.43\,\text{cm}\) to \(2.57\,\text{cm}\)
Correct Answer: \(2.43\,\text{cm}\) to \(2.57\,\text{cm}\)
Explanation: \( \textbf{Reported value:} \) \(2.50\pm0.07\,\text{cm}\).
The central value is \(2.50\,\text{cm}\), and the uncertainty is \(0.07\,\text{cm}\).
\( \textbf{Lower limit:} \)
\[
2.50\,\text{cm}-0.07\,\text{cm}=2.43\,\text{cm}
\]
\( \textbf{Upper limit:} \)
\[
2.50\,\text{cm}+0.07\,\text{cm}=2.57\,\text{cm}
\]
So the suggested interval is from \(2.43\,\text{cm}\) to \(2.57\,\text{cm}\).
The \(\pm\) notation represents spread on both sides of the central value.
\( \textbf{Final answer:} \) The interval is approximately \(2.43\,\text{cm}\) to \(2.57\,\text{cm}\).
204. The mean absolute error has the same unit as
ⓐ. the reciprocal of the measured quantity
ⓑ. the percentage error only
ⓒ. the number of observations
ⓓ. the measured quantity
Correct Answer: the measured quantity
Explanation: Mean absolute error is found from absolute deviations of readings from the mean. Each deviation is a difference between two values of the same physical quantity. Therefore the deviation has the same unit as the original measurement. If the readings are lengths in \(\text{cm}\), then the mean absolute error is also in \(\text{cm}\). It is not a percentage unless it is later divided by the measured value and multiplied by \(100\%\).
205. Four mass readings are \(50.1\,\text{g}\), \(50.3\,\text{g}\), \(50.2\,\text{g}\), and \(50.4\,\text{g}\). The mean value is \(50.25\,\text{g}\). The absolute deviations are \(0.15\,\text{g}\), \(0.05\,\text{g}\), \(0.05\,\text{g}\), and \(0.15\,\text{g}\). The result should be reported as
ⓐ. \(0.10\pm50.25\,\text{g}\)
ⓑ. \(50.25\pm0.40\,\text{g}\)
ⓒ. \(50.25\pm0.10\,\text{g}\)
ⓓ. \(50.25\pm50.25\,\text{g}\)
Correct Answer: \(50.25\pm0.10\,\text{g}\)
Explanation: \( \textbf{Given mean:} \) \(a_{\text{mean}}=50.25\,\text{g}\).
\( \textbf{Absolute deviations:} \) \(0.15\,\text{g}\), \(0.05\,\text{g}\), \(0.05\,\text{g}\), and \(0.15\,\text{g}\).
\( \textbf{Mean absolute error:} \)
\[
\Delta a_{\text{mean}}=\frac{0.15+0.05+0.05+0.15}{4}\,\text{g}
\]
\[
\Delta a_{\text{mean}}=\frac{0.40}{4}\,\text{g}
\]
\[
\Delta a_{\text{mean}}=0.10\,\text{g}
\]
\( \textbf{Report form:} \)
\[
a=a_{\text{mean}}\pm\Delta a_{\text{mean}}
\]
So the result is \(50.25\pm0.10\,\text{g}\).
The uncertainty is the average deviation, not the sum of all deviations.
\( \textbf{Final answer:} \) \(50.25\pm0.10\,\text{g}\).
206. For readings \(4.8\,\text{mm}\), \(5.0\,\text{mm}\), and \(5.2\,\text{mm}\), the reported value using mean and mean absolute error is approximately
ⓐ. \(5.0\pm0.13\,\text{mm}\)
ⓑ. \(4.8\pm5.2\,\text{mm}\)
ⓒ. \(15.0\pm0.13\,\text{mm}\)
ⓓ. \(5.0\pm0.40\,\text{mm}\)
Correct Answer: \(5.0\pm0.13\,\text{mm}\)
Explanation: \( \textbf{Readings:} \) \(4.8\,\text{mm}\), \(5.0\,\text{mm}\), and \(5.2\,\text{mm}\).
\( \textbf{Mean value:} \)
\[
a_{\text{mean}}=\frac{4.8+5.0+5.2}{3}\,\text{mm}
\]
\[
a_{\text{mean}}=\frac{15.0}{3}\,\text{mm}=5.0\,\text{mm}
\]
\( \textbf{Absolute deviations:} \)
\[
|4.8-5.0|=0.2\,\text{mm}
\]
\[
|5.0-5.0|=0\,\text{mm}
\]
\[
|5.2-5.0|=0.2\,\text{mm}
\]
\( \textbf{Mean absolute error:} \)
\[
\Delta a_{\text{mean}}=\frac{0.2+0+0.2}{3}\,\text{mm}
\]
\[
\Delta a_{\text{mean}}\approx0.13\,\text{mm}
\]
\( \textbf{Final answer:} \) The result is approximately \(5.0\pm0.13\,\text{mm}\).
207. A data record lists repeated readings and deviations from the mean.
| Reading | Value | Absolute deviation from mean |
|---|
| P | \(12.1\,\text{s}\) | \(0.1\,\text{s}\) |
| Q | \(12.0\,\text{s}\) | \(0\,\text{s}\) |
| R | \(11.9\,\text{s}\) | \(-0.1\,\text{s}\) |
The entry that needs correction is
ⓐ. P
ⓑ. none of them
ⓒ. Q
ⓓ. R
Correct Answer: R
Explanation: An absolute deviation cannot be negative because it is the magnitude of a difference. If the mean is \(12.0\,\text{s}\), the reading \(11.9\,\text{s}\) has signed deviation \(11.9\,\text{s}-12.0\,\text{s}=-0.1\,\text{s}\). Its absolute deviation is \(|-0.1\,\text{s}|=0.1\,\text{s}\). Rows P and Q are consistent with magnitude-based deviations. Row R has written the signed deviation instead of the absolute deviation.
208. A set of repeated observations is finally written as \(a_{\text{mean}}\pm\Delta a_{\text{mean}}\) rather than only \(a_{\text{mean}}\). This is done to show
ⓐ. that the unit is no longer needed
ⓑ. best estimate and typical uncertainty
ⓒ. that all individual readings were identical
ⓓ. that the measured quantity has become dimensionless
Correct Answer: best estimate and typical uncertainty
Explanation: The mean value \(a_{\text{mean}}\) gives the best estimate from repeated observations. The mean absolute error \(\Delta a_{\text{mean}}\) gives a measure of the typical spread of the readings around that mean. Writing both parts shows not only the central result but also how reliable the measurement is likely to be. The unit remains necessary for both the measured value and the uncertainty. A reported result with uncertainty gives more information than a bare mean value.
209. Relative error of a measurement is expressed as
ⓐ. \(\frac{\text{absolute error}}{\text{measured value}}\)
ⓑ. \(\frac{\text{measured value}}{\text{absolute error}}\)
ⓒ. \(\text{measured value}+\text{absolute error}\)
ⓓ. \(\text{absolute error}\times\text{measured value}\)
Correct Answer: \(\frac{\text{absolute error}}{\text{measured value}}\)
Explanation: Relative error compares the size of the absolute error with the size of the measured quantity. If the absolute error is small compared with the measured value, the relative error is small. This makes it useful for comparing the quality of measurements of different sizes. The ratio has no unit because both numerator and denominator have the same unit. A \(1\,\text{cm}\) error is serious in a \(5\,\text{cm}\) measurement but much less serious in a \(500\,\text{cm}\) measurement.
210. Percentage error is obtained from relative error by
ⓐ. subtracting it from the measured value
ⓑ. adding \(100\) to it
ⓒ. dividing it by \(100\)
ⓓ. multiplying it by \(100\%\)
Correct Answer: multiplying it by \(100\%\)
Explanation: Percentage error expresses relative error on a percentage scale. If relative error is \(\frac{\Delta a}{a}\), then percentage error is \(\frac{\Delta a}{a}\times100\%\). The factor \(100\%\) changes the fraction into a percent form without changing the physical meaning. Since relative error is dimensionless, percentage error is also dimensionless. Percentage form is often easier to compare between different measurements.
211. A length is measured as \(40.0\,\text{cm}\) with an absolute error of \(0.2\,\text{cm}\). The percentage error is
ⓐ. \(5.0\%\)
ⓑ. \(2.0\%\)
ⓒ. \(0.5\%\)
ⓓ. \(0.2\%\)
Correct Answer: \(0.5\%\)
Explanation: \( \textbf{Given data:} \) Measured length \(a=40.0\,\text{cm}\), absolute error \(\Delta a=0.2\,\text{cm}\).
\( \textbf{Required quantity:} \) Percentage error.
\( \textbf{Relative error:} \)
\[
\frac{\Delta a}{a}=\frac{0.2\,\text{cm}}{40.0\,\text{cm}}
\]
The unit \(\text{cm}\) cancels because the numerator and denominator are both lengths.
\( \textbf{Calculation:} \)
\[
\frac{0.2}{40.0}=0.005
\]
\( \textbf{Percentage error:} \)
\[
0.005\times100\%=0.5\%
\]
The result is a percentage because it compares the uncertainty with the measured value.
\( \textbf{Final answer:} \) The percentage error is \(0.5\%\).
212. Two mass measurements are reported as \(20.0\pm0.2\,\text{g}\) and \(200.0\pm0.5\,\text{g}\). The better measurement by percentage error is
ⓐ. both are equally good because both use \(\text{g}\)
ⓑ. \(20.0\pm0.2\,\text{g}\)
ⓒ. the one with the larger absolute error
ⓓ. \(200.0\pm0.5\,\text{g}\)
Correct Answer: \(200.0\pm0.5\,\text{g}\)
Explanation: \( \textbf{First measurement:} \) \(a_1=20.0\,\text{g}\), \(\Delta a_1=0.2\,\text{g}\).
\[
\frac{\Delta a_1}{a_1}\times100\%=\frac{0.2}{20.0}\times100\%=1.0\%
\]
\( \textbf{Second measurement:} \) \(a_2=200.0\,\text{g}\), \(\Delta a_2=0.5\,\text{g}\).
\[
\frac{\Delta a_2}{a_2}\times100\%=\frac{0.5}{200.0}\times100\%=0.25\%
\]
A smaller percentage error means better relative quality of measurement.
Although \(0.5\,\text{g}\) is a larger absolute error than \(0.2\,\text{g}\), it is small compared with \(200.0\,\text{g}\).
\( \textbf{Final answer:} \) \(200.0\pm0.5\,\text{g}\) has the smaller percentage error.
213. Relative error has no unit because it is formed by
ⓐ. dividing quantities with the same unit
ⓑ. changing every unit into \(\text{kg}\)
ⓒ. adding two values with different units
ⓓ. multiplying error by the measured value
Correct Answer: dividing quantities with the same unit
Explanation: Relative error is \(\frac{\Delta a}{a}\), where \(\Delta a\) is the absolute error and \(a\) is the measured value. Both \(\Delta a\) and \(a\) refer to the same physical quantity, so they have the same unit. When one is divided by the other, the unit cancels. This makes relative error a pure ratio. Percentage error is also unit-free because it is just relative error multiplied by \(100\%\).
214. The expression for percentage error is completed as \(\frac{\Delta a}{a}\times\_\_\_\_\).
ⓐ. \(\Delta a\)
ⓑ. \(100\%\)
ⓒ. \(a^2\)
ⓓ. \(\frac{1}{100}\,\text{cm}\)
Correct Answer: \(100\%\)
Explanation: Percentage error is a percentage form of relative error. The relative error is \(\frac{\Delta a}{a}\). Multiplying by \(100\%\) converts this fraction into percent notation. The factor does not introduce a unit of length, mass, or time. The measured value \(a\) and the absolute error \(\Delta a\) already appear in the ratio, so they are not repeated in the blank.
215. A table compares measurements of different quantities.
| Record | Measurement | Percentage error |
|---|
| P | \(10.0\pm0.1\,\text{cm}\) | \(1.0\%\) |
| Q | \(50.0\pm0.5\,\text{s}\) | \(1.0\%\) |
| R | \(100\pm1\,\text{g}\) | \(1.0\%\) |
| S | \(200\pm2\,\text{m}\) | \(10\%\) |
The record that needs correction is
ⓐ. Q
ⓑ. P
ⓒ. S
ⓓ. R
Correct Answer: S
Explanation: For record P, \(\frac{0.1}{10.0}\times100\%=1.0\%\), so it is correct. For record Q, \(\frac{0.5}{50.0}\times100\%=1.0\%\), so it is also correct. For record R, \(\frac{1}{100}\times100\%=1.0\%\). For record S, \(\frac{2}{200}\times100\%=1.0\%\), not \(10\%\). The percentage error depends on the ratio of error to measured value, not on the absolute error alone.
216. A diameter is measured as \(2.00\,\text{mm}\) with an uncertainty of \(0.01\,\text{mm}\). A length is measured as \(20.0\,\text{cm}\) with an uncertainty of \(0.1\,\text{cm}\). The comparison of percentage errors is
ⓐ. diameter has \(0.5\%\), length has \(5\%\)
ⓑ. both are \(0.5\%\)
ⓒ. diameter has \(5\%\), length has \(0.5\%\)
ⓓ. both are \(5\%\)
Correct Answer: both are \(0.5\%\)
Explanation: \( \textbf{Diameter measurement:} \) \(d=2.00\,\text{mm}\), \(\Delta d=0.01\,\text{mm}\).
\[
\frac{\Delta d}{d}\times100\%=\frac{0.01}{2.00}\times100\%=0.5\%
\]
\( \textbf{Length measurement:} \) \(l=20.0\,\text{cm}\), \(\Delta l=0.1\,\text{cm}\).
\[
\frac{\Delta l}{l}\times100\%=\frac{0.1}{20.0}\times100\%=0.5\%
\]
The units do not need to be the same across the two measurements because each error is divided by its own measured value.
Both measurements therefore have the same relative quality.
\( \textbf{Final answer:} \) Both percentage errors are \(0.5\%\).
217. When two measured quantities are added, the maximum absolute error in the result is found by
ⓐ. adding their absolute errors
ⓑ. multiplying their measured values
ⓒ. dividing the smaller error by the larger error
ⓓ. subtracting their absolute errors
Correct Answer: adding their absolute errors
Explanation: For addition of measured quantities, the maximum possible absolute errors add. If \(x=a+b\), then \(\Delta x=\Delta a+\Delta b\) for the maximum uncertainty. This rule is used because one measurement may be high by its error while the other is also high by its error. The worst-case combined deviation is therefore the sum of the separate absolute deviations. The signs in the measured values do not allow uncertainty to be cancelled in a maximum-error estimate.
218. Two lengths are measured as \(12.4\pm0.1\,\text{cm}\) and \(8.2\pm0.1\,\text{cm}\). Their sum with maximum absolute error is
ⓐ. \(4.2\pm0.2\,\text{cm}\)
ⓑ. \(20.6\pm0.2\,\text{cm}\)
ⓒ. \(20.6\pm0.1\,\text{cm}\)
ⓓ. \(20.6\pm1.0\,\text{cm}\)
Correct Answer: \(20.6\pm0.2\,\text{cm}\)
Explanation: \( \textbf{Given lengths:} \) \(a=12.4\pm0.1\,\text{cm}\), \(b=8.2\pm0.1\,\text{cm}\).
\( \textbf{Required quantity:} \) Sum and maximum absolute error.
\( \textbf{Value of sum:} \)
\[
x=a+b=12.4\,\text{cm}+8.2\,\text{cm}
\]
\[
x=20.6\,\text{cm}
\]
\( \textbf{Error rule for addition:} \)
\[
\Delta x=\Delta a+\Delta b
\]
\( \textbf{Substitution:} \)
\[
\Delta x=0.1\,\text{cm}+0.1\,\text{cm}=0.2\,\text{cm}
\]
The absolute errors add because both individual readings may deviate in the same direction.
\( \textbf{Final answer:} \) The sum is \(20.6\pm0.2\,\text{cm}\).
219. For a difference \(x=a-b\), the maximum absolute error is
ⓐ. \(\Delta a+\Delta b\)
ⓑ. \(\Delta a-\Delta b\)
ⓒ. \(\Delta b-\Delta a\)
ⓓ. \(\frac{\Delta a}{\Delta b}\)
Correct Answer: \(\Delta a+\Delta b\)
Explanation: In subtraction, the measured values are subtracted, but their uncertainties still add in the maximum-error estimate. If \(x=a-b\), then \(\Delta x=\Delta a+\Delta b\). This is because the largest possible error in the difference can occur when one reading is high and the other is low. These deviations make the difference shift more than either single error alone. The minus sign in the formula for the value does not mean the error is found by subtracting uncertainties.
220. A rod length is found by subtracting two scale readings: \(x=(18.6\pm0.1\,\text{cm})-(5.2\pm0.1\,\text{cm})\). The result with maximum absolute error is
ⓐ. \(13.4\pm0.2\,\text{cm}\)
ⓑ. \(13.4\pm0.0\,\text{cm}\)
ⓒ. \(23.8\pm0.2\,\text{cm}\)
ⓓ. \(13.4\pm0.1\,\text{cm}\)
Correct Answer: \(13.4\pm0.2\,\text{cm}\)
Explanation: \( \textbf{Given expression:} \) \(x=(18.6\pm0.1\,\text{cm})-(5.2\pm0.1\,\text{cm})\).
\( \textbf{Value of difference:} \)
\[
x=18.6\,\text{cm}-5.2\,\text{cm}=13.4\,\text{cm}
\]
\( \textbf{Error rule for subtraction:} \)
\[
\Delta x=\Delta a+\Delta b
\]
\( \textbf{Substitution:} \)
\[
\Delta x=0.1\,\text{cm}+0.1\,\text{cm}=0.2\,\text{cm}
\]
The uncertainties add even though the readings themselves are subtracted.
The difference could be too large or too small depending on the directions of the two individual errors.
\( \textbf{Final answer:} \) The result is \(13.4\pm0.2\,\text{cm}\).
