Chapter 3, Problem 3.BE

(a)

Interpretation Introduction

Interpretation:

The absolute and percent relative uncertainty with a reasonable number of significant figures has to be written

Concept Introduction:

Uncertainty:

Uncertainty means state of not certain in predicting a value.  In a measured value, the last digit will have associated uncertainty. Uncertainty has two values, absolute uncertainty and relative uncertainty.

Absolute uncertainty:

Expressed the marginal value associated with a measurement

Relative uncertainty:

Compares the size of absolute uncertainty with the size of its associated measurement

$\text{Relative uncertainty = }\frac{\text{absolute uncertainty}}{\text{magnitude of measurement}}$

For a set measurements having uncertainty values as $e_1,e_2\text{ and }{e}_3$, the uncertainty $(e_4)$ in addition and subtraction can be calculated as follows,

$e_4=\sqrt{e_1^2+e_2^2+e_3^2}$

Percent relative uncertainty:

$\text{Percent relative uncertainty = 100 }\times \text{ relative uncertainty}$

To Write: The absolute and percent relative uncertainty with a reasonable number of significant figures

Answer to Problem 3.BE

The absolute and percent relative uncertainty with a reasonable number of significant figures is $2.1\pm 0.2$  and $2.1\pm 0.8\%$ respectively.

Explanation of Solution

Given data:

$[12.41(\pm 0.09)÷4.16(\pm 0.01)]\times 7.0682(\pm 0.0004)=?$

Calculation of absolute and percent relative uncertainty:

For multiplication and division, convert absolute uncertainty to percent relative uncertainty.

For $12.41(\pm 0.09)$, percent relative uncertainty is $(\pm 0.09/12.4)\times 100=\pm 0.725\%$

For $4.16(\pm 0.01)$, percent relative uncertainty is $(\pm 0.01/4.16)\times 100=\pm 0.240\%$

For $7.0682(\pm 0.0004)$, percent relative uncertainty is $(\pm 0.0004/7.0682)\times 100=\pm 0.0057\%$

The uncertainty (e) is calculated as follows,

$\begin{array}{l}\text{uncertainty}\text{=}\sqrt{{0.725}^2+{0.0057}^2+{0.240}^2}\\ =0.764\%\\ \text{=0}\text{.8\% (rounded to correct significant figure)}\end{array}$

Substitute all the values in the given problem.

$\begin{array}{l}[12.41(\pm 0.09)÷4.16(\pm 0.01)]\times 7.0682(\pm 0.0004)\\ =\frac{12.41(\pm 0.725\%)\times 7.0682(\pm 0.0057\%)}{4.16(\pm 0.240\%)}\\ =21.086(\pm 0.764\%)\\ \text{Since }\sqrt{{0.725}^2+{0.0057}^2+{0.240}^2}=0.764\%\end{array}$

Convert relative uncertainty into absolute uncertainty as follows,

$\begin{array}{l}\text{Absolute uncertainty}\text{=}0.00764\times 21.086\\ =0.16\\ =0.2(\text{rounded to correct significant figure)}\end{array}$

Therefore,

The absolute uncertainty is given as $2.1\pm 0.2$

The percent relative uncertainty is $2.1\pm 0.8\%$

Conclusion

The absolute and percent relative uncertainty with a reasonable number of significant figures is written as $2.1\pm 0.2$ and $2.1\pm 0.8\%$ respectively.

(b)

Interpretation Introduction

Interpretation:

The absolute and percent relative uncertainty with a reasonable number of significant figures has to be written

Concept Introduction:

Uncertainty:

Uncertainty means state of not certain in predicting a value.  In a measured value, the last digit will have associated uncertainty. Uncertainty has two values, absolute uncertainty and relative uncertainty.

Absolute uncertainty:

Expressed the marginal value associated with a measurement

Relative uncertainty:

Compares the size of absolute uncertainty with the size of its associated measurement

$\text{Relative uncertainty = }\frac{\text{absolute uncertainty}}{\text{magnitude of measurement}}$

For a set measurements having uncertainty values as $e_1,e_2\text{ and }{e}_3$, the uncertainty $(e_4)$ in addition and subtraction can be calculated as follows,

$e_4=\sqrt{e_1^2+e_2^2+e_3^2}$

Percent relative uncertainty:

$\text{Percent relative uncertainty = 100 }\times \text{ relative uncertainty}$

To Write: The absolute and percent relative uncertainty with a reasonable number of significant figures

Answer to Problem 3.BE

The absolute and percent relative uncertainty with a reasonable number of significant figures is $27.4(\pm 0.9)$ and $27.4(\pm 3.2\%)$ respectively.

Explanation of Solution

Given data:

$[3.26(\pm 0.10)\times 8.47(\pm 0.05)]-0.18(\pm 0.06)=?$

Calculation of absolute and percent relative uncertainty:

First solve the terms in bracket.

For multiplication and division, convert absolute uncertainty to percent relative uncertainty.

For $3.26(\pm 0.10)$, percent relative uncertainty is $(\pm 0.10/3.26)\times 100=\pm 3.07\%$

For $8.47(\pm 0.05)$, percent relative uncertainty is $(\pm 0.05/8.47)\times 100=\pm 0.59\%$

The uncertainty (e) is calculated as follows,

$\begin{array}{l}\text{uncertainty}\text{=}\sqrt{{3.07}^2+{0.59}^2}\\ =\pm 3.12\%\end{array}$

Substitute all the values in the given problem.

$\begin{array}{l}[3.26(\pm 0.10)\times 8.47(\pm 0.05)]-0.18(\pm 0.06)\\ =[3.26(\pm 3.07\%)\times 8.47(\pm 0.59\%)]-0.18(\pm 0.06)\\ =[27.612(\pm 3.12\%)]-0.18(\pm 0.06)(\text{Since }\sqrt{{3.07}^2+{0.59}^2}=3.12\%)\\ =[27.612(\pm 0.863)]-0.18(\pm 0.06)[\text{Since (3}\text{.12/100)}\times \text{27}\text{.612=0}\text{.863]}\\ \text{=27}\text{.4}(\pm 0.86)\\ =27.4(\pm 0.9)(\text{rounded to correct significant figures)}\end{array}$

Therefore, the absolute uncertainty is $27.4(\pm 0.9)$

The percent relative uncertainty is calculated as follows,

$\begin{array}{l}\text{Percent Relative uncertainty }\text{=}\frac{0.863}{27.43}\times 100\\ =3.2\%\end{array}$

Therefore,

The absolute uncertainty is given as $27.4(\pm 0.9)$

The percent relative uncertainty is $27.4(\pm 3.2\%)$

Conclusion

The absolute and percent relative uncertainty with a reasonable number of significant figures is written as $27.4(\pm 0.9)$ and $27.4(\pm 3.2\%)$ respectively.

(c)

Interpretation Introduction

Interpretation:

The absolute and percent relative uncertainty with a reasonable number of significant figures has to be written

Concept Introduction:

Uncertainty:

Uncertainty means state of not certain in predicting a value.  In a measured value, the last digit will have associated uncertainty. Uncertainty has two values, absolute uncertainty and relative uncertainty.

Absolute uncertainty:

Expressed the marginal value associated with a measurement

Relative uncertainty:

Compares the size of absolute uncertainty with the size of its associated measurement

$\text{Relative uncertainty = }\frac{\text{absolute uncertainty}}{\text{magnitude of measurement}}$

For a set measurements having uncertainty values as $e_1,e_2\text{ and }{e}_3$, the uncertainty $(e_4)$ in addition and subtraction can be calculated as follows,

$e_4=\sqrt{e_1^2+e_2^2+e_3^2}$

Percent relative uncertainty:

$\text{Percent relative uncertainty = 100 }\times \text{ relative uncertainty}$

To Write: The absolute and percent relative uncertainty with a reasonable number of significant figures

Answer to Problem 3.BE

The absolute and percent relative uncertainty with a reasonable number of significant figures is $1.49(\pm 0.1)\times {10}^5$ and $1.49(\pm 9.0\%)\times {10}^5$ respectively.

Explanation of Solution

Given data:

$6.843(\pm 0.008)\times {10}^4÷[2.09(\pm 0.04)-1.63(\pm 0.01)]=?$

Calculation of absolute and percent relative uncertainty:

First solve the terms in bracket.

$\begin{array}{l}[2.09(\pm 0.04)-1.63(\pm 0.01)]\\ =0.46(\pm 0.0412)(\text{Since }\sqrt{{(0.04)}^2+{(0.01)}^2}=0.0412)\end{array}$

For multiplication and division, convert absolute uncertainty to percent relative uncertainty.

Substitute the above value and rewrite the given equation.

$6.843(\pm 0.008)\times {10}^4÷[0.46(\pm 0.0412)]=?$

For $6.843(\pm 0.008)$, percent relative uncertainty is $(\pm 0.008/6.843)\times 100=\pm 0.117\%$

For $0.46(\pm 0.0412)$, percent relative uncertainty is $(\pm 0.0412/0.46)\times 100=\pm 8.96\%$

Substitute all the values in the given problem.

$\begin{array}{l}6.843(\pm 0.008)\times {10}^4÷[0.46(\pm 0.0412)]\\ =6.843(\pm 0.117\%)\times {10}^4÷[0.46(\pm 8.96\%)]\\ =1.49(\pm 8.96\%)\times {10}^5(\text{Since }\sqrt{{0.117}^2+{8.96}^2}=8.96\%)\\ =1.49(\pm 9\%)\times {10}^5(\text{rounded to correct significant figure)}\end{array}$

Convert relative uncertainty into absolute uncertainty as follows,

$\begin{array}{l}\text{Absolute uncertainty}\text{=}0.0896\times 1.49\\ =0.1335\\ =0.1(\text{rounded to correct significant figure)}\end{array}$

Therefore,

The absolute uncertainty is given as $1.49(\pm 0.1)\times {10}^5$

The percent relative uncertainty is $1.49(\pm 9.0\%)\times {10}^5$

Conclusion

The absolute and percent relative uncertainty with a reasonable number of significant figures is written as $1.49(\pm 0.1)\times {10}^5$ and $1.49(\pm 9.0\%)\times {10}^5$ respectively.

(d)

Interpretation Introduction

Interpretation:

The absolute and percent relative uncertainty with a reasonable number of significant figures has to be written

Concept Introduction:

Uncertainty for powers and roots:

For the function $y=x^a$ , the relative uncertainty in $y(\%e_y)$ is $a$ times the relative uncertainty in $x(\%e_x)$

$y=x^a\Rightarrow \%e_y=a(\%e_x)$

To Write: The absolute and percent relative uncertainty with a reasonable number of significant figures

Answer to Problem 3.BE

The absolute and percent relative uncertainty with a reasonable number of significant figures is $1.80\pm 0.02$ and $1.80\pm 1\%$ respectively.

Explanation of Solution

Given data:

$\sqrt{3.24\pm 0.08}=?$

Calculation of absolute and percent relative uncertainty:

Use $y=x^a\Rightarrow \%e_y=a(\%e_x)$

$\begin{array}{l}\%e_y=\frac{1}{2}\%e_x\\ =\frac{1}{2}\left(\frac{0.08}{3.24}\times 100\right)\\ =1.235\%\end{array}$

Now,

$\begin{array}{l}{(}^{3.24}=1.80\pm 1.235\%(\text{Since}\sqrt{3.24}=1.80)\\ =1.80\pm 1.2\%(\text{rounded to correct significant figure) }\end{array}$

Convert relative uncertainty into absolute uncertainty as follows,

$\begin{array}{l}\text{Absolute uncertainty}\text{=}0.01235\times 1.80\\ =0.02223\\ =0.02(\text{rounded to correct significant figure)}\end{array}$

Therefore,

The absolute uncertainty is given as $1.80\pm 0.02$

The percent relative uncertainty is $1.80\pm 1\%$

Conclusion

The absolute and percent relative uncertainty with a reasonable number of significant figures is written $1.80\pm 0.02$ and $1.80\pm 1\%$ respectively.

(e)

Interpretation Introduction

Interpretation:

The absolute and percent relative uncertainty with a reasonable number of significant figures has to be written

Concept Introduction:

Uncertainty for powers and roots:

For the function $y=x^a$ , the relative uncertainty in $y(\%e_y)$ is $a$ times the relative uncertainty in $x(\%e_x)$

$y=x^a\Rightarrow \%e_y=a(\%e_x)$

To Write: The absolute and percent relative uncertainty with a reasonable number of significant figures

Answer to Problem 3.BE

The absolute and percent relative uncertainty with a reasonable number of significant figures is $1.1(\pm 0.1)\times {10}^2$ and $1.1(\pm 9.9\%)\times {10}^2$ respectively.

Explanation of Solution

Given data:

${(3.24\pm 0.08)}^4=?$

Calculation of absolute and percent relative uncertainty:

Use $y=x^a\Rightarrow \%e_y=a(\%e_x)$

$\begin{array}{l}\%e_y=4\%e_x\\ =4\left(\frac{0.08}{3.24}\times 100\right)\\ =9.877\%\end{array}$

Now,

$\begin{array}{l}{(}^{3.24}=110.20\pm 9.877\%(\text{Since (3}{\text{.24)}}^4=110.20)\\ =110.20\pm 9.9\%\\ \text{=}1.1\times {10}^2\pm 9.9\%(\text{rounded to correct significant figure)}\end{array}$

Convert relative uncertainty into absolute uncertainty as follows,

$\begin{array}{l}\text{Absolute uncertainty}\text{=}0.099\times 1.1\\ =0.1089\\ =0.11\\ \text{=0}\text{.1 (rounded to correct significant figure)}\end{array}$

Therefore,

The absolute uncertainty is given as $1.1(\pm 0.1)\times {10}^2$

The percent relative uncertainty is $1.1(\pm 9.9\%)\times {10}^2$

Conclusion

The absolute and percent relative uncertainty with a reasonable number of significant figures is written as $1.1(\pm 0.1)\times {10}^2$ and $1.1(\pm 9.9\%)\times {10}^2$ respectively.

(f)

Interpretation Introduction

Interpretation:

The absolute and percent relative uncertainty with a reasonable number of significant figures has to be written

Concept Introduction:

Exponents and logarithm:

Consider, $y=\log x$ .

Here, the absolute uncertainty in $y(e_y)$ is proportional to the relative uncertainty in x, which is $e_x/x$

$\begin{array}{l}\text{Uncertainty for logarithm:}\\ y=1\text{og}x\\ \Rightarrow e_y=\frac{1}{\text{ln 10 }x}\frac{e_x}{x}\\ \approx 0.43429\frac{e_x}{x}\end{array}$

To Write: The absolute and percent relative uncertainty with a reasonable number of significant figures

Answer to Problem 3.BE

The absolute and percent relative uncertainty with a reasonable number of significant figures is $0.51(\pm 0.01)$ and $0.51\pm 2\%$ respectively.

Explanation of Solution

Given data:

$\log (3.24\pm 0.08)=?$

Calculation of absolute and percent relative uncertainty:

Use $y=\log x\Rightarrow e_y=0.43429\frac{e_x}{x}$

$\begin{array}{l}{e}_y=0.43429\frac{e_x}{x}\\ =0.43429\left(\frac{0.08}{3.24}\right)\\ =0.0107\end{array}$

Now,

$\begin{array}{l}\log (3.24\pm 0.08)=0.5105\pm 0.0107(\text{Since log 3}\text{.24}=0.5101)\\ =0.51\pm 0.01(\text{rounded to correct significant figure)}\end{array}$

Calculate percent relative uncertainty as follows,

$\begin{array}{l}\text{Percent Relative uncertainty}\text{=(0}\text{.0107/0}\text{.5105)}\times \text{100}\\ =2.095\%\\ =2.1\%\\ \text{=2\% (rounded to correct significant figure)}\end{array}$

Therefore,

The absolute uncertainty is given as $0.51(\pm 0.01)$

The percent relative uncertainty is $0.51\pm 2\%$

Conclusion

The absolute and percent relative uncertainty with a reasonable number of significant figures is written as $0.51(\pm 0.01)$ and $0.51\pm 2\%$ respectively.

(g)

Interpretation Introduction

Interpretation:

The absolute and percent relative uncertainty with a reasonable number of significant figures has to be written

Concept Introduction:

Exponents and logarithm:

Consider, $y=\text{antilog }x\Rightarrow y={10}^x$ .

Here, the relative uncertainty in y is proportional to the absolute uncertainty in x.

$\begin{array}{l}{\text{Uncertainty for 10}}^{x:}\\ y={10}^x\\ \Rightarrow \frac{e_y}{y}=(\text{ln 10)}{e}_x\\ \approx 2.3026e_x\end{array}$

To Write: The absolute and percent relative uncertainty with a reasonable number of significant figures

Answer to Problem 3.BE

The absolute and percent relative uncertainty with a reasonable number of significant figures is $1.7(\pm 0.3)\times {10}^3$ and $1.7(\pm 18\%)\times {10}^3$ respectively.

Explanation of Solution

Given data:

${10}^{3.24\pm 0.08}=?$

Calculation of absolute and percent relative uncertainty:

Use $y=\text{antilog }x\Rightarrow y={10}^x$

$\begin{array}{l}\frac{e_y}{y}=2.3026e_x\\ =2.3026(0.08)\\ =0.184\\ =18.4\%\end{array}$

Now,

$\begin{array}{l}{10}^{3.24\pm 0.08}=1.74\times {10}^3\pm 18.4\%({\text{Since 10}}^{3.24}=1734.8\approx 1.74\times {10}^3)\\ \text{=}1.7\times {10}^3\pm 18\%(\text{rounded to correct significant figure)}\end{array}$

Convert percent relative uncertainty into absolute uncertainty as follows,

$\begin{array}{l}\text{Absolute uncertainty}\text{=}0.184\times 1.74\\ =0.320\\ =0.32\%\\ \text{=0}\text{.3\% (rounded to correct significant figure)}\end{array}$

Therefore,

The absolute uncertainty is given as $1.7(\pm 0.3)\times {10}^3$

The percent relative uncertainty is $1.7(\pm 18\%)\times {10}^3$

Conclusion

The absolute and percent relative uncertainty with a reasonable number of significant figures is written as $1.7(\pm 0.3)\times {10}^3$ and $1.7(\pm 18\%)\times {10}^3$ respectively.