Chapter 3, Problem 3.16P

(a)

Interpretation Introduction

Interpretation:

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

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

Answer to Problem 3.16P

The absolute and percent relative uncertainty with a reasonable number of significant figures is $10.18\pm 0.07$  and $10.18\pm 0.7\%$ respectively.

Explanation of Solution

Given data:

$9.23(\pm 0.03)+4.12(\pm 0.02)-3.26(\pm 0.06)=?$

Calculation of absolute and percent relative uncertainty:

$\begin{array}{l}+9.23(\pm 0.03)\\ +4.21(\pm 0.02)\\ \frac{-3.26(\pm 0.06)}{10.18(\pm 0.07)}\end{array}$

The uncertainty (e) is calculated as follows,

$\begin{array}{l}\text{uncertainty}\text{=}\sqrt{{0.03}^2+{0.02}^2+{0.06}^2}\\ =0.07\%\end{array}$

The percent relative uncertainty is calculated as follows,

$\begin{array}{l}\text{Percent relative uncertainty }\text{= 100 }\times \text{ relative uncertainty}\\ \text{=100}\times \frac{0.07}{10.18}\\ \text{=0}\text{.687}\\ \text{=0}\text{.7\% (rounded to correct significant figure)}\end{array}$

Therefore,

The absolute uncertainty is given as $10.18\pm 0.07$

The percent relative uncertainty is $10.18\pm 0.7\%$

Conclusion

The absolute and percent relative uncertainty with a reasonable number of significant figures is found out as $10.18\pm 0.07$ and $10.18\pm 0.7\%$ respectively.

(b)

Interpretation Introduction

Interpretation:

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

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

Answer to Problem 3.16P

The absolute and percent relative uncertainty with a reasonable number of significant figures is $174\pm 3$ and $174\pm 2\%$ respectively.

Explanation of Solution

Given data:

$91.3(\pm 1.0)\times 40.3(\pm 0.2)/21.1(\pm 0.2)=?$

Calculation of absolute and percent relative uncertainty:

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

For $91.3(\pm 1.0)$, percent relative uncertainty is $(\pm 1.0/91.3)\times 100=\pm 1.10\%$

For $40.3(\pm 0.2)$, percent relative uncertainty is $(\pm 0.2/40.3)\times 100=\pm 0.50\%$

For $21.1(\pm 0.2)$, percent relative uncertainty is $(\pm 0.2/21.1)\times 100=\pm 0.95\%$

The percentage uncertainty (e) is calculated as follows,

$\begin{array}{l}\text{\%uncertainty}\text{=}\sqrt{{1.10}^2+{0.50}^2+{0.95}^2}\\ =1.54\%\\ \text{=2\% (rounded to correct significant figure)}\end{array}$

Substitute all the values in the given problem.

$\begin{array}{l}91.3(\pm 1.0)\times 40.3(\pm 0.2)/21.1(\pm 0.2)\\ =\frac{91.3(\pm 1.10\%)\times 40.3(\pm 0.50\%)}{21.1(\pm 0.95\%)}\\ =174.37(\pm 1.54\%)\\ =174(\pm 2\%)(\text{rounded to correct significant figures)}\end{array}$

Convert relative uncertainty into absolute uncertainty as follows,

$\begin{array}{l}\text{Absolute uncertainty}\text{=}\pm 0.0154\times 174.37\\ =\pm 2.685\\ =\pm 3(\text{rounded to correct significant figure)}\end{array}$

Therefore,

The absolute uncertainty is given as $174\pm 3$

The percent relative uncertainty is $174\pm 2\%$

Conclusion

The absolute and percent relative uncertainty with a reasonable number of significant figures is found out as $174\pm 3$ and $174\pm 2\%$ respectively.

(c)

Interpretation Introduction

Interpretation:

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

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

Answer to Problem 3.16P

The absolute and percent relative uncertainty with a reasonable number of significant figures is $0.147\pm 0.003$ and $0.147\pm 2\%$ respectively.

Explanation of Solution

Given data:

$[4.97(\pm 0.05)-1.86(\pm 0.01)]/21.1(\pm 0.2)=?$

Calculation of absolute and percent relative uncertainty:

Solve the bracket first.

$\begin{array}{l}+4.97(\pm 0.05)\\ \frac{-1.86(\pm 0.01)}{3.11(\pm 0.0510)}\end{array}$

The uncertainty (e) is calculated as follows,

$\begin{array}{l}\text{uncertainty}\text{=}\sqrt{{0.05}^2+{0.1}^2}\\ =\pm 0.0510\end{array}$

Substitute the obtained value in the given problem and solve further.

$\begin{array}{l}[4.97(\pm 0.05)-1.86(\pm 0.01)]/21.1(\pm 0.2)\\ \Rightarrow [3.11(\pm 0.0510)]/21.1(\pm 0.2)\end{array}$

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

For $3.11(\pm 0.0510)$, percent relative uncertainty is $(\pm 0.0510/3.11)\times 100=\pm 1.64\%$

For $21.1(\pm 0.2)$, percent relative uncertainty is $(\pm 0.2/21.1)\times 100=\pm 0.95\%$

Thus,

$\begin{array}{l}3.11(\pm 0.0510)]/21.1(\pm 0.2)\\ =3.11(\pm 1.64\%)/21.1(\pm 0.95\%)\\ =0.147(\pm 1.90\%)(\text{since }\sqrt{{(1.64)}^2+{(0.95)}^2}=1.90\%\\ =0.147(\pm 2\%)(\text{rounded to correct significant figure)}\end{array}$

Convert relative uncertainty into absolute uncertainty as follows,

$\begin{array}{l}\text{Absolute uncertainty}\text{=}\pm 0.02\times 0.147\\ =\pm 0.00294\\ =\pm 0.003(\text{rounded to correct significant figure)}\end{array}$

Therefore,

The absolute uncertainty is given as $10.18\pm 0.07$

The percent relative uncertainty is $10.18\pm 0.7\%$

Conclusion

The absolute and percent relative uncertainty with a reasonable number of significant figures is found as $0.147\pm 0.003$ and $0.147\pm 2\%$ respectively.

(d)

Interpretation Introduction

Interpretation:

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

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

Answer to Problem 3.16P

The absolute and percent relative uncertainty with a reasonable number of significant figures is $7.86\pm 0.01$ and $7.86\pm 0.1\%$ respectively.

Explanation of Solution

Given data:

$2.0164(\pm 0.0008)+1.233(\pm 0.002)+4.61(\pm 0.01)=?$

Calculation of absolute and percent relative uncertainty:

$\begin{array}{l}+2.0164(\pm 0.0008)\\ +1.233(\pm 0.002)\\ \frac{+4.61(\pm 0.01)}{7.86(\pm 0.01)}\end{array}$

The uncertainty (e) is calculated as follows,

$\begin{array}{l}\text{uncertainty}\text{=}\sqrt{{0.0008}^2+{0.002}^2+{0.01}^2}\\ =0.01\end{array}$

The percent relative uncertainty is calculated as follows,

$\begin{array}{l}\text{Percent relative uncertainty }\text{= 100 }\times \text{ relative uncertainty}\\ \text{=100}\times \frac{0.01}{7.86}\\ \text{=0}\text{.127}\\ \text{=0}\text{.1\% (rounded to correct significant figure)}\end{array}$

Therefore,

The absolute uncertainty is given as $7.86\pm 0.01$

The percent relative uncertainty is $7.86\pm 0.1\%$

Conclusion

The absolute and percent relative uncertainty with a reasonable number of significant figures is found as $7.86\pm 0.01$ and $7.86\pm 0.1\%$ respectively.

(e)

Interpretation Introduction

Interpretation:

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

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

Answer to Problem 3.16P

The absolute and percent relative uncertainty with a reasonable number of significant figures is $2185.8\pm 0.8$ and $2185.8\pm 0.04\%$ respectively.

Explanation of Solution

Given data:

$2.0164(\pm 0.0008)\times {10}^3+1.233(\pm 0.002)\times {10}^2+4.61(\pm 0.01)\times {10}^1=?$

Calculation of absolute and percent relative uncertainty:

$\begin{array}{l}+2016.4(\pm 0.8)\\ +123.3(\pm 0.2)\\ \frac{+46.1(\pm 0.1)}{2185.8(\pm 0.8)}\end{array}$

The uncertainty (e) is calculated as follows,

$\begin{array}{l}\text{uncertainty}\text{=}\sqrt{{0.8}^2+{0.2}^2+{0.1}^2}\\ =0.8\end{array}$

The percent relative uncertainty is calculated as follows,

$\begin{array}{l}\text{Percent relative uncertainty }\text{= 100 }\times \text{ relative uncertainty}\\ \text{=100}\times \frac{0.8}{2185.8}\\ \text{=0}\text{.0365}\\ \text{=0}\text{.04\% (rounded to correct significant figure)}\end{array}$

Therefore,

The absolute uncertainty is given as $7.86\pm 0.01$

The percent relative uncertainty is $7.86\pm 0.1\%$

Conclusion

The absolute and percent relative uncertainty with a reasonable number of significant figures is found as $2185.8\pm 0.8$ and $2185.8\pm 0.04\%$ respectively.

(f)

Interpretation Introduction

Interpretation:

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

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

Answer to Problem 3.16P

The absolute and percent relative uncertainty with a reasonable number of significant figures is $1.464\pm 0.007$ and $1.464\pm 0.5\%$ respectively.

Explanation of Solution

Given data:

${[3.14(\pm 0.05)]}^{1/3}=?$

Calculation of absolute and percent relative uncertainty:

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

$\begin{array}{l}\Rightarrow x=3.14\pm 0.05\\ \Rightarrow \%e_x=(0.05/3.14)\times 100=1.592\%\end{array}$

$\begin{array}{l}\%e_y=\frac{1}{3}\%e_x\\ =\frac{1}{3}\left(1.592\%\right)\\ =0.531\%\end{array}$

Now,

$\begin{array}{l}{(}^{3.24}=1.4643\pm 0.531\%(\text{Since}\sqrt{3.14}=1.4643)\\ =1.464\pm 0.5\%(\text{rounded to correct significant figure) }\end{array}$

Convert relative uncertainty into absolute uncertainty as follows,

$\begin{array}{l}\text{Absolute uncertainty}\text{=}0.00531\times 1.464\\ =0.007773\\ =0.0078\\ =0.007(\text{rounded to correct significant figure)}\end{array}$

Therefore,

The absolute uncertainty is given as $1.464\pm 0.007$

The percent relative uncertainty is $1.464\pm 0.5\%$

Conclusion

The absolute and percent relative uncertainty with a reasonable number of significant figures is found out as $1.464\pm 0.007$ and $1.464\pm 0.5\%$ respectively.

(g)

Interpretation Introduction

Interpretation:

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

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

Answer to Problem 3.16P

The absolute and percent relative uncertainty with a reasonable number of significant figures is $0.496(\pm 0.006)$ and $0.496(\pm 1.3\%)$ respectively.

Explanation of Solution

Given data:

$\log [3.14\pm 0.05]=?$

Calculation of absolute and percent relative uncertainty:

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

$\begin{array}{l}{e}_y=0.43429\frac{e_x}{x}\\ =0.43429\left(\frac{0.05}{3.14}\right)\\ =0.006915\\ =0.006\text{ (round to correct significant figure)}\end{array}$

Now,

$\begin{array}{l}\log (3.14\pm 0.05)=0.4969\pm 0.006915(\text{Since log 3}\text{.14}=0.5101)\\ =0.496\pm 0.006(\text{rounded to correct significant figure)}\end{array}$

Calculate percent relative uncertainty as follows,

$\begin{array}{l}\text{Percent Relative uncertainty}\text{=(0}\text{.000069/0}\text{.496)}\times \text{100}\\ =0.0139\%\\ =1.39\%\\ \text{=1}\text{.3\% (rounded to correct significant figure)}\end{array}$

Therefore,

The absolute uncertainty is given as $0.496(\pm 0.006)$

The percent relative uncertainty is $0.496(\pm 1.3\%)$

Conclusion

The absolute and percent relative uncertainty with a reasonable number of significant figures is found out as $0.496(\pm 0.006)$ and $0.496(\pm 1.3\%)$ respectively.