Chapter 3, Problem 3.13P

(a)

Interpretation Introduction

Interpretation:

The absolute and percent relative uncertainty with a reasonable number of significant figures has to be found out for $\left[9.23(\pm 0.03)+4.21(\pm 0.02)\right]-3.26(\pm 0.06)=?$

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:

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

Answer to Problem 3.13P

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

Explanation of Solution

Given data:

  $\left[9.23(\pm 0.03)+4.21(\pm 0.02)\right]-3.26(\pm 0.06)=?$

Calculation of absolute and percent relative uncertainty:

• The first term in the bracket $\left[9.23(\pm 0.03)+4.21(\pm 0.02)\right]$ is solved as follows,

On solving the given subtraction, we get

  $\begin{array}{l}9.23(\pm 0.03)\\ \frac{+4.21(\pm 0.02)}{13.44\left(\pm e\right)}\end{array}$

The uncertainty (e) is calculated as follows,

  $\begin{array}{l}{e}^2={0.03}^2\text{ + 0}{\text{.02}}^{\text{2}}\\ \\ \text{ =}\sqrt{\text{0}\text{.0013}}=0.0360\\ \\ \text{e =}\text{0}\text{.04 (rounded to correct significant figure)}\text{.}\end{array}$

The absolute uncertainty for the above addition part $\left[9.23(\pm 0.03)+4.21(\pm 0.02)\right]$ obtained as $13.44\pm 0.04.$

• On solving $\left[\text{13}\text{.44±0}\text{.04}\text{.}\right]\text{-}\left[\text{3}\text{.26(±0}\text{.06)}\right]\text{=?}$ as follows,

  $\begin{array}{l}13.44(\pm 0.04)\\ \frac{-3.26(\pm 0.06)}{10.18\left(\pm e\right)}\end{array}$

The uncertainty (e) is calculated as follows,

  $\begin{array}{l}{e}^2={0.04}^2\text{ + 0}{\text{.06}}^{\text{2}}\\ \\ \text{ =}\sqrt{\text{0}\text{.0052}}=0.07211\\ \\ \text{e =}\text{0}\text{.07 (rounded to correct significant figure)}\text{.}\end{array}$

Therefore, the absolute uncertainty of $\left[9.23(\pm 0.03)+4.21(\pm 0.02)\right]$ is given as $\left(10.18\pm 0.07\right).$

Convert the absolute uncertainty to relative uncertainty as $\frac{\text{0}\text{.07}}{\text{10}\text{.18}}\text{×100=}\text{0}\text{.7 \%}$

Therefore, the relative uncertainty is given as $10.18\pm \left(0.7\%\right).$

Conclusion

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

(b)

Interpretation Introduction

Interpretation:

The absolute and percent relative uncertainty with a reasonable number of significant figures has to be found out for $\left[91.3(\pm 1.0)\times 40.3(\pm 0.2)\right]÷21.2\left(\pm 0.2\right)=?$

Concept Introduction:

Refer part (a).

Answer to Problem 3.13P

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:

  $\left[91.3(\pm 1.0)\times 40.3(\pm 0.2)\right]÷21.2\left(\pm 0.2\right)=?$

Calculation of absolute and percent relative uncertainty:

• On solving the given multiplication part, we get

  $\begin{array}{l}91.3(\pm 1.0)\\ \frac{\times 40.3(\pm 0.2)}{3680(\pm e)}\end{array}$

For multiplication of $91.3(\pm 1.0)\times 40.3(\pm 0.2)=?$ , convert absolute uncertainty to percent relative uncertainty.

For $91.3(\pm 1.0)$, percent relative uncertainty is $(\pm 1.0/91.3)\times 100=\pm 1.1\%$

For $40.3(\pm 0.2)$, percent relative uncertainty is $(\pm 0.2/40.3)\times 100=\pm 0.5\%$

The uncertainty (e) is calculated as follows,

  $\begin{array}{l}\%e^2={1.1}^2\text{ + 0}{\text{.5}}^{\text{2}}\\ \\ \text{ =}\sqrt{\text{1}\text{.46}}=1.2\%\\ \\ \text{\%e =}\text{1}\text{.2\% (rounded to correct significant figure)}\text{.}\end{array}$

Therefore, the relative uncertainty is given as $3680\pm 1.2\%$

• For division $\left[\text{3680±1}\text{.2\%}\right]\text{÷}\text{21}\text{.2}\left(\text{±0}\text{.2}\right)\text{=?}$, convert absolute uncertainty to percent relative uncertainty.

For $\text{21}\text{.2}\left(\text{±0}\text{.2}\right)$, percent relative uncertainty is $(\pm 0.2/21.2)\times 100=\pm 0.9\%$

Therefore,

  $\begin{array}{l}\text{3679}\left(\text{±1}\text{.2\%}\right)\\ \frac{÷\text{21}\text{.2}\left(\text{±0}\text{.9\%}\right)}{\text{174}\left(\text{\%e}\right)}\\ \\ \text{Since, }\\ \\ {\text{\%e}}^{\text{2}}\text{=}\text{1}{\text{.2}}^{\text{2}}\text{ + 0}{\text{.9}}^{\text{2}}\\ \\ \text{\%e =}\sqrt{\text{2}\text{.25}}\text{=}\text{1}\text{.5}\\ \\ \text{=}\text{2\% }\text{(rounded to correct significant figure)}\text{.}\end{array}$

Convert relative uncertainty into absolute uncertainty as $174\times \frac{2}{100}=3$

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 founds 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 for $\left[4.97(\pm 0.05)-1.86(\pm 0.01)\right]÷21.2(\pm 0.2)=?.$

Concept Introduction:

Refer part (a).

Answer to Problem 3.13P

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

Explanation of Solution

Given data:

  $\left[4.97(\pm 0.05)-1.86(\pm 0.01)\right]÷21.2(\pm 0.2)=?.$

Calculation of absolute and percent relative uncertainty:

• On solving the given subtraction part, we get

  $\begin{array}{l}4.97(\pm 0.05)\\ \frac{-1.86(\pm 0.01)}{3.11\left(\pm e\right)}\end{array}$

The uncertainty (e) is calculated as follows,

  $\begin{array}{l}{e}^2={0.05}^2\text{ + 0}{\text{.01}}^{\text{2}}\\ \\ \text{ =}\sqrt{\text{0}\text{.0026}}=0.05099\\ \\ \text{e =}\text{0}\text{.05 (rounded to correct significant figure)}\text{.}\end{array}$

Therefore, the absolute uncertainty of $\left[4.97(\pm 0.05)-1.86(\pm 0.01)\right]$ is given as $3.11\pm 0.05$

• For division of $\left[\text{3}\text{.11±0}\text{.05}\right]\text{÷}\left[\text{21}\text{.2(±0}\text{.2)}\right]\text{=?}$, convert absolute uncertainty to percent relative uncertainty.

For $\text{3}\text{.11±0}\text{.05}$, percent relative uncertainty is $(\pm 0.05/3.11)\times 100=\pm 2\%.$

For $\text{21}\text{.2(±0}\text{.2)}$, percent relative uncertainty is $(\pm 0.2/21.2)\times 100=\pm 1\%.$

On solving the given division, we get

  $\begin{array}{l}\text{3}\text{.11}\left(\text{±2\%}\right)\\ \frac{÷\text{ 21}\text{.2(±1\%)}}{0.147\left(\pm e\%\right)}\end{array}$

The uncertainty (e) is calculated as below,

  $\begin{array}{l}\%e^2=2^2{\text{ + 1}}^{\text{2}}\\ \\ \text{ e}\text{=}\sqrt{5}\text{=}\text{2}\text{.23\%}\\ \\ \text{=}\text{2\%}\text{(rounded to correct significant figure)}\text{.}\end{array}$

Convert relative uncertainty into absolute uncertainty as $0.147\times \frac{2}{100}=0.00294\simeq 0.003$

Therefore,

The absolute uncertainty is given as $\left(0.147\pm 0.003\right).$

The percent relative uncertainty is $\left(0.147\pm 2\%\right).$

Conclusion

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

(d)

Interpretation Introduction

Interpretation:

The absolute and percent relative uncertainty with a reasonable number of significant figures has to be found out for $2.0164(\pm 0.0008)+1.233(\pm 0.002)+4.61(\pm 0.01)=?$

Concept Introduction:

Refer part (a).

Answer to Problem 3.13P

The absolute and percent relative uncertainty with a reasonable number of significant figures is $\left(10.18\pm 0.07\right)$ and  $10.18\pm \left(0.7\%\right)$ 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:

On solving the given addition, we get

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

The uncertainty (e) is calculated as follows,

  $\begin{array}{l}{e}^2={0.0008}^2\text{ + 0}{\text{.002}}^{\text{2}}\text{+ }\text{0}{\text{.01}}^{\text{2}}\\ \\ \text{ =}\sqrt{\text{0}\text{.00010464}}=0.01023\\ \\ \text{e =}\text{0}\text{.01 (rounded to correct significant figure)}\text{.}\end{array}$

Therefore, the absolute uncertainty of given data is $\left(7.86\pm 0.01\right).$

Convert the absolute uncertainty to relative uncertainty as $\frac{\text{0}\text{.01}}{\text{7}\text{.86}}\text{×100=}\text{0}\text{.1\%}$

Therefore, the relative uncertainty is given as $7.86\pm \left(0.1\%\right).$

Conclusion

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

(e)

Interpretation Introduction

Interpretation:

The absolute and percent relative uncertainty with a reasonable number of significant figures has to be found out for $2.0164(\pm 0.0008)\times {10}^3+1.233(\pm 0.002)\times {10}^2+4.61(\pm 0.01)\times {10}^1=?$

Concept Introduction:

Refer part (a).

Answer to Problem 3.13P

The absolute and percent relative uncertainty with a reasonable number of significant figures is $2185.8\left(\pm 0.8\right)$ and $2185.8\left(\pm 0.04\%\right)$ 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:

By reducing scientific notion into $2016.4(\pm 0.8)+123.3(\pm 0.2)+46.1(\pm 0.1)=?$

On solving the given addition, we get

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

The uncertainty (e) is calculated as follows,

  $\begin{array}{l}{e}^2=\left({0.8}^2\text{ +}\text{0}{\text{.2}}^{\text{2}}\text{+ 0}{\text{.1}}^{\text{2}}\right)\\ \\ \text{ =}\sqrt{\text{0}\text{.69}}=0.830\\ \\ \text{e =}\text{0}\text{.8 (rounded to correct significant figure)}\text{.}\end{array}$

Therefore, the absolute uncertainty of given data is $2185.8\left(\pm 0.8\right).$

Convert the absolute uncertainty to relative uncertainty as $\frac{\text{0}\text{.8}}{\text{2185}\text{.8}}\text{×100}\text{=}\text{0}\text{.0365\%}\text{=}\text{0}\text{.04\%}$

Therefore, the relative uncertainty is given as $2185.8\left(\pm 0.04\%\right).$

Conclusion

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