Chapter 3, Problem 3.12P

(a)

Interpretation Introduction

Interpretation:

The absolute and percent relative uncertainty with a reasonable number of significant figures has to be found out for $\left[12.41(\pm 0.09)÷4.16(\pm 0.01)\right]\times 7.0682(\pm 0.0004)=?$

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 3.12P

The absolute and percent relative uncertainty with a reasonable number of significant figures is $21.06\pm \left(0.16\right)$ and $21.06\pm 0.6\%$ respectively.

Explanation of Solution

Given data:

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

Calculation of absolute and percent relative uncertainty:

On solving the given division, we get

  $\begin{array}{l}12.41(\pm 0.09)\\ \frac{÷4.16(\pm 0.01)}{2.98(\pm e)}\end{array}$

For division of $\left[12.41(\pm 0.09)\right]÷\left[4.16(\pm 0.01)\right]\text{=}\text{?}$, convert absolute uncertainty to percent relative uncertainty.

For $12.41(\pm 0.09)$, percent relative uncertainty is $(\pm 0.09/12.41)\times 100=\pm 0.7\%$

For $4.16(\pm 0.01)$, percent relative uncertainty is $(\pm 0.01/4.16)\times 100=\pm 0.2\%$

Therefore,

  $\begin{array}{l}12.41(\pm 0.7\%)\\ \frac{÷4.16(\pm 0.2\%)}{2.98(\pm e\%)}\\ \\ \text{Since, }\\ \\ \%e^2={0.7}^2\text{ + 0}{\text{.2}}^{\text{2}}\\ \\ \%e\text{ =}\sqrt{\text{0}\text{.53}}=0.73\\ \\ \text{=}\text{0}\text{.7\% }\text{(rounded to correct significant figure)}\text{.}\end{array}$

Convert relative uncertainty into absolute uncertainty as $2.98\times \frac{0.7}{100}=\mathrm{0.02.}$

Therefore,

The absolute uncertainty for the above division part obtained as $2.98\left(\pm 0.02\right).$

For the multiplication of $\left[\text{2}\text{.98}\left(\text{±0}\text{.02}\right)\right]\text{×}\left(\text{7}\text{.0682(±0}\text{.0004)}\right)$ is calculated as follows,

  $\begin{array}{l}\text{2}\text{.98}\left(\text{±0}\text{.02}\right)\\ \frac{\times \text{7}\text{.0682(±0}\text{.0004)}}{21.06(\pm e)}\end{array}$

For multiplication of $\left[\text{2}\text{.98}\left(\text{±0}\text{.02}\right)\right]\text{×}\left(\text{7}\text{.0682(±0}\text{.0004)}\right)\text{=}\text{?}$ , convert absolute uncertainty to percent relative uncertainty.

For $\text{2}\text{.98}\left(\text{±0}\text{.02}\right)$, percent relative uncertainty is $(\pm 0.02/2.98)\times 100=\pm 0.6\%$

For $\text{7}\text{.0682(±0}\text{.0004)}$, percent relative uncertainty is $(\pm 0.0004/7.0682)\times 100=\pm 0.005\%$

The uncertainty (e) is calculated as follows,

  $\begin{array}{l}\%e^2={0.005}^2\text{ + 0}{\text{.6}}^{\text{2}}\\ \\ \text{ =}\sqrt{\text{0}\text{.360025}}=0.6\%\\ \\ \text{\%e =}0.6\text{\% (rounded to correct significant figure)}\text{.}\end{array}$

Therefore, the relative uncertainty is given as $21.06\pm 0.6\%$

Convert the relative uncertainty to absolute uncertainty as $\text{21}\text{.06×}\frac{\text{0}\text{.6}}{\text{100}}\text{=}\text{0}\text{.13}$

Therefore, the absolute uncertainty is given as $21.06\pm \left(0.16\right).$

Conclusion

The absolute and percent relative uncertainty with a reasonable number of significant figures is found as $21.06\pm \left(0.16\right)$ and $21.06\pm 0.6\%$ respectively.

(b)

Interpretation Introduction

Interpretation:

The absolute and percent relative uncertainty with a reasonable number of significant figures has to be found out for $\left[3.26(\pm 0.10)\times 8.47(\pm 0.05)\right]-0.18\left(\pm 0.06\right)=?$

Concept Introduction:

Refer part (a).

Answer to Problem 3.12P

The absolute and percent relative uncertainty with a reasonable number of significant figures is $\left(27.42\pm 0.85\right)$ and $27.42\pm 3.1\%.$ respectively.

Explanation of Solution

Given data:

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

Calculation of absolute and percent relative uncertainty:

On solving the given multiplication, we get

  $\begin{array}{l}\text{3}\text{.26 }(\pm 0.10)\\ \frac{\times 8.47(\pm 0.05)}{27.6122(\pm e)}\end{array}$

For multiplication of $\left[\text{3}\text{.26 }(\pm 0.10)\right]\text{×}\left[8.47(\pm 0.05)\right]\text{=}\text{?}$ , convert absolute uncertainty to percent relative uncertainty.

For $\text{3}\text{.26 }(\pm 0.10)$, percent relative uncertainty is $(\pm 0.10/3.26)\times 100=\pm 3.0\%$

For $8.47(\pm 0.05)$, percent relative uncertainty is $(\pm 0.05/8.47)\times 100=\pm 0.6\%$

The uncertainty (e) is calculated as follows,

  $\begin{array}{l}\%e^2={3.0}^2\text{ + 0}{\text{.6}}^{\text{2}}\\ \\ \text{ =}\sqrt{\text{9}\text{.36}}=3.1\%\\ \\ \text{\%e =}\text{3}\text{.1\% (rounded to correct significant figure)}\text{.}\end{array}$

Therefore, the relative uncertainty is given as $27.6\pm 3.1\%$

Convert $27.6\pm 3.1\%$ into absolute uncertainty as $27.6\times \frac{3.1}{100}=\pm 0.85$

For subtraction of $\left[\text{27}\text{.6}\left(\text{±0}\text{.85}\right)\right]\text{-}\text{0}\text{.18}\left(\text{±0}\text{.06}\right)\text{=}\text{?}$ is calculated as follows,

  $\begin{array}{l}\text{27}\text{.6}\left(\text{±0}\text{.85}\right)\\ \frac{\text{-}\text{0}\text{.18}\left(\text{±0}\text{.06}\right)}{27.42\left(e\right)}\\ \\ \text{Since, }\\ \\ e^2={0.85}^2\text{ + 0}{\text{.06}}^{\text{2}}\\ \\ e\text{ =}\sqrt{\text{0}\text{.7261}}=0.85\\ \\ \text{=}\text{0}\text{.8 }\text{(rounded to correct significant figure)}\text{.}\end{array}$

Convert absolute uncertainty $\left(\text{27}\text{.42±0}\text{.85}\right)$ of into relative uncertainty as $\frac{0.86}{27.42}\times 100=3.1\%.$

Therefore,

The absolute uncertainty is given as $\left(27.42\pm 0.85\right)$

The percent relative uncertainty is $27.42\pm 3.1\%.$

Conclusion

The absolute and percent relative uncertainty with a reasonable number of significant figures is founds as $\left(27.42\pm 0.85\right)$ and $27.42\pm 3.1\%.$ respectively.

(c)

Interpretation Introduction

Interpretation:

The absolute and percent relative uncertainty with a reasonable number of significant figures has to be found out for $6.843(\pm 0.008)\times {10}^4÷\left[2.09(\pm 0.04)-1.63(\pm 0.01)\right]=?.$

Concept Introduction:

Refer part (a).

Answer to Problem 3.12P

The absolute and percent relative uncertainty with a reasonable number of significant figures is $\left(14.9\pm 1.34\right)\times 10^4$ and $\left(14.9\pm 9\%\right)\times 10^4$ respectively.

Explanation of Solution

Given data:

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

Calculation of absolute and percent relative uncertainty:

On solving the given subtraction, we get

  $\begin{array}{l}2.09(\pm 0.04)\\ \frac{-1.63(\pm 0.01)}{0.46\pm e}\end{array}$

The uncertainty (e) is calculated as follows,

  $\begin{array}{l}{e}^2={0.04}^2\text{ + 0}{\text{.01}}^{\text{2}}\\ \\ \text{ =}\sqrt{\text{0}\text{.0017}}=0.04\\ \\ \text{e =}\text{0}\text{.04 (rounded to correct significant figure)}\text{.}\end{array}$

Therefore, the absolute uncertainty of $\left[2.09(\pm 0.04)-1.63(\pm 0.01)\right]$ is given as $0.46\pm 0.04.$

For division, convert absolute uncertainty to percent relative uncertainty.

$\left[\text{6}\text{.843(±0}{\text{.008)×10}}^{\text{4}}\right]\text{÷}\left[\text{0}\text{.46±0}\text{.04}\text{.}\right]\text{=?}$

For $\text{6}\text{.843(±0}{\text{.008)×10}}^{\text{4}}$, percent relative uncertainty is $(\pm 0.008/6.843)\times 100=\pm 0.1\%.$

For $\text{0}\text{.46±0}\text{.04}\text{.}$, percent relative uncertainty is $(\pm 0.04/0.46)\times 100=\pm 9\%.$

On solving the given division, we get

  $\begin{array}{l}\text{6}\text{.843(±0}{\text{.1)×10}}^{\text{4}}\\ \frac{÷\text{ 0}\text{.46}\left(\text{±9}\right)\text{.}}{14.9\left(\pm e\%\right){\text{×10}}^{\text{4}}}\end{array}$

The uncertainty (e) is calculated as below,

  $\begin{array}{l}\%e^2={0.1}^2{\text{ + 9}}^{\text{2}}\\ \\ \text{ e}\text{=}\sqrt{81.01}\text{=}\text{9\%}\\ \\ \text{=}\text{9\%}\text{(rounded to correct significant figure)}\text{.}\end{array}$

Convert relative uncertainty into absolute uncertainty as $14.9\times \frac{9}{100}=1.34$

Therefore,

The absolute uncertainty is given as $\left(14.9\pm 1.34\right)\times 10^4.$

The percent relative uncertainty is $\left(14.9\pm 9\%\right)\times 10^4.$

Conclusion

The absolute and percent relative uncertainty with a reasonable number of significant figures is calculated as $\left(14.9\pm 1.34\right)\times 10^4$ and $\left(14.9\pm 9\%\right)\times 10^4$ respectively.