Write each answer with a reasonable number of figures. Find the absolute and percent relative uncertainty for each answer.

$\left(a\right)\mathbf{}\mathbf{[}\mathbf{12}\mathbf{.}\mathbf{41}\left(\pm 0.09\right)\mathbf{÷}\mathbf{4}\mathbf{.}\mathbf{16}\left(\pm 0.01\right)\mathbf{]}\mathbf{\times }\mathbf{7}\mathbf{.}\mathbf{0682}\mathbf{(}\mathbf{\pm }\mathbf{0}\mathbf{.}\mathbf{0004}\mathbf{)}\mathbf{=}\mathbf{?}$

$$\begin{gathered} \mathbf{\left(}\mathbf{b}\mathbf{\right)}\mathbf{}\mathbf{[}\mathbf{3}\mathbf{.}\mathbf{26}\mathbf{(}\mathbf{\pm }\mathbf{0}\mathbf{.}\mathbf{10}\mathbf{)}\mathbf{\times }\mathbf{8}\mathbf{.}\mathbf{47}\mathbf{(}\mathbf{\pm }\mathbf{0}\mathbf{.}\mathbf{05}\mathbf{)}\mathbf{]}\mathbf{-}\mathbf{0}\mathbf{.}\mathbf{18}\mathbf{(}\mathbf{\pm }\mathbf{0}\mathbf{.}\mathbf{06}\mathbf{)}\mathbf{=}\mathbf{?} \\ \mathbf{\left(}\mathbf{c}\mathbf{\right)}\mathbf{}\mathbf{6}\mathbf{.}\mathbf{843}\mathbf{(}\mathbf{\pm }\mathbf{0}\mathbf{.}\mathbf{008}\mathbf{)}\mathbf{\times }{\mathbf{10}}^{\mathbf{4}}\mathbf{÷}\left[2.09(\pm 0.04)-1.63(\pm 0.01)\right]\mathbf{=}\mathbf{?} \\ \mathbf{\left(}\mathbf{d}\mathbf{\right)}\mathbf{}\sqrt{\mathbf{3}\mathbf{.}\mathbf{24}\mathbf{\pm }\mathbf{0}\mathbf{.}\mathbf{08}}\mathbf{=}\mathbf{?} \\ \mathbf{\left(}\mathbf{e}\mathbf{\right)}\mathbf{}{\left(3.24\pm 0.08\right)}^{\mathbf{4}}\mathbf{=}\mathbf{?} \\ \mathbf{\left(}\mathbf{f}\mathbf{\right)}\mathbf{}\mathbf{log}\left(3.24\pm 0.08\right)\mathbf{=}\mathbf{?} \\ \mathbf{\left(}\mathbf{g}\mathbf{\right)}\mathbf{}{\mathbf{10}}^{\mathbf{3}\mathbf{.}\mathbf{24}\mathbf{\pm }\mathbf{0}\mathbf{.}\mathbf{08}}\mathbf{=}\mathbf{?} \end{gathered}$$

(a)

$$\begin{gathered} \left[12.41\right]\left(\pm 0.09\right)÷4.16\left(0.01\right)\times 7.\times 0682\left(\pm 0.0004\right) \\ =\left[12.41\left(\pm 0.725\%\right)÷4.16\left(0.240\%\right)\right]\times 7.\times 0682\left(\pm 0.0057\%\right) \\ \frac{12.41}{4.16}\times 7.0682=21.09 \\ \sqrt{0.{725}^2+0.{0057}^2+0.{24}^2}=0.764 \\ \left[12.41\left(\pm 0.725\%\right)÷4.16\left(\pm 0.240\%\right)\right]\times 7.0682\left(\pm 0.0057\%\right) \\ =21.086\left(\pm 0.764\%\right) \\ =21.086\left(\pm 0.16\right) \end{gathered}$$

Therefore, the absolute uncertainty is ±0.16

The relative uncertainty will be

$\frac{0.16}{21.09}\times 100=0.8\%$

So, the answer is 21.09±(0.16 or 0.8%)

(b)

$$\begin{gathered} \left[3.26\left(\pm 0.1\right)\times 8.47\left(\pm 0.05\right)\right]-0.18(\pm 0.06) \\ =\left[3.26\left(\pm 3.07\right)\times 8.47\left(\pm 0.59\%\right)\right]-0.18(\pm 0.06) \\ 3.26\times 8.47=27.612 \\ \left[27.612\left(\pm 3.12\%\right)\right]-0.18\left(\pm 0.06\right) \\ =\left[27.612\left(\pm 0.863\%\right)\right]-0.18\left(\pm 0.06\right) \\ =27.4\left(\pm 0.86\right) \end{gathered}$$$$\begin{gathered} \left[3.26\left(\pm 0.1\right)\times 8.47\left(\pm 0.05\right)\right]-0.18\left(\pm 0.06\right) \\ =\left[3.26\left(\pm 3.07\%\right)\times 8.47\left(\pm 0.59\%\right)\right]-0.18\left(\pm 0.06\right) \\ 3.26\times 8.47=27.612 \\ \left[27.612\left(\pm 3.12\%\right)\right]-0.18(\pm 0.06) \\ =\left[27.612\left(\pm 0.683\%\right)\right]-0.18(\pm 0.06) \\ =27.4(\pm 0.86) \end{gathered}$$

Therefore, the absolute uncertainty is ±0.86

The relative uncertainty will be

$\frac{0.86}{27.4}\times 100=3.2\%$

So, the answer is 27.4±(0.86 or 3.2%)

(c)

$$\begin{gathered} 6.843\left(\pm 0.008\right)\times {10}^4÷\left[2.09\left(\pm 0.04\right)-1.63\left(\pm 0.01\right)\right] \\ =6.843\left(\pm 0.008\right)\times {10}^4\left[0.46\left(\pm 0.0412\right)\right] \\ =1.49\left(\pm 8.96\%\right)\times {10}^5 \\ =1.49\left(\pm 0.13\%\right)\times {10}^5 \end{gathered}$$

Therefore, the absolute uncertainty is ±0.13

The relative uncertainty will be

role="math" localid="1663312769591" $\frac{0.13}{1.49}\times 100=8.7\%$

So, the answer is 1.49×10⁵±(0.13 or 8.7%)

(d)

$$\begin{gathered} \%e_y=\frac{1}{2}\%e_x \\ =\frac{1}{2}\left(\frac{0.08}{3.24}\times 100\right) \\ =1.235\% \\ \sqrt{3.24\pm \left(0.08\right)} \\ =1.80\pm 1.235\% \\ =1.80\pm 0.022 \end{gathered}$$

Therefore, the absolute uncertainty is ±0.022

The relative uncertainty will be

$\frac{0.022}{1.80}\times 100=1.2\%$

So, the answer is 1.80± (0.022 or 1.2%)

(e)

$$\begin{gathered} \%e_y=4\%e_x \\ =4\left(\frac{0.08}{3.24}\times 100\right) \\ =9.877\% \\ {(3.24\pm \left(0.008\right))}^4=110.2\pm 9.877\% \\ =110.2\pm 0.11 \end{gathered}$$

Therefore, the absolute uncertainty is ±0.13

The relative uncertainty will be

$\frac{0.11}{110.2}\times 100=10\%$

So, the answer is 110.2±(0.11 or 10%)

(f)

With two digits after decimal, the number with the least precision acquires the final value after calculation and writes it with two decimal places. It may be seen from log table base e.

$\log (3.24)=1.175573=1.18$

The absolute uncertainty is:

$\frac{1}{\log \left(10\right)}\times \left(\frac{0.08}{3.24}\right)=1\times \left(\frac{0.08}{3.24}\right)=0.0154321=0.02$

The percentage of relative uncertainty

$\frac{0.0154321\times 100}{1.175573}=\frac{1.54321}{1.175573}=1.31273=1.31\%$

So,the answer is $1.18\pm (0.02$or 1.13%)

(g)

With two digits after decimal, the number with the least precision acquires the final value after calculation and writes it with two decimal places.

${10}^{3.24}={10}^3\times {10}^{0.24}=1000\times 1.737801=1737.801=1737.80$

The absolute uncertainty is:

${10}^{0.08}=1.202264=1.20$

The percentage of relative uncertainty

$\frac{1.2022264\times 100}{1737.801}=\frac{120.22264}{1737.801}=0.0691831=0.07\%$

So, the answer is $1737.80\pm (1.20or0.07\%)$