Tag Archives: Significant Figures

Problem 1-20: Solving for the percent uncertainty of a given blood pressure of 120±2 mmHg

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PROBLEM:

(a) A person’s blood pressure is measured to be 120 \pm 2 mmHg. What is its percent uncertainty?

(b) Assuming the same percent uncertainty, what is the uncertainty in a blood pressure measurement of 80 mmHg?

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SOLUTION:

Part A

The percent uncertainty is computed as

\text{\% uncertainty}=\frac{2\:\text{mmHg}}{120\:\text{mmHg}}\times 100\%=1.7\% \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right)

Part B

The uncertainty in the blood pressure is

\delta _{bp}=\frac{1.7\:\%}{100\:\%}\times 80\:\text{mmHg}=1.3\:\text{mmHg} \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right)
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Problem 1-19: Calculating the percent uncertainty and range of speed with the same percent uncertainty

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PROBLEM:

(a) If your speedometer has an uncertainty of 2.0 km/h at a speed of 90 km/h, what is the percent uncertainty?

(b) If it has the same percent uncertainty when it reads 60 km/ h, what is the range of speeds you could be going?

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SOLUTION:

Part a

The percent uncertainty is computed as

\text{\% uncertainty}=\frac{2.0\ \text{km/hr}}{90\:\text{km/hr}}\times 100\%=2.2\% \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right)

Part b

The tolerance of the velocity is

\delta _v=\frac{2.2\:\%}{100\:\%}\times 60\:\text{km/hr}=1\:\text{km/hr}

Therefore, the range of the velocity is 60±1km/h, or that is 59 to 61 km/h. \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right)

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Problem 1-18: Significant figures, uncertainty, and accuracy of the numbers 99 and 100

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PROBLEM:

(a) How many significant figures are in the numbers 99 and 100?

(b) If the uncertainty in each number is 1, what is the percent uncertainty in each?

(c) Which is a more meaningful way to express the accuracy of these two numbers, significant figures or percent uncertainties?

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SOLUTION:

Part A

99 has 2 significant figures

100 has 3 significant figures

Part B

\begin{align*}
\frac{1}{99}\times 100\% & =1.0\:\%  \\ 
\frac{1}{100}\times 100\% & =1.00\%
\end{align*}

Part C

Based on the results of parts a and b, the percent uncertainties are more meaningful.

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Problem 1-17: Stating the correct significant figures in a given calculation

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PROBLEM:

State how many significant figures are proper in the results of the following calculations:

\begin{align*}
\left(a\right)\:\frac{\left(106.7\right)\left(98.2\right)}{\left(46.210\right)\left(1.01\right)} \ \qquad \ \left(b\right)\:\left(18.7\right)^2 \ \qquad \ \left(c\right)\:\left(1.60\times 10^{-19}\right)\left(3712\right)
\end{align*}
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SOLUTION:

Part A

The answer is limited by 98.2 and 1.01. They are both 3 significant figures. So the result should be 3 significant figures.

Part B

The answer is limited by 18.7. The answer should be 3 significant figures. So the result should be 3 significant figures.

Part C

The answer is limited by 1.60 which is 3 significant figures. So the result should be 3 significant figures.

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Problem 1-16: Solving for the remaining soda after the removal of some volume

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PROBLEM:

A can contains 375 mL of soda. How much is left after 308 mL is removed?

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SOLUTION:

What is left is calculated as

\begin{align*}
\text{Volume left} & =375 \ \text{mL}-308 \ \text{mL} \\
& = 67 \ \text{mL} \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right)
\end{align*}

Therefore, 67 mL of soda is left.

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Problem 1-15: Solving for the number of heartbeats of a person with the correct significant figure

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PROBLEM:

(a) Suppose that a person has an average heart rate of 72.0 beats/min. How many beats does he or she have in 2.0 y?
(b) In 2.00 y?
(c) In 2.000 y?

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SOLUTION:

Part a

This is a problem on rounding off. Take note that 72.0 has three significant figures, 2.0 has two significant figures, 2.00 has three significant figures, and 2.000 has four significant figures.

Convert 72 beats/min to beats/year.

\begin{align*}
72 \ \text{beats/min} & = \left(\frac{72.0\:\text{beats}}{1\:\bcancel{\text{min}}}\right)\left(\frac{60.0\:\bcancel{\text{min}}}{1.00\:\bcancel{\text{hr}}}\right)\left(\frac{24.0\:\bcancel{\text{hr}}}{1.00\:\bcancel{\text{day}}}\right)\left(\frac{365.25\:\bcancel{\text{days}}}{1\:\text{year}}\right)\\
& =3.7869\times 10^7\:\text{beats/year}
\end{align*}

In 2 years, the number of beats of an average person is

\begin{align*}
& = 3.7869 \times 10^{7} \ \text{beats/year} \times 2 \ \text{years} \\
& = 7.5738 \times 10^7 \ \text{beats} 
\end{align*}

For part a, the answer is limited to 2.0 which is 2 significant figures. The answer is 7.6 \times 10^{7} \ \text{beats} \ \color{DarkOrange} \left( \text{Answer} \right).

Part b

For item letter b), the answer is limited by 2.00, which is 3 significant figures. The answer is 7.57\times 10^7\:\text{beats} \ \color{DarkOrange} \left( \text{Answer} \right).

Part c

For item letter c), the answer is limited by 72.0, which is 3 significant figures. The answer is  7.57\times 10^7\:\text{beats} \ \color{DarkOrange} \left( \text{Answer} \right).

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