Chapter 1, Problem 1.58P

(a)

Interpretation Introduction

Interpretation:

The following calculation is to be solved to the correct number of significant figures.

$\frac{2.795\text{ m}\times 3.10\text{ m}}{6.48\text{ m}}$

Concept introduction:

Significant figures of a number are the digits which carry meaningful contribution to its measurement resolution. The rightmost digit of the quantity is the most uncertain digit. The number of certain and uncertain digit in a quantity is considered as significant figures. The digit with a higher number of significant figures has a higher certainty of measurement.

To determine the number of significant figures in a quantity following steps is followed.

1. The quantity must has a decimal point.

2. Start counting from the left and proceed towards the right until the first nonzero digit is encountered. All nonzero digit and the zeroes between two nonzero digits are considered as significant figures. For example, 0.0000765 has three significant figures and 7009 has four significant figures.

3. Zeroes after a decimal point are significant figures. For example, 42.0 have three significant figures.

4. Trailing zeroes that do nothing but are used to set a decimal point are non-significant figures. However, exponential notation can be used to avoid confusion. For example, 4300 has 3 significant figures. It can be expressed in scientific notation as $4.30\times {10}^3$ or

$4.300\times {10}^3$. The number of significant figures in $4.30\times {10}^3$ and $4.300\times {10}^3$ is 3 and 4 respectively.

5. Zeroes present before a trailing decimal point are significant figures. For example, 3200 has only two significant figures but 3200. has 4 significant figures.

Rules to determine significant figures in calculations are as follows:

(1) In multiplication and division operations the result carries the same number of significant figures as the operand or measurement with the fewest significant figures.

(2) In addition and subtraction operations, the result carries the same number of decimal places as the operand or measurement with fewest decimal places.

(3) Exact numbers do not affect the number of significant digits in the final answer.

In mathematical expression which involves mixed operations the result of each intermediate step with proper significant figures. Avoid rounding of the result at intermediate steps. Round off the final answer of the calculation. The rules to round off are as follows:

(1) If the last dropped digit is greater than 5 then increase the preceding digit by 1.

(2) If the last dropped digit is less than 5 then the preceding digit does not change.

(3) If the last digit dropped is 5, then the preceding digit is increased by 1 if it is odd and remains the same if it is even. Also, if 5 is followed by zeroes only then rule (3) is applicable and if it is followed by non-zero digit then rule (1) is applicable.

Answer to Problem 1.58P

The answer of the calculation to a correct number of significant figures is $1.34\text{ m}$.

Explanation of Solution

The given expression is,

$\frac{2.795\text{ m}\times 3.10\text{ m}}{6.48\text{ m}}$

The least significant digit in each number is underlined as follows:

$\frac{2.79\underset{\_}{5}\text{ m}\times 3.1\underset{\_}{0}\text{ m}}{6.4\underset{\_}{8}\text{ m}}$

Multiply $2.79\underset{\_}{5}\text{ m}$ and $3.1\underset{\_}{0}\text{ m}$ in the numerator as:

$2.79\underset{\_}{5}\text{ m}\times 3.1\underset{\_}{0}\text{ m}=\text{8}\text{.6}\underset{\_}{\text{6}}{\text{45 m}}^2$

The result of the multiplication operation must carry the same number of significant figures as the measurement with the fewest significant figures. $2.795\text{ m}$ has 4 significant figures and $3.10\text{ m}$ has 3 significant figures. Therefore, the result must have three significant figures. The least significant digit in the result is 6.

Divide $\text{8}\text{.6}\underset{\_}{\text{6}}{\text{45 m}}^2$ by $6.4\underset{\_}{8}\text{ m}$ as:

$\begin{array}{c}\frac{\text{8}\text{.6}\underset{\_}{\text{6}}{\text{45 m}}^2}{6.4\underset{\_}{8}\text{ m}}=1.3\underset{\_}{3}71\text{ m}\\ =1.34\text{ m}\end{array}$

In division operation the result carries the same number of significant figures as the measurement with the fewest significant figures. $\text{8}\text{.6}\underset{\_}{\text{6}}{\text{45 m}}^2$ has 5 significant figures and $6.4\underset{\_}{8}\text{ m}$ has 3 significant figures. The result must have three significant figures. Therefore, the final answer must be rounded off to the 3 significant figures.

Conclusion

The final answer is rounded off to 3 significant figures. The answer of the calculation to a correct number of significant figures is $1.34\text{ m}$.

(b)

Interpretation Introduction

Interpretation:

The following calculation is to be solved to the correct number of significant figures.

$V=\frac{4}{3}\pi r^3$

Concept introduction:

Significant figures of a number are the digits which carry meaningful contribution to its measurement resolution. The rightmost digit of the quantity is the most uncertain digit.

The number of certain and uncertain digit in a quantity is considered as significant figures. The digit with a higher number of significant figures has a higher certainty of measurement.

To determine the number of significant figures in a quantity following steps is followed.

1. The quantity must has a decimal point.

2. Start counting from the left and proceed towards the right until the first nonzero digit is encountered. All nonzero digit and the zeroes between two nonzero digits are considered as significant figures. For example, 0.0000765 has three significant figures and 7009 has four significant figures.

3. Zeroes after a decimal point are significant figures. For example, 42.0 have three significant figures.

4. Trailing zeroes that do nothing but are used to set a decimal point are non-significant figures. However, exponential notation can be used to avoid confusion. For example, 4300 has 3 significant figures. It can be expressed in scientific notation as $4.30\times {10}^3$ or

$4.300\times {10}^3$. The number of significant figures in $4.30\times {10}^3$ and $4.300\times {10}^3$ is 3 and 4 respectively.

5. Zeroes present before a trailing decimal point are significant figures. For example, 3200 has only two significant figures but 3200. has 4 significant figures.

Rules to determine significant figures in calculations are as follows:

(1) In multiplication and division operations the result carries the same number of significant figures as the operand or measurement with the fewest significant figures.

(2) In addition and subtraction operations, the result carries the same number of decimal places as the operand or measurement with fewest decimal places.

(3) Exact numbers do not affect the number of significant digits in the final answer.

In mathematical expression which involves mixed operations the result of each intermediate step with proper significant figures. Avoid rounding of the result at intermediate steps. Round off the final answer of the calculation. The rules to round off are as follows:

(1) If the last dropped digit is greater than 5 then increase the preceding digit by 1.

(2) If the last dropped digit is less than 5 then the preceding digit does not change.

(3) If the last digit dropped is 5, then the preceding digit is increased by 1 if it is odd and remains the same if it is even. Also, if 5 is followed by zeroes only then rule (3) is applicable and if it is followed by non-zero digit then rule (1) is applicable.

Answer to Problem 1.58P

The answer of the calculation to a correct number of significant figures is $21621{\text{ mm}}^3$.

Explanation of Solution

The value of $r$ is $17.282\text{ mm}$.

The formula to calculate the volume is,

$V=\frac{4}{3}\pi r^3$

Substitute $17.282\text{ mm}$ for $r$ and $3.14159$ for $\pi$ in the above equation.

$\begin{array}{c}V=\frac{4}{3}\left(3.1415\underset{\_}{9}\right){\left(17.28\underset{\_}{2}\text{ mm}\right)}^3\\ =2162\underset{\_}{0}.74{\text{ mm}}^3\\ =21621{\text{ mm}}^3\end{array}$

The result of the multiplication and division operation must carry the same number of significant figures as the measurement with the fewest significant figures. $17.282\text{ mm}$ has 5 significant figures and $3.14159$ has 6 significant figures. The result must have 5 significant figures. Therefore, the final answer must be rounded off to the 5 significant figures. The last digit to be dropped is 7. 7 is greater than 5, therefore, the preceding digit is increased by 1.

Conclusion

The final answer is rounded off to 5 significant figures. The answer of the calculation to a correct number of significant figures is $21621{\text{ mm}}^3$.

(c)

Interpretation Introduction

Interpretation:

The following calculation is to be solved to the correct number of significant figures.

$1.110\text{ cm}+17.3\text{ cm}+108.2\text{ cm}+\text{316 cm}$

Concept introduction:

Significant figures of a number are the digits which carry meaningful contribution to its measurement resolution. The rightmost digit of the quantity is the most uncertain digit.

The number of certain and uncertain digit in a quantity is considered as significant figures. The digit with a higher number of significant figures has a higher certainty of measurement.

To determine the number of significant figures in a quantity following steps is followed.

1. The quantity must has a decimal point.

2. Start counting from the left and proceed towards the right until the first nonzero digit is encountered. All nonzero digit and the zeroes between two nonzero digits are considered as significant figures. For example, 0.0000765 has three significant figures and 7009 has four significant figures.

3. Zeroes after a decimal point are significant figures. For example, 42.0 have three significant figures.

4. Trailing zeroes that do nothing but are used to set a decimal point are non-significant figures. However, exponential notation can be used to avoid confusion. For example, 4300 has 3 significant figures. It can be expressed in scientific notation as $4.30\times {10}^3$ or

$4.300\times {10}^3$. The number of significant figures in $4.30\times {10}^3$ and $4.300\times {10}^3$ is 3 and 4 respectively.

5. Zeroes present before a trailing decimal point are significant figures. For example, 3200 has only two significant figures but 3200. has 4 significant figures.

Rules to determine significant figures in calculations are as follows:

(1) In multiplication and division operations the result carries the same number of significant figures as the operand or measurement with the fewest significant figures.

(2) In addition and subtraction operations, the result carries the same number of decimal places as the operand or measurement with fewest decimal places.

(3) Exact numbers do not affect the number of significant digits in the final answer.

In mathematical expression which involves mixed operations the result of each intermediate step with proper significant figures. Avoid rounding of the result at intermediate steps. Round off the final answer of the calculation. The rules to round off are as follows:

(1) If the last dropped digit is greater than 5 then increase the preceding digit by 1.

(2) If the last dropped digit is less than 5 then the preceding digit does not change.

(3) If the last digit dropped is 5, then the preceding digit is increased by 1 if it is odd and remains the same if it is even. Also, if 5 is followed by zeroes only then rule (3) is applicable and if it is followed by non-zero digit then rule (1) is applicable.

Answer to Problem 1.58P

The answer of the calculation to a correct number of significant figures is $443\text{ cm}$.

Explanation of Solution

The given expression is,

$1.110\text{ cm}+17.3\text{ cm}+108.2\text{ cm}+\text{316 cm}$

The result of the addition operation must carry the same number of decimal places as the measurement with fewest decimal places.

Add $1.110\text{ cm}$, $17.3\text{ cm}$, $108.2\text{ cm}$, and $\text{316 cm}$ as follows,

$\begin{array}{c}1.110\text{ cm}+17.3\text{ cm}+108.2\text{ cm}+\text{316 cm}=44\underset{\_}{2}.61\text{ cm}\\ =443\text{ cm}\end{array}$

Measurement $\text{316 cm}$ has no digit after the decimal point. Therefore, the final answer is rounded off to 3 significant figures. The last digit to be dropped is 6. 6 is greater than 5, therefore, the preceding digit is increased by 1.

Conclusion

The final answer is rounded off to 3 significant figures. The answer of the calculation to a correct number of significant figures is $443\text{ cm}$.