Chapter 1, Problem 22QRT

(a)

Interpretation Introduction

Interpretation:

The following calculations has to be performed and the result with the proper number of significant figures has to be expressed.

  $\text{2221}\text{.05-}\frac{\text{3256}\text{.5}}{\text{3}\text{.20}}$

Concept Introduction:

Significant digits:

In a number, the digits which contribute to the precision of the number are said to be significant digits.

Rules for determination of significant digits in a number:

• All non-zero digits are significant.
• The zero’s appearing between two non-zero digits are significant.
• The zero’s before the non-zero numbers are not significant.
• Zero’s after non-zero number without decimal are non-significant.
• Zero’s after non-zero number with decimal are significant.
• Zero’s after the decimal point are significant.
• Any numbers with scientific notation are significant.

The sum of the significant digits has to be given in the lowest decimal present in the values.

The significant digit in multiplication and division has to be given as the quantity with the fewest significant figures.

Explanation of Solution

The given calculation is

  $\text{2221}\text{.05-}\frac{\text{3256}\text{.5}}{\text{3}\text{.20}}$

The calculation is division.  The numerator consists of five significant figures as all non-zero digits are significant.  The denominator consists of three significant figures as zero’s present after the non-zero digit with decimal point are significant.  The result has three significant figures ($\text{1,}\text{0}\text{and}\text{2}$).

  $\text{2}{\text{.22105×10}}^{\text{3}}\text{ - 1}{\text{.02×10}}^{\text{3}}$

The resulted value is subtracted with $\text{2}{\text{.22105×10}}^{\text{3}}$.  The value consists of six significant figures as zero’s present after the non-zero digit with decimal point are significant.  The resulting value contains three significant figures.

  $\text{1}{\text{.20×10}}^{\text{3}}$

The result contains three significant figures ($\text{1,}\text{2}\text{and}\text{0}$).

(b)

Interpretation Introduction

Interpretation:

The following calculations has to be performed and the result with the proper number of significant figures has to be expressed.

  $\text{343}\text{.2}\text{×}\text{(2}{\text{.01×10}}^{\text{-3}}\text{)}$

Concept Introduction:

Refer part (a).

Explanation of Solution

Given, $\text{343}\text{.2}\text{×}\text{(2}{\text{.01×10}}^{\text{-3}}\text{)}$.

The calculation is of multiplication.  The first value ($\text{343}\text{.2}$) consists of four significant figures and the second value ($\text{2}{\text{.01×10}}^{\text{-3}}$) consists of three significant figures as all non-zero digits and zero’s after a decimal point are significant.  The resulting value has to be given with three significant figures.

  $\begin{array}{l}\text{343}\text{.2 × (2}{\text{.01×10}}^{\text{-3}}\text{)}\\ \\ \text{= 343}\text{.2× 0}\text{.00201}\\ \\ \text{= 0}\text{.690}\end{array}$

As the zero’s present before the non-zero digits are non-significant, the resulted value contains three significant figures ($\text{6,}\text{9}\text{and}\text{0}$).

(c)

Interpretation Introduction

Interpretation:

The following calculations has to be performed and the result with the proper number of significant figures has to be expressed.

  $\text{S=}{\text{4πr}}^{\text{2}}\text{where}\text{r = 2}\text{.55}\text{cm}$

Concept Introduction:

Refer part (a).

Explanation of Solution

Given, $\text{S=}{\text{4πr}}^{\text{2}}\text{where}\text{r = 2}\text{.55}\text{cm}$.

The value of $\text{r}$ is given and the value of $\text{π = 3}\text{.1415926}\mathrm{...}$.  The value of $\text{r}$ contains three significant figures and the value of $\text{π}$ contains eight significant figures as all non-zero digits are significant.  The resulting value has to be given with three significant figures.

Substitute the values in the formula.

  $\begin{array}{l}\text{S=}{\text{4πr}}^{\text{2}}\\ \\ \text{S=4(3}\text{.1415926)(2}\text{.55}{\text{cm)}}^{\text{2}}\\ \\ \text{S=4(3}\text{.1415926)(6}\text{.50}{\text{cm)}}^{\text{2}}\\ \\ \text{S=4×20}\text{.4}{\text{cm}}^{\text{2}}\\ \\ \text{S=81}\text{.7}{\text{cm}}^{\text{2}}\end{array}$

The result contains three significant figures ($\text{8,}\text{1}\text{and}\text{7}$).

(d)

Interpretation Introduction

Interpretation:

The following calculations has to be performed and the result with the proper number of significant figures has to be expressed.

  $\frac{\text{2802}}{\text{15}}\text{-}\text{(0}\text{.0025×10,000)}$

Concept Introduction:

Refer part (a).

Explanation of Solution

Given, $\frac{\text{2802}}{\text{15}}\text{-}\text{(0}\text{.0025×10,000)}$.

The calculation is division.  The numerator consists of four significant figures as all non-zero digits are significant.  The denominator consists of two significant figures as all non-zero digits are significant.  The result has to be with two significant figures.

  $\begin{array}{l}\frac{\text{2802}}{\text{15}}\text{-}\text{(0}\text{.0025×10,000)}\\ \\ \text{=1}{\text{.9×10}}^{\text{2}}\text{-}\text{(0}\text{.0025×10,000)}\end{array}$

The multiplication part gives two significant figures.

  $\begin{array}{l}\frac{\text{2802}}{\text{15}}\text{-}\text{(0}\text{.0025×10,000)}\\ \\ \text{=1}{\text{.9×10}}^{\text{2}}\text{-}\text{0}{\text{.25×10}}^{\text{2}}\end{array}$

The result consist of two significant figures.  The value obtained is rounded off.

  $\begin{array}{l}\frac{\text{2802}}{\text{15}}\text{-(0}\text{.0025×10,000)}\\ \\ \text{=1}{\text{.9×10}}^{\text{2}}\text{-}\text{0}{\text{.25×10}}^{\text{2}}\\ \\ \text{=1}\text{.65}{\text{×10}}^{\text{2}}\\ \\ \text{=1}\text{.7}{\text{×10}}^{\text{2}}\end{array}$

The result consists of two significant figures ($\text{1}\text{and}\text{7}$).