Chapter 28, Problem 7A

To determine

(a)

The measurement of $\text{a}$.

Answer to Problem 7A

The measurement of $\text{a}$ is ${\left(\frac{3}{32}\right)}^{\prime \text{}\prime }$.

Explanation of Solution

Given information:

The below figure represent the Metric steel Rule.

Figure-(1)

Write the expression for the measurement by the enlarged fractional rule.

  $\text{M}=\text{F}+\text{T}{\left(\frac{1}{32}\right)}^{\prime \text{}\prime }$ ........ (I)

Here, the measurement by the enlarged fractional rule is $\text{M}$, the total measurement of scale nearest of the point is $\text{F}$ and the fractional measurement of scale nearest of the point is $\text{T}$.

Here, the total measurement of scale nearest of $\text{a}$ is $0$ and the fractional measurement of scale nearest of $\text{a}$ is $3$ as shown in Figure-(1).

Calculation:

Substitute $3$ for $\text{T}$ and $0$ for $\text{F}$ in Equation (I).

  $\begin{array}{c}\text{M}=0+\left(3\right){\left(\frac{1}{32}\right)}^{\prime \text{}\prime }\\ =\left(3\right){\left(\frac{1}{32}\right)}^{\prime \text{}\prime }\\ ={\left(\frac{3}{32}\right)}^{\prime \text{}\prime }\end{array}$

Conclusion:

The measurement of $\text{a}$ is ${\left(\frac{3}{32}\right)}^{\prime \text{}\prime }$.

To determine

(b)

The measurement of $\text{b}$.

Answer to Problem 7A

The measurement of $\text{b}$ is ${\left(\frac{5}{16}\right)}^{\prime \text{}\prime }$.

Explanation of Solution

Given information:

Here, the total measurement of scale nearest of $\text{b}$ is $0$ and the fractional measurement of scale nearest of $\text{b}$ is $10$ as shown in Figure-(1).

Calculation:

Substitute $0$ for $\text{T}$ and $10$ for $\text{F}$ in Equation (I).

  $\begin{array}{c}\text{M}=\left(0\right)+\left(10\right){\left(\frac{1}{32}\right)}^{\prime \text{}\prime }\\ =\left(10\right){\left(\frac{1}{32}\right)}^{\prime \text{}\prime }\\ ={\left(\frac{5}{16}\right)}^{\prime \text{}\prime }\end{array}$

Conclusion:

The measurement of $\text{b}$ is ${\left(\frac{5}{16}\right)}^{\prime \text{}\prime }$.

To determine

(c)

The measurement of $\text{c}$.

Answer to Problem 7A

The measurement of $\text{c}$ is $0.5^{″}$.

Explanation of Solution

Given information:

Here, the total measurement of scale nearest of $\text{c}$ is $0$ and the fractional measurement of scale nearest of $\text{c}$ is $16$ as shown in Figure-(1).

Calculation:

Substitute $0$ for $\text{T}$ and $16$ for $\text{F}$ in Equation (I).

  $\begin{array}{c}\text{M}=\left(0\right)+\left(16\right){\left(\frac{1}{32}\right)}^{\prime \text{}\prime }\\ =\left(16\right){\left(\frac{1}{32}\right)}^{\prime \text{}\prime }\\ ={\left(\frac{1}{2}\right)}^{\prime \text{}\prime }\\ =0.5^{″}\end{array}$

Conclusion:

The measurement of $\text{c}$ is ${0.5}^{\prime \text{}\prime }$.

To determine

(d)

The measurement of $\text{d}$.

Answer to Problem 7A

The measurement of $\text{d}$ is ${\left(\frac{5}{8}\right)}^{\prime \text{}\prime }$.

Explanation of Solution

Given information:

Here, the total measurement of scale nearest of $\text{d}$ is $0.5^{″}$ and the fractional measurement of scale nearest of $\text{d}$ is $4$ as shown in Figure-(1).

Calculation:

Substitute $0.5^{″}$ for $\text{T}$ and $4$ for $\text{F}$ in Equation (I).

  $\begin{array}{c}\text{M}=0.5^{″}+\left(4\right){\left(\frac{1}{32}\right)}^{\prime \text{}\prime }\\ ={\left(\frac{1}{2}\right)}^{\prime \text{}\prime }+{\left(\frac{1}{8}\right)}^{\prime \text{}\prime }\\ ={\left(\frac{5}{8}\right)}^{\prime \text{}\prime }\end{array}$

Conclusion:

The measurement of $\text{d}$ is ${\left(\frac{5}{8}\right)}^{\prime \text{}\prime }$.

To determine

(e)

The measurement of $\text{e}$.

Answer to Problem 7A

The measurement of $\text{e}$ is ${\left(\frac{3}{4}\right)}^{\prime \text{}\prime }$.

Explanation of Solution

Given information:

Here, the total measurement of scale nearest of $\text{e}$ is $0.5^{″}$ and the fractional measurement of scale nearest of $\text{e}$ is $8$ as shown in Figure-(1).

Calculation:

Substitute $0.5^{″}$ for $\text{T}$ and $8$ for $\text{F}$ in Equation (I).

  $\begin{array}{c}\text{M}=\left(0.5^{″}\right)+\left(8\right){\left(\frac{1}{32}\right)}^{\prime \text{}\prime }\\ ={\left(\frac{1}{2}\right)}^{\prime \text{}\prime }+{\left(\frac{1}{4}\right)}^{\prime \text{}\prime }\\ ={\left(\frac{3}{4}\right)}^{\prime \text{}\prime }\end{array}$

Conclusion:

The measurement of $\text{e}$ is ${\left(\frac{3}{4}\right)}^{\prime \text{}\prime }$.

To determine

(f)

The measurement of $\text{f}$.

Answer to Problem 7A

The measurement of $\text{f}$ is ${\left(\frac{29}{32}\right)}^{\prime \text{}\prime }$.

Explanation of Solution

Given information:

Here, the total measurement of scale nearest of $\text{f}$ is $0.5^{″}$ and the fractional measurement of scale nearest of $\text{f}$ is $13$ as shown in Figure-(1).

Calculation:

Substitute $0.5^{″}$ for $\text{T}$ and $13$ for $\text{F}$ in Equation (I).

  $\begin{array}{c}\text{M}=\left(0.5^{″}\right)+\left(13\right){\left(\frac{1}{32}\right)}^{\prime \text{}\prime }\\ ={\left(\frac{1}{2}\right)}^{\prime \text{}\prime }+{\left(\frac{13}{32}\right)}^{\prime \text{}\prime }\\ ={\left(\frac{29}{32}\right)}^{\prime \text{}\prime }\end{array}$

Conclusion:

The measurement of $\text{f}$ is ${\left(\frac{29}{32}\right)}^{\prime \text{}\prime }$.

To determine

(g)

The measurement of $\text{g}$.

Answer to Problem 7A

The measurement of $\text{g}$ is ${\left(\frac{35}{32}\right)}^{\prime \text{}\prime }$.

Explanation of Solution

Given information:

Here, the total measurement of scale nearest of $\text{g}$ is $1^{″}$ and the fractional measurement of scale nearest of $\text{g}$ is $3$ as shown in Figure-(1).

Calculation:

Substitute $1^{″}$ for $\text{T}$ and $3$ for $\text{F}$ in Equation (I).

  $\begin{array}{c}\text{M}=\left(1^{″}\right)+\left(3\right){\left(\frac{1}{32}\right)}^{\prime \text{}\prime }\\ =\left(1^{″}\right)+{\left(\frac{3}{32}\right)}^{\prime \text{}\prime }\\ ={\left(\frac{35}{32}\right)}^{\prime \text{}\prime }\end{array}$

Conclusion:

The measurement of $\text{g}$ is ${\left(\frac{35}{32}\right)}^{\prime \text{}\prime }$.

To determine

(h)

The measurement of h.

Answer to Problem 7A

The measurement of h is ${\left(\frac{21}{16}\right)}^{\prime \text{}\prime }$.

Explanation of Solution

Given information:

Here, the total measurement of scale nearest of h is $1^{″}$ and the fractional measurement of scale nearest of h is $10$ as shown in Figure-(1).

Calculation:

Substitute $1^{″}$ for $\text{T}$ and $10$ for $\text{F}$ in Equation (I).

  $\begin{array}{c}\text{M}=\left(1^{″}\right)+\left(10\right){\left(\frac{1}{32}\right)}^{\prime \text{}\prime }\\ =\left(1^{″}\right)+{\left(\frac{5}{16}\right)}^{\prime \text{}\prime }\\ ={\left(\frac{21}{16}\right)}^{\prime \text{}\prime }\end{array}$

Conclusion:

The measurement of h is ${\left(\frac{21}{16}\right)}^{\prime \text{}\prime }$.

To determine

(i)

The measurement of i.

Answer to Problem 7A

The measurement of i is ${\left(\frac{5}{64}\right)}^{\prime \text{}\prime }$.

Explanation of Solution

Given information:

Write the expression of measurement of the enlarged decimal rule.

  $\text{M}=\text{F}{\left(\frac{1}{32}\right)}^{\prime \text{}\prime }+\text{T}{\left(\frac{1}{64}\right)}^{\prime \text{}\prime }$ ........ (II)

Here, the total measurement of scale nearest of $\text{i}$ is $2$ and the fractional measurement of scale nearest of $\text{i}$ is $1$ as shown in Figure-(1).

Calculation:

Substitute $2$ for $\text{F}$ and $1$ for $\text{T}$ in Equation (II).

  $\begin{array}{c}\text{M}=2{\left(\frac{1}{32}\right)}^{\prime \text{}\prime }+{\left(\frac{1}{64}\right)}^{\prime \text{}\prime }\\ ={\left(\frac{1}{6}\right)}^{\prime \text{}\prime }+{\left(\frac{1}{64}\right)}^{\prime \text{}\prime }\\ ={\left(\frac{5}{64}\right)}^{\prime \text{}\prime }\end{array}$

Conclusion:

The measurement of i is ${\left(\frac{5}{64}\right)}^{\prime \text{}\prime }$.

To determine

(j)

The measurement of $\text{j}$.

Answer to Problem 7A

The measurement of $\text{j}$ is ${\left(\frac{7}{32}\right)}^{\prime \text{}\prime }$.

Explanation of Solution

Given information:

Here, the total measurement of scale nearest of $\text{j}$ is $6$ and the fractional measurement of scale nearest of $\text{c}$ is $2$ as shown in Figure-(1).

Calculation:

Substitute $6$ for $\text{F}$ and $2$ for $\text{T}$ in Equation (II).

  $\begin{array}{c}\text{M}=\left(6\right){\left(\frac{1}{32}\right)}^{\prime \text{}\prime }+\left(2\right){\left(\frac{1}{64}\right)}^{\prime \text{}\prime }\\ ={\left(\frac{3}{16}\right)}^{\prime \text{}\prime }+{\left(\frac{1}{32}\right)}^{\prime \text{}\prime }\\ ={\left(\frac{7}{32}\right)}^{\prime \text{}\prime }\end{array}$

Conclusion:

The measurement of $\text{j}$ is ${\left(\frac{7}{32}\right)}^{\prime \text{}\prime }$.

To determine

(k)

The measurement of $\text{k}$.

Answer to Problem 7A

The measurement of $\text{k}$ is ${\left(\frac{3}{8}\right)}^{\prime \text{}\prime }$.

Explanation of Solution

Given information:

Here, the total measurement of scale nearest of k is $0$ and the fractional measurement of scale nearest of k is $12$ as shown in Figure-(1).

Calculation:

Substitute $0$ for $\text{T}$ and $12$ for $\text{F}$ in Equation (I).

  $\begin{array}{c}\text{M}=\left(0\right)+\left(12\right){\left(\frac{1}{32}\right)}^{\prime \text{}\prime }\\ ={\left(\frac{3}{8}\right)}^{\prime \text{}\prime }\end{array}$

Conclusion:

The measurement of $\text{k}$ is ${\left(\frac{3}{8}\right)}^{\prime \text{}\prime }$.

To determine

(l)

The measurement of l.

Answer to Problem 7A

The measurement of l is ${\left(\frac{35}{64}\right)}^{\prime \text{}\prime }$.

Explanation of Solution

Given information:

Here, the total measurement of scale nearest of l is $5$ and the fractional measurement of scale nearest of l is $1$ as shown in Figure-(1).

Write the expression for the enlarged decimal rule.

  $\text{M}={\text{T}}_{\text{1}}+\text{F}{\left(\frac{1}{32}\right)}^{\prime \text{}\prime }+\text{T}{\left(\frac{1}{64}\right)}^{\prime \text{}\prime }$ ...... (III)

Here, the total measurement is ${\text{T}}_{\text{1}}$.

Calculation:

Substitute ${\left(\frac{3}{8}\right)}^{\prime \text{}\prime }$ for ${\text{T}}_{\text{1}}$, $5$ for $\text{F}$ and $1$ for $\text{T}$ in Equation (II).

  $\begin{array}{c}\text{M}={\left(\frac{3}{8}\right)}^{\prime \text{}\prime }+5{\left(\frac{1}{32}\right)}^{\prime \text{}\prime }+\left(1\right){\left(\frac{1}{64}\right)}^{\prime \text{}\prime }\\ ={\left(\frac{3}{8}\right)}^{\prime \text{}\prime }+{\left(\frac{5}{32}\right)}^{\prime \text{}\prime }+{\left(\frac{1}{64}\right)}^{\prime \text{}\prime }\\ ={\left(\frac{35}{64}\right)}^{\prime \text{}\prime }\end{array}$

Conclusion:

The measurement of l is ${\left(\frac{35}{64}\right)}^{\prime \text{}\prime }$.

To determine

(m)

The measurement of m.

Answer to Problem 7A

The measurement of m is ${\left(\frac{51}{64}\right)}^{\prime \text{}\prime }$.

Explanation of Solution

Given information:

Here, the total measurement of scale nearest of m is $24$ and the fractional measurement of scale nearest of $\text{m}$ is $3$ as shown in Figure-(1).

Calculation:

Substitute $24$ for $\text{F}$ and $3$ for $\text{T}$ in Equation (I).

  $\begin{array}{c}\text{M}=\left(24\right){\left(\frac{1}{32}\right)}^{\prime \text{}\prime }+\left(3\right){\left(\frac{1}{64}\right)}^{\prime \text{}\prime }\\ ={\left(\frac{51}{64}\right)}^{\prime \text{}\prime }\end{array}$

Conclusion:

The measurement of m is ${\left(\frac{51}{64}\right)}^{\prime \text{}\prime }$.

To determine

(n)

The measurement of n.

Answer to Problem 7A

The measurement of n is ${\left(\frac{31}{32}\right)}^{\prime \text{}\prime }$.

Explanation of Solution

Given information:

Write the expression of measurement of the enlarged decimal rule.

  $\text{M}={\text{T}}_{\text{1}}-\text{T}{\left(\frac{1}{64}\right)}^{\prime \text{}\prime }$ ........ (IV)

Here, the total measurement is ${\text{T}}_{\text{1}}$.

Here, the fractional measurement of scale nearest of n is $2$ and the total measurement is $1^{″}$ as shown in Figure-(1).

Calculation:

Substitute $1^{″}$ for ${\text{T}}_{\text{1}}$ and $2$ for $\text{T}$ in Equation (IV).

  $\begin{array}{c}\text{M}=\left(1^{″}\right)-2{\left(\frac{1}{64}\right)}^{\prime \text{}\prime }\\ =\left(1^{″}\right)-{\left(\frac{1}{32}\right)}^{\prime \text{}\prime }\\ ={\left(\frac{31}{32}\right)}^{\prime \text{}\prime }\end{array}$

Conclusion:

The measurement of n is ${\left(\frac{31}{32}\right)}^{\prime \text{}\prime }$.

To determine

(o)

The measurement of o.

Answer to Problem 7A

The measurement of o is ${\left(1\frac{15}{64}\right)}^{\prime \text{}\prime }$.

Explanation of Solution

Given information:

Here, the total measurement of scale nearest of m is $7$, the fractional measurement of scale nearest of m is $1$ and the total measurement is $1^{″}$ as shown in Figure-(1).

Calculation:

Substitute $1^{″}$ for ${\text{T}}_{\text{1}}$, $7$ for $\text{F}$ and $1$ for $\text{T}$ in Equation (III).

  $\begin{array}{c}\text{M}=\left(1^{″}\right)+\left(7\right){\left(\frac{1}{32}\right)}^{\prime \text{}\prime }+\left(1\right){\left(\frac{1}{64}\right)}^{\prime \text{}\prime }\\ =\left(1^{″}\right)+{\left(\frac{7}{32}\right)}^{\prime \text{}\prime }+{\left(\frac{1}{64}\right)}^{\prime \text{}\prime }\\ ={\left(1\frac{15}{64}\right)}^{\prime \text{}\prime }\end{array}$

Conclusion:

The measurement of o is ${\left(1\frac{15}{64}\right)}^{\prime \text{}\prime }$.

To determine

(o)

The measurement of p.

Answer to Problem 7A

The measurement of p is ${\left(1\frac{29}{64}\right)}^{\prime \text{}\prime }$.

Explanation of Solution

Given information:

Here, the total measurement of scale nearest of m is $4$, the fractional measurement of scale nearest of m is $1$ and the total measurement is ${\left(\frac{21}{16}\right)}^{\prime \text{}\prime }$ as shown in Figure-(1).

Calculation:

Substitute ${\left(\frac{21}{16}\right)}^{\prime \text{}\prime }$ for ${\text{T}}_{\text{1}}$, $4$ for $\text{F}$ and $1$ for $\text{F}$ in Equation (IV).

  $\begin{array}{c}\text{M}={\left(\frac{21}{16}\right)}^{\prime \text{}\prime }+\left(4\right){\left(\frac{1}{32}\right)}^{\prime \text{}\prime }+\left(1\right){\left(\frac{1}{64}\right)}^{\prime \text{}\prime }\\ ={\left(\frac{21}{16}\right)}^{\prime \text{}\prime }+{\left(\frac{1}{8}\right)}^{\prime \text{}\prime }+{\left(\frac{1}{64}\right)}^{\prime \text{}\prime }\\ ={\left(\frac{29}{64}\right)}^{\prime \text{}\prime }\end{array}$

Conclusion:

The measurement of p is ${\left(1\frac{29}{64}\right)}^{\prime \text{}\prime }$.