Chapter 29, Problem 9A

To determine

(a)

The value of measurement A.

The value of measurement B.

The value of measurement C.

Answer to Problem 9A

The value of measurement A is $3.225^{″}$.

The value of measurement B is $3.250^{″}$.

The value of measurement C is $17$.

Explanation of Solution

Given information:

The vernier caliper setting is $3.242^{″}$.

Write the expression for the lower limit of vernier scale reading.

  $\text{A}=\text{MSR}+\text{L}+\text{D}$ ..... (I)

Here, the lower limit of vernier scale reading is $\text{A}$, the main scale reading is $\text{MSR}$, the reading of $0.1^{″}$ division on the main scale is $\text{L}$ and the reading of $0.025^{″}$ divisions on the main scale is $\text{D}$.

Write the expression for the upper limit of vernier scale reading.

  $\text{B}=\text{A}+0.025^{″}$ ..... (II)

Here, the upper limit of vernier scale reading is $\text{B}$.

Write the expression for the verneir coincidence.

  $\text{C}=\frac{\text{VCS}-\text{A}}{\text{LC}}$ ..... (III)

Here, the verneir coincidence is $\text{C}$, the vernier caliper setting is $\text{VCS}$ and the least count is $\text{LC}$.

Calculation:

Substitute $3^{″}$ for $\text{MSR}$, $0.2^{″}$ for $\text{L}$ and $0.025^{″}$ for $\text{D}$ in Equation (I).

  $\begin{array}{c}\text{A}=\text{<msup><mo>3</mo><mo>″</mo></msup>}+\text{0}\text{.<msup><mo>2</mo><mo>″</mo></msup>}+\text{0}\text{.02<msup><mo>5</mo><mo>″</mo></msup>}\\ \text{=3}\text{.22<msup><mo>5</mo><mo>″</mo></msup>}\end{array}$

Substitute $3.225^{″}$ for $\text{A}$ in Equation (II).

  $\begin{array}{c}\text{B}=\text{3}\text{.22<msup><mo>5</mo><mo>″</mo></msup>}+0.025^{″}\\ =3.250^{″}\end{array}$

Substitute $3.242^{″}$ for $\text{VCS}$, $3.225^{″}$ for $\text{A}$ and $0.001^{″}$ for $\text{LC}$ in Equation (III).

  $\begin{array}{c}\text{C}=\frac{\text{3}\text{.24<msup><mo>2</mo><mo>″</mo></msup>}-\text{3}\text{.22<msup><mo>5</mo><mo>″</mo></msup>}}{\text{0}\text{.00<msup><mo>1</mo><mo>″</mo></msup>}}\\ =\frac{0.\text{01<msup><mo>7</mo><mo>″</mo></msup>}}{\text{0}\text{.00<msup><mo>1</mo><mo>″</mo></msup>}}\\ =17\end{array}$

Conclusion:

The value of measurement A is $3.225^{″}$.

The value of measurement B is $3.250^{″}$.

The value of measurement C is $17$.

To determine

(b)

The value of measurement A is $2.875^{″}$.

The value of measurement B is $2.900^{″}$.

The value of measurement C is $2$.

Explanation of Solution

Given information:

The vernier caliper setting is $2.877^{″}$.

Calculation:

Substitute $2^{″}$ for $\text{MSR}$, $0.8^{″}$ for $\text{L}$ and $0.075^{″}$ for $\text{D}$ in Equation (I).

  $\begin{array}{c}\text{A}=\text{<msup><mo>2</mo><mo>″</mo></msup>}+\text{0}\text{.<msup><mo>8</mo><mo>″</mo></msup>}+\text{0}\text{.07<msup><mo>5</mo><mo>″</mo></msup>}\\ =\text{2}\text{.87<msup><mo>5</mo><mo>″</mo></msup>}\end{array}$

Substitute $2.875^{″}$ for $\text{A}$ in Equation (II).

  $\begin{array}{c}\text{B}=\text{2}\text{.87<msup><mo>5</mo><mo>″</mo></msup>}+0.025^{″}\\ =2.900^{″}\end{array}$

Substitute $2.877^{″}$ for $\text{VCS}$, $2.875^{″}$ for $\text{A}$ and $0.001^{″}$ for $\text{LC}$ in Equation (III).

  $\begin{array}{c}\text{C}=\frac{\text{2}\text{.87<msup><mo>7</mo><mo>″</mo></msup>}-\text{2}\text{.87<msup><mo>5</mo><mo>″</mo></msup>}}{\text{0}\text{.00<msup><mo>1</mo><mo>″</mo></msup>}}\\ =\frac{0.\text{00<msup><mo>2</mo><mo>″</mo></msup>}}{\text{0}\text{.00<msup><mo>1</mo><mo>″</mo></msup>}}\\ =2\end{array}$

Conclusion:

The value of measurement A is $2.875^{″}$.

The value of measurement B is $2.900^{″}$.

The value of measurement C is $2$.

To determine

(c)

The value of measurement A is $4.825^{″}$.

The value of measurement B is $4.850^{″}$.

The value of measurement C is $14$.

Explanation of Solution

Given information:

The vernier caliper setting is $4.839^{″}$.

Calculation:

Substitute $4^{″}$ for $\text{MSR}$, $0.8^{″}$ for $\text{L}$ and $0.025^{″}$ for $\text{D}$ in Equation (I).

  $\begin{array}{c}\text{A}=\text{<msup><mo>4</mo><mo>″</mo></msup>}+\text{0}\text{.<msup><mo>8</mo><mo>″</mo></msup>}+\text{0}\text{.02<msup><mo>5</mo><mo>″</mo></msup>}\\ =\text{4}\text{.82<msup><mo>5</mo><mo>″</mo></msup>}\end{array}$

Substitute $4.825^{″}$ for $\text{A}$ in Equation (II).

  $\begin{array}{c}\text{B}=\text{4}\text{.82<msup><mo>5</mo><mo>″</mo></msup>}+0.025^{″}\\ =4.850^{″}\end{array}$

Substitute $4.839^{″}$ for $\text{VCS}$, $4.825^{″}$ for $\text{A}$ and $0.001^{″}$ for $\text{LC}$ in Equation (III).

  $\begin{array}{c}\text{C}=\frac{\text{4}\text{.83<msup><mo>9</mo><mo>″</mo></msup>}-\text{4}\text{.82<msup><mo>5</mo><mo>″</mo></msup>}}{\text{0}\text{.00<msup><mo>1</mo><mo>″</mo></msup>}}\\ =\frac{0.\text{01<msup><mo>4</mo><mo>″</mo></msup>}}{\text{0}\text{.00<msup><mo>1</mo><mo>″</mo></msup>}}\\ =14\end{array}$

Conclusion:

The value of measurement A is $4.825^{″}$.

The value of measurement B is $4.850^{″}$.

The value of measurement C is $14$.

To determine

(d)

The value of measurement A is $0.600^{″}$.

The value of measurement B is $0.625^{″}$.

The value of measurement C is $11$.

Explanation of Solution

Given information:

The vernier caliper setting is $0.611^{″}$.

Calculation:

Substitute $0^{″}$ for $\text{MSR}$, $0.6^{″}$ for $\text{L}$ and $0.000^{″}$ for $\text{D}$ in Equation (I).

  $\begin{array}{c}\text{A}=\text{<msup><mo>0</mo><mo>″</mo></msup>}+\text{0}\text{.<msup><mo>6</mo><mo>″</mo></msup>}+\text{0}\text{.00<msup><mo>0</mo><mo>″</mo></msup>}\\ =\text{0}\text{.60<msup><mo>0</mo><mo>″</mo></msup>}\end{array}$

Substitute $0.600^{″}$ for $\text{A}$ in Equation (II).

  $\begin{array}{c}\text{B}=\text{0}\text{.60<msup><mo>0</mo><mo>″</mo></msup>}+0.025^{″}\\ =0.625^{″}\end{array}$

Substitute $0.611^{″}$ for $\text{VCS}$, $0.600^{″}$ for $\text{A}$ and $0.001^{″}$ for $\text{LC}$ in Equation (III).

  $\begin{array}{c}\text{C}=\frac{\text{0}\text{.61<msup><mo>1</mo><mo>″</mo></msup>}-0.600^{″}}{\text{0}\text{.00<msup><mo>1</mo><mo>″</mo></msup>}}\\ =\frac{0.\text{01<msup><mo>1</mo><mo>″</mo></msup>}}{\text{0}\text{.00<msup><mo>1</mo><mo>″</mo></msup>}}\\ =11\end{array}$

Conclusion:

The value of measurement A is $0.600^{″}$.

The value of measurement B is $0.625^{″}$.

The value of measurement C is $11$.

To determine

(e)

The value of measurement A is $4.350^{″}$.

The value of measurement B is $4.375^{″}$.

The value of measurement C is $19$.

Explanation of Solution

Given information:

The vernier caliper setting is $4.369^{″}$.

Calculation:

Substitute $4^{″}$ for $\text{MSR}$, $0.3^{″}$ for $\text{L}$ and $0.050^{″}$ for $\text{D}$ in Equation (I).

  $\begin{array}{c}\text{A}=\text{<msup><mo>4</mo><mo>″</mo></msup>}+\text{0}\text{.<msup><mo>3</mo><mo>″</mo></msup>}+\text{0}\text{.05<msup><mo>0</mo><mo>″</mo></msup>}\\ =\text{4}\text{.35<msup><mo>0</mo><mo>″</mo></msup>}\end{array}$

Substitute $4.350^{″}$ for $\text{A}$ in Equation (II).

  $\begin{array}{c}\text{B}=\text{4}\text{.35<msup><mo>0</mo><mo>″</mo></msup>}+0.025^{″}\\ =4.375^{″}\end{array}$

Substitute $4.369^{″}$ for $\text{VCS}$, $4.350^{″}$ for $\text{A}$ and $0.001^{″}$ for $\text{LC}$ in Equation (III).

  $\begin{array}{c}\text{C}=\frac{\text{4}\text{.36<msup><mo>9</mo><mo>″</mo></msup>}-4.350^{″}}{\text{0}\text{.00<msup><mo>1</mo><mo>″</mo></msup>}}\\ =\frac{0.\text{01<msup><mo>9</mo><mo>″</mo></msup>}}{\text{0}\text{.00<msup><mo>1</mo><mo>″</mo></msup>}}\\ =19\end{array}$

Conclusion:

The value of measurement A is $4.350^{″}$.

The value of measurement B is $4.375^{″}$.

The value of measurement C is $19$.

To determine

(f)

The value of measurement A is $0.075^{″}$.

The value of measurement B is $0.100^{″}$.

The value of measurement C is $9$.

Explanation of Solution

Given information:

The vernier caliper setting is $0.084^{″}$.

Calculation:

Substitute $0^{″}$ for $\text{MSR}$, $0.0^{″}$ for $\text{L}$ and $0.075^{″}$ for $\text{D}$ in Equation (I).

  $\begin{array}{c}\text{A}=\text{<msup><mo>0</mo><mo>″</mo></msup>}+\text{0}\text{.<msup><mo>0</mo><mo>″</mo></msup>}+\text{0}\text{.07<msup><mo>5</mo><mo>″</mo></msup>}\\ =\text{0}\text{.07<msup><mo>5</mo><mo>″</mo></msup>}\end{array}$

Substitute $0.075^{″}$ for $\text{A}$ in Equation (II).

  $\begin{array}{c}\text{B}=\text{0}\text{.07<msup><mo>5</mo><mo>″</mo></msup>}+0.025^{″}\\ =0.100^{″}\end{array}$

Substitute $0.084^{″}$ for $\text{VCS}$, $0.075^{″}$ for $\text{A}$ and $0.001^{″}$ for $\text{LC}$ in Equation (III).

  $\begin{array}{c}\text{C}=\frac{0.084^{″}-0.075^{″}}{\text{0}\text{.00<msup><mo>1</mo><mo>″</mo></msup>}}\\ =\frac{0.\text{00<msup><mo>9</mo><mo>″</mo></msup>}}{\text{0}\text{.00<msup><mo>1</mo><mo>″</mo></msup>}}\\ =9\end{array}$

Conclusion:

The value of measurement A is $0.075^{″}$.

The value of measurement B is $0.100^{″}$.

The value of measurement C is $9$.

To determine

(g)

The value of measurement A is $7.850^{″}$.

The value of measurement B is $7.875^{″}$.

The value of measurement C is $7$.

Explanation of Solution

Given information:

The vernier caliper setting is $7.857^{″}$.

Calculation:

Substitute $7^{″}$ for $\text{MSR}$, $0.8^{″}$ for $\text{L}$ and $0.050^{″}$ for $\text{D}$ in Equation (I).

  $\begin{array}{c}\text{A}=\text{<msup><mo>7</mo><mo>″</mo></msup>}+\text{0}\text{.<msup><mo>8</mo><mo>″</mo></msup>}+\text{0}\text{.05<msup><mo>0</mo><mo>″</mo></msup>}\\ =\text{7}\text{.85<msup><mo>0</mo><mo>″</mo></msup>}\end{array}$

Substitute $7.850^{″}$ for $\text{A}$ in Equation (II).

  $\begin{array}{c}\text{B}=\text{7}\text{.85<msup><mo>0</mo><mo>″</mo></msup>}+0.025^{″}\\ =7.875^{″}\end{array}$

Substitute $7.857^{″}$ for $\text{VCS}$, $7.850^{″}$ for $\text{A}$ and $0.001^{″}$ for $\text{LC}$ in Equation (III).

  $\begin{array}{c}\text{C}=\frac{7.857^{″}-7.850^{″}}{\text{0}\text{.00<msup><mo>1</mo><mo>″</mo></msup>}}\\ =\frac{0.\text{00<msup><mo>7</mo><mo>″</mo></msup>}}{\text{0}\text{.00<msup><mo>1</mo><mo>″</mo></msup>}}\\ =7\end{array}$

Conclusion:

The value of measurement A is $7.850^{″}$.

The value of measurement B is $7.875^{″}$.

The value of measurement C is $7$.

To determine

(h)

The value of measurement A is $1.625^{″}$.

The value of measurement B is $1.650^{″}$.

The value of measurement C is $21$.

Explanation of Solution

Given information:

The vernier caliper setting is $1.646^{″}$.

Calculation:

Substitute $1^{″}$ for $\text{MSR}$, $0.6^{″}$ for $\text{L}$ and $0.025^{″}$ for $\text{D}$ in Equation (I).

  $\begin{array}{c}\text{A}=\text{<msup><mo>1</mo><mo>″</mo></msup>}+\text{0}\text{.<msup><mo>6</mo><mo>″</mo></msup>}+\text{0}\text{.02<msup><mo>5</mo><mo>″</mo></msup>}\\ =\text{1}\text{.62<msup><mo>5</mo><mo>″</mo></msup>}\end{array}$

Substitute $1.625^{″}$ for $\text{A}$ in Equation (II).

  $\begin{array}{c}\text{B}=\text{1}\text{.62<msup><mo>5</mo><mo>″</mo></msup>}+0.025^{″}\\ =1.650^{″}\end{array}$

Substitute $1.646^{″}$ for $\text{VCS}$, $1.625^{″}$ for $\text{A}$ and $0.001^{″}$ for $\text{LC}$ in Equation (III).

  $\begin{array}{c}\text{C}=\frac{1.646^{″}-1.625^{″}}{\text{0}\text{.00<msup><mo>1</mo><mo>″</mo></msup>}}\\ =\frac{0.\text{02<msup><mo>1</mo><mo>″</mo></msup>}}{\text{0}\text{.00<msup><mo>1</mo><mo>″</mo></msup>}}\\ =21\end{array}$

Conclusion:

The value of measurement A is $1.625^{″}$.

The value of measurement B is $1.650^{″}$.

The value of measurement C is $21$.

To determine

(i)

The value of measurement A is $4.025^{″}$.

The value of measurement B is $4.050^{″}$.

The value of measurement C is $9$.

Explanation of Solution

Given information:

The vernier caliper setting is $4.034^{″}$.

Calculation:

Substitute $4^{″}$ for $\text{MSR}$, $0.0^{″}$ for $\text{L}$ and $0.025^{″}$ for $\text{D}$ in Equation (I).

  $\begin{array}{c}\text{A}=\text{<msup><mo>4</mo><mo>″</mo></msup>}+\text{0}\text{.<msup><mo>0</mo><mo>″</mo></msup>}+\text{0}\text{.02<msup><mo>5</mo><mo>″</mo></msup>}\\ =\text{4}\text{.02<msup><mo>5</mo><mo>″</mo></msup>}\end{array}$

Substitute $4.025^{″}$ for $\text{A}$ in Equation (II).

  $\begin{array}{c}\text{B}=\text{4}\text{.02<msup><mo>5</mo><mo>″</mo></msup>}+0.025^{″}\\ =4.050^{″}\end{array}$

Substitute $4.034^{″}$ for $\text{VCS}$, $4.025^{″}$ for $\text{A}$ and $0.001^{″}$ for $\text{LC}$ in Equation (III).

  $\begin{array}{c}\text{C}=\frac{4.034^{″}-4.025^{″}}{\text{0}\text{.00<msup><mo>1</mo><mo>″</mo></msup>}}\\ =\frac{0.\text{00<msup><mo>9</mo><mo>″</mo></msup>}}{\text{0}\text{.00<msup><mo>1</mo><mo>″</mo></msup>}}\\ =9\end{array}$

Conclusion:

The value of measurement A is $4.025^{″}$.

The value of measurement B is $4.050^{″}$.

The value of measurement C is $9$.

To determine

(j)

The value of measurement A is $0.000^{″}$.

The value of measurement B is $0.025^{″}$.

The value of measurement C is $22$.

Explanation of Solution

Given information:

The vernier caliper setting is $0.022^{″}$.

Calculation:

Substitute $0^{″}$ for $\text{MSR}$, $0.0^{″}$ for $\text{L}$ and $0.000^{″}$ for $\text{D}$ in Equation (I).

  $\begin{array}{c}\text{A}=\text{<msup><mo>0</mo><mo>″</mo></msup>}+\text{0}\text{.<msup><mo>0</mo><mo>″</mo></msup>}+\text{0}\text{.00<msup><mo>0</mo><mo>″</mo></msup>}\\ =\text{0}\text{.00<msup><mo>0</mo><mo>″</mo></msup>}\end{array}$

Substitute $0.000^{″}$ for $\text{A}$ in Equation (II).

  $\begin{array}{c}\text{B}=\text{0}\text{.00<msup><mo>0</mo><mo>″</mo></msup>}+0.025^{″}\\ =0.025^{″}\end{array}$

Substitute $0.022^{″}$ for $\text{VCS}$, $0.000^{″}$ for $\text{A}$ and $0.001^{″}$ for $\text{LC}$ in Equation (III).

  $\begin{array}{c}\text{C}=\frac{0.022^{″}-0.000^{″}}{\text{0}\text{.00<msup><mo>1</mo><mo>″</mo></msup>}}\\ =\frac{0.\text{02<msup><mo>2</mo><mo>″</mo></msup>}}{\text{0}\text{.00<msup><mo>1</mo><mo>″</mo></msup>}}\\ =22\end{array}$

Conclusion:

The value of measurement A is $0.000^{″}$.

The value of measurement B is $0.025^{″}$.

The value of measurement C is $22$.

To determine

(k)

The value of measurement A is $3.325^{″}$.

The value of measurement B is $3.250^{″}$.

The value of measurement C is $8$.

Explanation of Solution

Given information:

The vernier caliper setting is $3.333^{″}$.

Calculation:

Substitute $3^{″}$ for $\text{MSR}$, $0.3^{″}$ for $\text{L}$ and $0.025^{″}$ for $\text{D}$ in Equation (I).

  $\begin{array}{c}\text{A}=\text{<msup><mo>3</mo><mo>″</mo></msup>}+\text{0}\text{.<msup><mo>3</mo><mo>″</mo></msup>}+\text{0}\text{.02<msup><mo>5</mo><mo>″</mo></msup>}\\ =\text{3}\text{.32<msup><mo>5</mo><mo>″</mo></msup>}\end{array}$

Substitute $3.325^{″}$ for $\text{A}$ in Equation (II).

  $\begin{array}{c}\text{B}=\text{3}\text{.32<msup><mo>5</mo><mo>″</mo></msup>}+0.025^{″}\\ =3.350^{″}\end{array}$

Substitute $3.333^{″}$ for $\text{VCS}$, $3.325^{″}$ for $\text{A}$ and $0.001^{″}$ for $\text{LC}$ in Equation (III).

  $\begin{array}{c}\text{C}=\frac{3.333^{″}-3.325^{″}}{\text{0}\text{.00<msup><mo>1</mo><mo>″</mo></msup>}}\\ =\frac{0.\text{00<msup><mo>8</mo><mo>″</mo></msup>}}{\text{0}\text{.00<msup><mo>1</mo><mo>″</mo></msup>}}\\ =8\end{array}$

Conclusion:

The value of measurement A is $3.325^{″}$.

The value of measurement B is $3.250^{″}$.

The value of measurement C is $8$.

To determine

(l)

The value of measurement A is $5.975^{″}$.

The value of measurement B is $6.000^{″}$.

The value of measurement C is $24$.

Explanation of Solution

Given information:

The vernier caliper setting is $5.999^{″}$.

Calculation:

Substitute $5^{″}$ for $\text{MSR}$, $0.9^{″}$ for $\text{L}$ and $0.075^{″}$ for $\text{D}$ in Equation (I).

  $\begin{array}{c}\text{A}=\text{<msup><mo>5</mo><mo>″</mo></msup>}+\text{0}\text{.<msup><mo>9</mo><mo>″</mo></msup>}+\text{0}\text{.07<msup><mo>5</mo><mo>″</mo></msup>}\\ =\text{5}\text{.97<msup><mo>5</mo><mo>″</mo></msup>}\end{array}$

Substitute $5.975^{″}$ for $\text{A}$ in Equation (II).

  $\begin{array}{c}\text{B}=\text{5}\text{.97<msup><mo>5</mo><mo>″</mo></msup>}+0.025^{″}\\ =6.000^{″}\end{array}$

Substitute $5.999^{″}$ for $\text{VCS}$, $5.975^{″}$ for $\text{A}$ and $0.001^{″}$ for $\text{LC}$ in Equation (III).

  $\begin{array}{c}\text{C}=\frac{5.999^{″}-5.975^{″}}{\text{0}\text{.00<msup><mo>1</mo><mo>″</mo></msup>}}\\ =\frac{0.\text{02<msup><mo>4</mo><mo>″</mo></msup>}}{\text{0}\text{.00<msup><mo>1</mo><mo>″</mo></msup>}}\\ =24\end{array}$

Conclusion:

The value of measurement A is $5.975^{″}$.

The value of measurement B is $6.000^{″}$.

The value of measurement C is $24$.

To determine

(m)

The value of measurement A is $0.275^{″}$.

The value of measurement B is $0.300^{″}$.

The value of measurement C is $3$.

Explanation of Solution

Given information:

The vernier caliper setting is $0.278^{″}$.

Calculation:

Substitute $0^{″}$ for $\text{MSR}$, $0.2^{″}$ for $\text{L}$ and $0.075^{″}$ for $\text{D}$ in Equation (I).

  $\begin{array}{c}\text{A}=\text{<msup><mo>0</mo><mo>″</mo></msup>}+\text{0}\text{.<msup><mo>2</mo><mo>″</mo></msup>}+\text{0}\text{.07<msup><mo>5</mo><mo>″</mo></msup>}\\ =\text{0}\text{.27<msup><mo>5</mo><mo>″</mo></msup>}\end{array}$

Substitute $0.275^{″}$ for $\text{A}$ in Equation (II).

  $\begin{array}{c}\text{B}=\text{0}\text{.27<msup><mo>5</mo><mo>″</mo></msup>}+0.025^{″}\\ =0.300^{″}\end{array}$

Substitute $0.278^{″}$ for $\text{VCS}$, $0.275^{″}$ for $\text{A}$ and $0.001^{″}$ for $\text{LC}$ in Equation (III).

  $\begin{array}{c}\text{C}=\frac{0.278^{″}-0.275^{″}}{\text{0}\text{.00<msup><mo>1</mo><mo>″</mo></msup>}}\\ =\frac{0.\text{00<msup><mo>3</mo><mo>″</mo></msup>}}{\text{0}\text{.00<msup><mo>1</mo><mo>″</mo></msup>}}\\ =3\end{array}$

Conclusion:

The value of measurement A is $0.275^{″}$.

The value of measurement B is $0.300^{″}$.

The value of measurement C is $3$.

To determine

(n)

The value of measurement A is $0.950^{″}$.

The value of measurement B is $0.975^{″}$.

The value of measurement C is $15$.

Explanation of Solution

Given information:

The vernier caliper setting is $0.965^{″}$.

Calculation:

Substitute $0^{″}$ for $\text{MSR}$, $0.9^{″}$ for $\text{L}$ and $0.050^{″}$ for $\text{D}$ in Equation (I).

  $\begin{array}{c}\text{A}=\text{<msup><mo>0</mo><mo>″</mo></msup>}+\text{0}\text{.<msup><mo>9</mo><mo>″</mo></msup>}+\text{0}\text{.05<msup><mo>0</mo><mo>″</mo></msup>}\\ =\text{0}\text{.95<msup><mo>0</mo><mo>″</mo></msup>}\end{array}$

Substitute $0.950^{″}$ for $\text{A}$ in Equation (II).

  $\begin{array}{c}\text{B}=\text{0}\text{.95<msup><mo>0</mo><mo>″</mo></msup>}+0.025^{″}\\ =0.975^{″}\end{array}$

Substitute $0.965^{″}$ for $\text{VCS}$, $0.950^{″}$ for $\text{A}$ and $0.001^{″}$ for $\text{LC}$ in Equation (III).

  $\begin{array}{c}\text{C}=\frac{0.965^{″}-0.950^{″}}{\text{0}\text{.00<msup><mo>1</mo><mo>″</mo></msup>}}\\ =\frac{0.\text{01<msup><mo>5</mo><mo>″</mo></msup>}}{\text{0}\text{.00<msup><mo>1</mo><mo>″</mo></msup>}}\\ =15\end{array}$

Conclusion:

The value of measurement A is $0.950^{″}$.

The value of measurement B is $0.975^{″}$.

The value of measurement C is $15$.