Question

The pitch of a screw gauge is $0.5mm$ and there are $100$ divisions on it circular scale. The instrument reads $+2$ circular scale divisions when nothing is put in-between its jaws. In measuring the diameter of a wire, there are 8 divisions on the main scale and 83rd circular scale division coincides with the reference line. Then the diameter of the wire is

• A. $4.15mm$
• B. $4.405mm$
• C. $3.05mm$
• D. $1.25mm$

Answer 1

The correct option is B $4.405mm$

$\begin{array}{cc}\text{zero error}& =\frac{\text{pitch}}{\text{Total divisions}}\times 2CSD\\ & =\frac{0.5}{100}\times 2=0.01mm\end{array}$

$\begin{array}{c}\text{Reading Calculation-}\\ \text{linear scale Reading}\left(LSR\right)=8\times \text{Pitch}=4mm\text{.}\\ \text{Circular Scale Reading}\left(CSR\right)=\frac{\text{Pitch}}{\text{Total CSD}}\times \text{circular Scale div}\end{array}$

$\text{CSR}=\frac{0.5mm}{100}\times 83=0.415mm$

$\begin{array}{cc}\text{Actual Reading}& =LSR+CSR-\text{zers errore}\\ & =4mm+0.415mm-0.01mm\\ & =4.405mm.\end{array}$