Question

A screw gauge is 0.5 mm pitch and $5\mu m$ least count is used to determine the diameter of a wire. If the head scaling reading is 72 and the diameter of the wire measured is 350 microns, the zero error of the instrument is ____.

• A. $10\times {10}^{-3}\mu m$
• B. $0.01cm$
• C. $10\times {10}^{-6}m$
• D. None of these

Answer 1

The correct option is C $10\times {10}^{-6}m$

$\begin{array}{c}L\cdot C=5\times {10}^{-6}m=0.005mm\\ \text{Pitch}=0.5mm\\ L\cdot C=\frac{\text{Pith}}{\text{no of division on C.S.}}\\ \text{No. of division on}C\cdot S=\frac{0.5}{0.005}=100\end{array}$

$\begin{array}{cc}C\cdot S\cdot R& =72\times L\cdot C\\ & =72\times 0.005\\ & =0.36mm\\ \text{Final reading}& =350\times {10}^{-6}m\\ & =0.350mm\end{array}$

$\begin{array}{c}0.350=MSR+CSR-\text{Zero}\text{error}\\ \text{0.350}=0+0.36-\text{zero}\text{error}\\ \text{zero error}=0.01mm\\ =10\times {10}^{-6}m\end{array}$