Question

A screw gauge with a pitch of $0.5mm$ and a circular scale with $50$ divisions is used to measured the thickness of a thin sheet of Aluminium. Before staring the measurement, It is found that when the two jaws of the screw gauge are brought in contact, the ${45}^{th}$ division coincides with the main scale line and that the zero of the main scale is barely visible. What is the thickness of the sheet if the main scale reading is $0.5mm$ and the ${25}^{th}$ division coincides with the main scale line?

• A. $0.50mm$
• B. $0.75mm$
• C. $0.80mm$
• D. $0.70mm$

Answer 1

The correct option is C $0.80mm$

$\begin{array}{cc}\text{pitch}& =0.5mm\\ L\cdot C=& \text{pitch}/(no.of\text{div on}CS)\\ & =\frac{0.5}{50}\\ & =0.01mm\end{array}$

$\begin{array}{c}\text{(-) ive zero error}\\ =-1+(50-45)\times 0.01\end{array}$

$\begin{array}{c}=-1+0.a5\\ =\text{-0.05}mm\\ \text{Final reading =}MSR\text{+ VSR}\text{- zero eror}\end{array}$

$\begin{array}{c}=0.5+25\times 0.01-(-0.05)\\ =0.5+0.25+0.05\\ =0.5+0.83\\ =0.8mm\end{array}$