Question

A screw gauge with a positive zero error of 5 divisions is used to find the thickness of the glass plate. When the glass plate is held between the two studs, the main scale reading is 5 mm and the head scale coinciding division is 25. If the least count of the screw gauge is 0.01 mm, find the thickness of the glass plate.

• A. 5.2 mm
• B. 6.5 mm
• C. 6.8 mm
• D. 9.8 mm

Answer 1

The correct option is A 5.2 mm

$\begin{array}{c}\text{Given, Positive zero error =5 division}\\ \text{Head scale reading}=25\text{, least count}=0.01mm\\ \text{Error}=5\times 0.01=0.05mm\text{(Positive error)}\\ \text{Thickness of glass plate}=MSR+(HSR\times LC)\pm \text{error}\\ =5+25\times 0.01-0.05\\ =5+0.25-0.05\\ =5.20mm\end{array}$