Question

The pitch and the number of divisions, on the circular scale for a given screw gauge are $0.5\text{mm}$ and $100$ respectively. When the screw gauge is fully tightened without any object, the zero of its circular scale lies $3$ division below the mean line. The readings of the main scale and the circular scale, for a thin sheet, are $5.5\text{mm}$ and $48$ respectively, the thickness of the sheet is

• A. $\text{5.950 mm}$
• B. $\text{5.725 mm}$
• C. $\text{5.740 mm}$
• D. $\text{5.755 mm}$

Answer 1

The correct option is B $\text{5.725 mm}$

Thickness of sheet $=MSR+CSR-(+veerror)....\left(1\right)$
Least count of screw gauge,

$LC=\frac{\text{Pitch}}{\text{No. of division}}=\frac{0.5}{100}$
$=0.5\times {10}^{-2}\text{mm}$
$+$ve error$=3\times 0.5\times {10}^{-2}\text{mm}$
$=1.5\times {10}^{-2}\text{mm}=0.015\text{mm}$
Reading of circular scale is $48$
So,
$CSR=48\times LC$
$CSR=48\times 0.5\times {10}^{-2}$

From the equation $1$

Thickness of sheet
$=MSR+CSR-(+\text{ve error})$
$=5.5\text{mm}+48\times 0.5\times {10}^{-2}-0.015=5.725\text{mm}$