The pitch and the number of divisions, on the circular scale for a given screw gauge are $0.5 \mathrm{~mm}$ 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 $5.5 \mathrm{~mm}$ and 48 respectively, the thickness of the sheet is:
Correct Option: 3
(3) Least count of screw gauge,
$\mathrm{LC}=\frac{\text { Pitch }}{\text { No. of division }}$
$=0.5 \times 10^{-3}=0.5 \times 10^{-2} \mathrm{~mm}$
$+$ ve error $=3 \times 0.5 \times 10^{-2} \mathrm{~mm}$
$=1.5 \times 10^{-2} \mathrm{~mm}=0.015 \mathrm{~mm}$
Reading $=$ MSR $+$ CSR $-(+$ ve error $)$
$=5.5 \mathrm{~mm}+\left(48 \times 0.5 \times 10^{-2}\right)-0.015$
$=5.5+0.24-0.015=5.725 \mathrm{~mm}$