Question

The pitch of a screw gauge is $1mm$ and there are $100$ divisions on circular scale. When faces A and B are just touching each without putting anything between the studs $32nd$ division on the circular scale coincides with reference line. When a glass plate is placed between the studs, the linear scale reads $4$ divisions and the circular scale reads $16$ divisions. Find the thickness of the glass plate. Zero of linear scale is not hidden from circular scale when A and B touches each other.

• A. $3.841mm$
• B. $3.84mm$
• C. $3.85mm$
• D. $3.846mm$

Answer 1

The correct option is B $3.84mm$

Least count (LC) $=\frac{Pitch}{Numberofdivisionsoncircularscale}$
$=\frac{1}{100}$ mm
$=0.01$ mm

As zero is not hidden from cricular scale when A and B touches each other.

Hence, the screw gauge has positive error.
$e=+n\left(LC\right)=32\times 0.01=0.32$ mm
Linear scale reading $=4\times \left(1mm\right)=4mm$
Circular scale reading $=16\times \left(0.01mm\right)=0.16mm$
$\therefore$ Measured reading-e $=(4+0.16)$mm=4.16 mm
$\therefore$ Absolute reading $=$ Measured reading-e
$=(4.16-0.32)mm=3.84mm$

Therefore, thickness of the glass plate is $3.84mm$.