Question

In a screw gauge, 5 complete rotations of the screw cause it to move a linear distance of 0.25 cm. There are 100 circular scale divisions. The thickness of a wire measureed by this screw gauge gives a reading of 4 main scale divisions and 30 circular scale divisions. Assuming negligible zero error, the thickness of the wire is :

• A. 0.3150 cm
• B. 0.2150 cm
• C. 0.4300 cm
• D. 0.0430 cm

Answer 1

Answer: (B)

0.2150 cm

$\begin{array}{cc}\text{pitch}& =\text{distance moved in}\text{one rotation}\\ & =\frac{0.25cm}{5}\text{rotations}\end{array}$

$\begin{array}{cc}\text{pitch}& =0.05cm/rotation\\ L\cdot C& =\frac{\text{pitch}}{\text{no. of div on}C\cdot S}\\ & =\frac{0.05}{100}\\ & =0.0005cm\end{array}$

$\begin{array}{c}\text{Final reading}=MSR+CSR\\ \\ \begin{array}{cc}C\cdot S\cdot R& =30\times LC\\ & =30\times 0.0005\\ & =0.015cm\end{array}\end{array}$

$\text{1 M.S.D is assumed as equal to pitch.}$
$\begin{array}{c}M\cdot S\cdot R=0\cdot 05\times 4=0\cdot 2\\ \text{Final reading}=0.215cm\end{array}$