Question

A micrometer screw gauge having a positive zero error of 5 divisions is used to measure the diameter of the wire, when reading on the main scale is the 3rd division, and the 48th circular scale division coincides with the baseline. If the micrometer has 10 divisions to a centimetre on the main scale and 100 divisions on the circular scale, calculate the corrected diameter.

Answer 1

Step 1, Given,

1. Positive error = $5$ divisions
2. Number of divisions to a centimetre= $10$
3. Number of divisions on the main scale = $100$

Step 2 formula used,

$Pitch=\frac{unitofmainscale}{numberofdivisionsintheunit}$

$LeastCount\left(LC\right)=\frac{Pitch}{no.ofdivisiononthecircularscale}$

$$\begin{gathered} Observeddiameter=MainScalereading+LeastCount\times Circular.Scalereading \\ =0.3+0.001\times 48 \\ =0.3+0.048 \\ =0.348cm \end{gathered}$$

$$\begin{gathered} Correction=-(error\times L.C) \\ Correcteddiameter=observeddiamter+correction \end{gathered}$$

Step 3 solution:

Pitch = $$\begin{gathered} \frac{1}{10} \\ =0.1cm \end{gathered}$$

L C = $$\begin{gathered} \frac{0.1}{100} \\ =0.001cm \end{gathered}$$

Observed diameter = $0.348cm$

Correction = $$\begin{gathered} -(5+0.004) \\ =-0.005 \\ correcteddiameter=0.0348-0.005 \end{gathered}$$

Corrected diameter = 0.343 cm