Question

The pitch of the screw gauge is 1 mm and there are 100 divisions on the circular scale. When nothing is put in between the jaws, the zero of the circular scale lies 8 divisions below the reference line. When a wire is placed between the jaws, the first linear scale division is clearly visible while 72nd division on circular scale coincides with the reference line. The radius of the wire is :

• A. 1.64 mm
• B. 0.90 mm
• C. 0.82 mm
• D. 1.80 mm

Hint:

Always remember: {Correct Reading = Observed Reading $-$ Zero Error}. If the circular scale zero is below the reference line, the error is positive; if above, it is negative.

Answer 1

The Correct Option is C

Step 1: Calculate Least Count ($LC$) = $\text{Pitch} / \text{Number of divisions} = 1\text{ mm} / 100 = 0.01\text{ mm}$.
Step 2: Determine Zero Error. Since the zero of the circular scale is 8 divisions \itextbf{below} the reference line, it is a positive zero error. Zero Error = $+8 \times 0.01 = +0.08\text{ mm}$.
Step 3: Calculate Observed Reading = $MSR + (CSR \times LC)$. First linear scale division visible $\Rightarrow MSR = 1\text{ mm}$. Circular scale reading $= 72 \times 0.01 = 0.72\text{ mm}$. Observed Reading $= 1 + 0.72 = 1.72\text{ mm}$.
Step 4: Correct Diameter = Observed Reading $-$ Zero Error $= 1.72 - 0.08 = 1.64\text{ mm}$.
Step 5: Radius $= \text{Diameter} / 2 = 1.64 / 2 = 0.82\text{ mm}$.