There are 100 divisions on the circular scale of a screw gauge of pitch 1 ~mm . With no measuring quantity in between the jaws, the zero of the circular scale lies 5 divisions below the reference line. The diameter of a wire is then measured using this screw gauge. It is found that 4 linear scale divisions are clearly visible while 60 divisions on circular scale coincide with the reference line. The diameter of the wire is :

Units and Measurements PYQ — JEE Main 2024 · 8 Apr · Shift 2 · Q70 with Solution

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Correct answer: C

Given:

So,

\text{Least count} = \frac{\text{Pitch}}{\text{No. of divisions}} = \frac{1}{100} = 0.01\,\text{mm}

When no object is between the jaws, the zero of the circular scale is 5 divisions below the reference line.

This means the instrument shows a positive reading even when nothing is measured, so it has a positive zero error of:

5 \times 0.01 = 0.05\,\text{mm}

Hence, the zero correction is:

-0.05\,\text{mm}

Main scale reading = $4\,\text{mm}$ (since 4 linear scale divisions are visible)
Circular scale reading = $60$

Circular scale contribution:

60 \times 0.01 = 0.60\,\text{mm}

So observed reading is:

4.00 + 0.60 = 4.60\,\text{mm}

True diameter:

4.60 - 0.05 = 4.55\,\text{mm}

\boxed{4.55\,\text{mm}}

So the correct option is C.