In a screw gauge, 5 complete rotations of the screw cause it to move a linear distance of 0.25cm. There are 100 circular scale divisions. The thickness of a wire measured 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 :

Units and Measurements PYQ — JEE Main 2018 · 15 Apr · Shift 1 · Q54 with Solution

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Correct answer: B

Given that $5$ complete rotations move the screw by $0.25\,\text{cm}$ ,

$\text{Pitch} = \frac{0.25}{5} = 0.05\,\text{cm}$

So, in one complete rotation, the screw moves $0.05\,\text{cm}$ .

There are $100$ circular scale divisions.

$\text{Least Count} = \frac{\text{Pitch}}{\text{No. of circular scale divisions}} = \frac{0.05}{100} = 0.0005\,\text{cm}$

The reading is $4$ main scale divisions.
In a screw gauge, one main scale division equals the pitch of the screw, i.e. $0.05\,\text{cm}$ .

Hence,

$\text{Main scale reading} = 4 \times 0.05 = 0.20\,\text{cm}$

$\text{Circular scale reading} = 30 \times 0.0005 = 0.015\,\text{cm}$

$\text{Thickness} = \text{MSR} + \text{CSR} = 0.20 + 0.015 = 0.215\,\text{cm}$

Thus, the thickness of the wire is

$\boxed{0.2150\,\text{cm}}$

Therefore, the correct option is B.