The pitch and the number of divisions, on the circular scale, for a given screw gauge are 0.5 mm and 100 respectively. When the screw gauge is fully tightened without any object, the zero of its circular scale lies 3 divisions below the mean line. The readings of the main scale and the circular scale, for a thin sheet, are 5.5 mm and 48 respectively, the thickness of this sheet is :

Units and Measurements PYQ — JEE Main 2019 · 9 Jan · Shift 2 · Q55 with Solution

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Correct answer: C

Given:

So, least count is

$\text{L.C.} = \frac{\text{Pitch}}{\text{No. of divisions}} = \frac{0.5}{100} = 0.005\,\text{mm}$

When the screw gauge is fully tightened, the zero of circular scale lies 3 divisions below the mean line.

This means the instrument shows a positive zero error of

$\text{Zero error} = 3 \times 0.005 = 0.015\,\text{mm}$

Hence, zero correction is

$\text{Zero correction} = -0.015\,\text{mm}$

Given:

Circular scale contribution:

$48 \times 0.005 = 0.240\,\text{mm}$

Observed reading:

$5.5 + 0.240 = 5.740\,\text{mm}$

True thickness:

$\text{True reading} = \text{Observed reading} + \text{Zero correction}$

$= 5.740 - 0.015 = 5.725\,\text{mm}$

$\boxed{5.725\,\text{mm}}$

So the correct option is C.