In an experiment to find out the diameter of wire using screw gauge, the following observations were noted : (A) Screw moves 0.5 ~mm on main scale in one complete rotation (B) Total divisions on circular scale =50 (C) Main scale reading is 2.5 ~mm (D) 45^ th division of circular scale is in the pitch line (E) Instrument has 0.03 mm negative error Then the diameter of wire is :

Units and Measurements PYQ — JEE Main 2022 · 29 Jul · Shift 1 · Q66 with Solution

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Correct answer: C

Given:

So, least count is

$\text{L.C.} = \frac{\text{pitch}}{\text{number of divisions}} = \frac{0.5}{50} = 0.01\,\text{mm}$

Main scale reading:

$\text{MSR} = 2.5\,\text{mm}$

Circular scale reading:

$\text{CSR} = 45 \times 0.01 = 0.45\,\text{mm}$

Therefore, observed reading is

$\text{Observed reading} = 2.5 + 0.45 = 2.95\,\text{mm}$

The instrument has a negative zero error of $0.03\,\text{mm}$ .

For a negative zero error, the zero correction is positive:

$\text{Zero correction} = +0.03\,\text{mm}$

Hence, true diameter is

$\text{True diameter} = 2.95 + 0.03 = 2.98\,\text{mm}$

$2.98\,\text{mm}$

So the correct option is C.