A screw gauge with a pitch of 0.5 mm and a circular scale with 50 divisions is used to measure the thickness of a thin sheet of Aluminium. Before starting the measurement, it is found that when the two jaws of the screw gauge are brought in contact, the 45th division coincides with the main scale line and that the zero of the main scale is barely visible. What is the thickness of the sheet if the main scale reading is 0.5 mm and the 25th division coincides with the main scale line?

Units and Measurements PYQ — JEE Main 2016 · Shift 0 · Q62 with Solution

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Correct answer: D

Given:

So, least count is

$\text{L.C.} = \frac{\text{Pitch}}{\text{No. of divisions}} = \frac{0.5}{50} = 0.01\,\text{mm}$

When the jaws are in contact:

This means the zero of circular scale is above the reference line, so the screw gauge has a negative zero error.

Magnitude of zero error:

$\text{Zero error} = (50 - 45) \times 0.01 = 5 \times 0.01 = 0.05\,\text{mm}$

Thus,

$\text{Zero error} = -0.05\,\text{mm}$

Hence, zero correction is

$\text{Zero correction} = +0.05\,\text{mm}$

Given during measurement:

Circular scale contribution:

$25 \times 0.01 = 0.25\,\text{mm}$

Observed reading:

$0.5 + 0.25 = 0.75\,\text{mm}$

True thickness:

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

$= 0.75 + 0.05 = 0.80\,\text{mm}$

$0.80\,\text{mm}$

So the correct option is D.