Question

A screw gauge with a pitch of 0.5mm 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 45^th division coincides with the main scale line and that the zero of the main scale is barely visible. Find the thickness of the sheet if the main scale reading is $0 \cdot 5{\text{mm}}$ and the 25^th division coincides with the main scale.
A) $0 \cdot 50{\text{mm}}$
B) $0 \cdot 75{\text{mm}}$
C) $0 \cdot 80{\text{mm}}$
D) $0 \cdot 70{\text{mm}}$

Answer 1

Hint: A screw gauge is an apparatus which is used to measure lengths in the range of millimetres. It has a pitch scale which is the main scale and a circular scale which can be rotated above the pitch scale. An image of a screw gauge is given below.

Here, before measuring the thickness, we observed that the 0^th or 50^th division of the circular scale and the zero of the pitch scale did not coincide. So there exists a zero error. The zero correction will be the negative of the zero error. The actual thickness of the sheet will be the sum of the main scale reading, the product of the coinciding division and the least count of the screw gauge and its zero correction.

Formulas used:
-The least count of a screw gauge is given by, ${\text{L}}{\text{.C}} = \dfrac{{{\text{Pitch}}}}{{{\text{T}}{\text{.N}}{\text{.D}}}}$ where ${\text{T}}{\text{.N}}{\text{.D}}$ is the total number of divisions on the circular scale of the screw gauge.
-The zero error of the screw gauge is given be, ${\text{zero error}} = - \left( {{\text{Ref}}{\text{. no}} - {\text{C}}{\text{.S}}{\text{.D}}} \right) \times {\text{L}}{\text{.C}}$ where ${\text{C}}{\text{.S}}{\text{.D}}$ is the coinciding division on the circular scale, ${\text{Ref}}{\text{. no}}$ is the starting (or ending) number on the circular scale and ${\text{L}}{\text{.C}}$ is the least count of the screw gauge.
-The total reading of the screw gauge is given by, ${\text{T}}{\text{.R}} = {\text{M}}{\text{.S}}{\text{.R + }}\left( {{\text{C}}{\text{.S}}{\text{.R}} \times {\text{L}}{\text{.C}}} \right) + c$ where ${\text{M}}{\text{.S}}{\text{.R}}$ is the main scale reading, ${\text{C}}{\text{.S}}{\text{.R}}$ is the circular scale reading, ${\text{L}}{\text{.C}}$ is the least count of the screw gauge and $c$ is the zero correction of the screw gauge.

Complete step by step solution:
Step 1: List all the known readings given in the question.
The pitch of the screw gauge is given to be ${\text{pitch}} = 0 \cdot 5{\text{mm}}$ .
The total number of divisions on the circular scale is given to be ${\text{T}}{\text{.N}}{\text{.D}} = 50{\text{divisions}}$.
The division on the circular scale that coincides with the zero of the main scale when the two jaws are brought in contact before measuring the thickness is given to be ${\text{C}}{\text{.S}}{\text{.D}} = 45$ .
When measuring the thickness of the aluminium sheet, the main scale reading is given to be ${\text{M}}{\text{.S}}{\text{.R}} = 0 \cdot 5{\text{mm}}$ and the corresponding circular scale reading is given to be ${\text{C}}{\text{.S}}{\text{.R}} = 25$ .

Step 2: Express the least count and the zero correction of the screw gauge.
The least count of the screw gauge can be expressed as ${\text{L}}{\text{.C}} = \dfrac{{{\text{Pitch}}}}{{{\text{T}}{\text{.N}}{\text{.D}}}}$ -------- (1)
Substituting for ${\text{pitch}} = 0 \cdot 5{\text{mm}}$ and ${\text{T}}{\text{.N}}{\text{.D}} = 50{\text{divisions}}$ in equation (1) we get, ${\text{L}}{\text{.C}} = \dfrac{{0 \cdot 5}}{{50}} = 0 \cdot 01{\text{mm}}$
Thus the least count of the screw gauge is obtained to be ${\text{L}}{\text{.C}} = 0 \cdot 01{\text{mm}}$ .
Now the zero error of the screw gauge will be given by the relation, ${\text{zero error}} = - \left( {{\text{Ref}}{\text{. no}} - {\text{C}}{\text{.S}}{\text{.D}}} \right) \times {\text{L}}{\text{.C}}$ -------- (2)
Substituting for ${\text{Ref}}{\text{. no}} = 50$ , ${\text{L}}{\text{.C}} = 0 \cdot 01{\text{mm}}$ and ${\text{C}}{\text{.S}}{\text{.D}} = 45$ in equation (2) we get, ${\text{zero error}} = - \left( {50 - 45} \right) \times 0 \cdot 01 = - 0 \cdot 05{\text{mm}}$
Thus the zero error of the screw gauge is obtained to be $ - 0 \cdot 05{\text{mm}}$ .

Step 3: Express the zero correction and finally obtain the total reading or the actual reading of the screw gauge.
Since ${\text{zero error}} = - 0 \cdot 05{\text{mm}}$ , the zero correction will be the negative of the zero error.
i.e., $c = - {\text{zero error}} = - \left( { - 0 \cdot 05} \right) = 0 \cdot 05{\text{mm}}$
Now the total reading from the screw gauge is given by,
${\text{T}}{\text{.R}} = {\text{M}}{\text{.S}}{\text{.R + }}\left( {{\text{C}}{\text{.S}}{\text{.R}} \times {\text{L}}{\text{.C}}} \right) + c$ -------- (3)
Substituting for ${\text{M}}{\text{.S}}{\text{.R}} = 0 \cdot 5{\text{mm}}$ , ${\text{C}}{\text{.S}}{\text{.R}} = 25$ , ${\text{L}}{\text{.C}} = 0 \cdot 01{\text{mm}}$ and $c = 0 \cdot 05{\text{mm}}$ in equation (3) we get, ${\text{T}}{\text{.R}} = 0 \cdot 5{\text{ + }}\left( {25 \times 0 \cdot 01} \right) + 0 \cdot 05 = 0 \cdot 80{\text{mm}}$
$\therefore $ The thickness of the aluminium sheet is obtained to be $0 \cdot 80{\text{mm}}$ .

So the correct option is C.

Note: Here the coinciding division on the circular scale while determining the zero error is greater than half of the total number of divisions and so the zero on the circular scale lies above the zero line on the main scale. Thus we consider the last division i.e. ${\text{Ref}}{\text{. no}} = 50$ and obtain the zero error to be a negative value. If the coinciding division was less than half of the total number of divisions then we consider the starting division i.e., ${\text{Ref}}{\text{. no}} = 0$ as the zero on the circular scale lies below the zero line on the main scale and so the zero error will be obtained as a positive value using equation (2).