Question

Two full turns of the circular scale of a screw gauge covers a distance of 1 mm on its main scale. The total number of divisions on the circular scale is 50. Further, it is found that the screw gauge has a zero error of \[ - 0.03\,{\text{mm}}\]. While measuring the diameter of a thin wire, a student notes the main scale reading of 3 mm and the number of circular scale divisions in line with the main scale as 35. The diameter of the wire is:
A. 3.32 mm
B. 3.73 mm
C. 3.67 mm
D. 3.38 mm

Answer 1

Hint: Calculate the pitch of the screw gauge. Then calculate the L.C of the screw gauge by substituting the value of pitch and the number of divisions on the circular scale. Use the expression for reading on the screw gauge to determine the diameter of the wire.

Formula used:
\[{\text{pitch}} = \dfrac{{{\text{Distance covered by the spindle}}}}{{{\text{Number of rotations of the spindle}}}}\]
\[{\text{L}}{\text{.C}} = \dfrac{{{\text{Pitch}}}}{{{\text{Number of divisions on the circular scale}}}}\]
\[{\text{reading}} = \left[ {\left( {{\text{Main scale}}\,\,{\text{reading}}} \right) + \left( {{\text{circular scale reading}} \times {\text{L}}{\text{.C}}} \right)} \right] - {\text{zero error}}\]
Here, L.C is the least count.

Complete step by step answer:
To determine the diameter of the wire, we should know the least count of the screw gauge. To determine the least count of the screw gauge, we first have to determine the pitch of the screw gauge.
We have, the pitch of the screw gauge is given as,
\[{\text{pitch}} = \dfrac{{{\text{Distance covered by the spindle}}}}{{{\text{Number of rotations of the spindle}}}}\]
We have given the distance covered by the spindle of the screw gauge is 1 mm and the number rotations is 2. Therefore,
\[{\text{pitch}} = \dfrac{{{\text{1}}\,{\text{mm}}}}{2}\]
\[ \Rightarrow {\text{pitch}} = 0.5\,{\text{mm}}\]
Now, the least count of the screw gauge is given as,
\[{\text{L}}{\text{.C}} = \dfrac{{{\text{Pitch}}}}{{{\text{Number of divisions on the circular scale}}}}\]
\[{\text{L}}{\text{.C}} = \dfrac{{{\text{0}}{\text{.5}}\,{\text{mm}}}}{{50}}\]
\[ \Rightarrow {\text{L}}{\text{.C}} = 0.01\,{\text{mm}}\]

Now, we know the reading on the screw gauge is taken as,
\[{\text{reading}} = \left[ {\left( {{\text{Main scale}}\,\,{\text{reading}}} \right) + \left( {{\text{circular scale reading}} \times {\text{L}}{\text{.C}}} \right)} \right] - {\text{zero error}}\]
Substituting 3 mm for main scale reading, 35 for circular scale reading, 0.01 mm for L.C and \[ - 0.03\,{\text{mm}}\] for zero error in the above equation, we get,
\[{\text{reading}} = \left[ {3 + \left( {35 \times 0.01} \right)} \right] - \left( { - 0.03} \right)\]
\[ \Rightarrow {\text{reading}} = 3.35 + 0.03\]
\[ \Rightarrow {\text{reading}} = 3.38\,{\text{mm}}\]
Therefore, the diameter of the wire is 3.38 mm.

So, the correct answer is “Option D”.

Note:
To answer this question, students should know how the measurements of diameter are taken using the micrometer screw gauge. The spindle of the screw gauge is the head of the screw gauge which advances as we rotate it. The number of circular scale divisions in line with the main scale is 35 means the circular scale reading is 35.