Question

In a screw gauge, $5$ complete rotations of the screw cause it to move a linear distance of$0.25cm$. There are $100$ circular scale divisions. The thickness of a wire measured by this screw gauge gives a reading of $4$ main scale divisions and $30$circular scale divisions. Assuming negligible zero error, the thickness of the wire is:
A. $0.0430cm$
B. $0.3150cm$
C. $0.4300cm$
D. $0.2150cm$

Answer 1

Hint: Concept of least count and measurement of values like diameter of a wire using screw gauge from the main scale divisions and the circular scale coinciding division.

Formula used:
(1). Pitch of screw gauge ${\text{ = }}\dfrac{{{\text{Distance moved by screw gauge}}}}{{{\text{Number of fuel rotations given}}}}$
(2). Least count ${\text{ = }}\dfrac{{{\text{Pitch}}}}{{{\text{Total number of divisions on circular scale}}}}$
(3). Total reading $ = $ Linear scale reading $ + \,n \times $least count
Where n is the division on circular scale coinciding with the reference line.

Complete step by step answer:
$ \bullet $ A screw gauge is a device which is used to measure accurately the diameter of a wire or the thickness of a sheet of metal.
$ \bullet $ We know that, pitch is given by
Pitch ${\text{ = }}\dfrac{{{\text{Distance moved by screw}}}}{{{\text{Number of full rotations given}}}}$
Here, number of full rotations $ = 5$
Distance moved by screw $ = 0.25cm$
So, pitch $ = \dfrac{{0.25}}{5} = 0.05cm$
$ \bullet $ Least count ${\text{ = }}\dfrac{{{\text{Pitch}}}}{{{\text{Total number of divisions on circular scale}}}}$
Total number of divisions on circular scale $ = 100$
So, least count $ = \dfrac{{0.05}}{{100}}cm$
$ = 0.05 \times {10^{ - 2}}cm$
$ \bullet $ Total reading $ = $ Linear scale reading $ + n \times $Least count.
Where n is the division on circular scale coinciding with reference line
Here, least count $ = 0.05 \times {10^{ - 2}}cm$
$n = 30$
Linear scale reading $ = 4 \times $pitch
$ = 4 \times 0.05 = 0.2cm$
So, total reading $ = 0.2 + \left( {30 \times 0.05 \times {{10}^{ - 2}}} \right)cm$
$
   = 0.2 + 0.0150 \\
   = 0.2150cm \\
 $
So, the thickness of wire is $0.2150cm$

So, the correct answer is “Option D”.

Note:
 Pitch is linear distance moved per rotation not per cm. The word pitch is used in circular motion.
$ \bullet $ Both screw gauge and Vernier callipers are length measuring devices.
$ \bullet $ Screw gauge is used to measure external measurements only while Vernier callipers can be used for both external as well as internal measurements.