Question

A screw gauge gives the following reading when used to measure the diameter of a wire.
Main scale reading: \[0\,{\text{mm}}\]
Circular scale reading: \[52\] divisions
Given that \[1\,{\text{mm}}\] on the main scale corresponds to \[100\] divisions of the circular scale.
The diameter of wire from the above data is:
A. \[0.52\,{\text{cm}}\]
B. \[0.052\,{\text{cm}}\]
C. \[0.026\,{\text{cm}}\]
D. \[0.005\,{\text{cm}}\]

Answer 1

Hint: First of all, we will find the pitch from the given question followed by least count by using a simple formula, as the number of divisions in the circular scale is given. After that as we have main scale reading too, we will calculate the actual diameter of the wire.

Complete step by step solution:
In the given question, we are supplied with the following data:
There is a screw which is used to measure the thickness of smaller objects.Here it measures the diameter of a wire.The main scale reading is found to be \[0\,{\text{mm}}\] .The circular scale reading is found to be \[52\] divisions.We are also given that \[1\,{\text{mm}}\] on the main scale corresponds to \[100\] divisions of the circular scale.We are asked to find the diameter of the thin wire using the above data.

To begin with, we need to discuss a bit about the pitch, least count, etc. of the screw gauge. This will help us to correctly answer the question.
The screw gauge pitch is defined as the distance per revolution passed by the spindle, which is determined by rolling the head scale over the pitch scale to complete one full rotation.
In measurement theory, the lowest and most precise value in the calculated quantity that can be resolved on the instrument's scale is the least count of a measuring instrument.
Let us first calculate the least count of the screw gauge, which is given by the formula:
\[LC = \dfrac{p}{n}\] …… (1)
Where,
\[LC\] indicates the least count of the screw gauge.
\[p\] indicates the pitch of the screw.
\[n\] indicates the number of circular divisions in one complete revolution.
Now, we will substitute the required values in the equation (1) and we get:
$LC = \dfrac{p}{n} \\
\Rightarrow LC = \dfrac{1}{{100}} \\
\Rightarrow LC = 0.01\,{\text{mm}} \\
\Rightarrow LC = 0.001\,{\text{cm}} \\$
Therefore, the least count of the screw gauge is found to be \[0.001\,{\text{cm}}\] .
So, the diameter of the wire can be calculated by the formula, which is given below:
\[D = MSR + \left( {CSR \times LC} \right)\] …… (2)
Where,
\[D\] indicates the diameter of the wire.
\[MSR\] indicates the main scale reading.
\[CSR\] indicates the circular scale reading.
Now, we substitute the required values in the equation (2) and we get:
$D = MSR + \left( {CSR \times LC} \right) \\
\Rightarrow D = 0 + \left( {52 \times 0.001} \right) \\
\therefore D = 0.052\,{\text{cm}}$
Hence, the thickness of the wire is found to be \[0.052\,{\text{cm}}\] .

The correct option is B.

Note: In the given question, every time there won’t be \[100\] divisions in the circular scale. In that case, the least count will change too. Least count purely depends on the use. Instruments with higher accuracy will have a greater number of divisions in the circular scale.