Question

The pitch of the screw gauge is 0.5mm. Its circular scale contains 50 divisions. The least count of the screw gauge is:
a) 0.001mm
b) 0.01mm
c) 0.02mm
d) 0.025mm

Answer 1

Hint: The least count of an instrument is the smallest measurement it can make. For a screw gauge we have a pitch scale which can be considered as the main scale and the circular scale which divides the main scale into smaller units of its smallest measurements. Therefore using the definition of least count, we can obtain the correct answer of the above question.
Formula used:
$L.C=\dfrac{n(P)}{n(C)}$

Complete answer:
A screw gauge basically contains a pitch scale equivalent to the main scale of any instrument. In the question is given that the smallest division on the pitch scale i.e. the pitch of the instrument is 0.5mm. Let us say the instrument has no zero error and the circular scale just overlaps with the pitch scale zero. In the question it is also given to us that the circular scale is divided into 50 divisions.
Let us say the screw gauge has a pitch of n(P)and the number of divisions on the circular scale be n(C). Further let us say we give one complete rotation to the circular scale. Since the instrument has no zero error, it will just overlap with one division on the pitch scale. Hence the number of divisions moved on the circular scale will be n(C) and that on the pitch scale will be n(P). It is not that this is the least measurement we can make. If we just rotate the circular scale by half rotation, the number of divisions on the circular scale would be $\dfrac{n(C)}{2}$ and that on the pitch scale would be $\dfrac{n(P)}{2}$ . Now if we observe we can see the smallest measurement that we can make is by dividing the pitch of the screw gauge by the number of divisions on the circular scale. Hence the least (L.C) count is given by,
$\begin{align}
  & L.C=\dfrac{n(P)}{n(C)} \\
 & \because n(P)=0.5mm,\text{ }n(C)=50div \\
 & \Rightarrow L.C=\dfrac{0.5mm}{50} \\
 & \therefore L.C=0.01mm \\
\end{align}$

Therefore the correct answer of the above question is option b.

Note:
It is to be noted that the least count for different screw gauges can be different. It depends on how the instrument is calibrated. The name micrometer screw gauge comes from the principle of micrometer screw.