Question

If the pitch of a screw gauge is $ 0.5mm $ and its least count is $ 0.01mm $ . Find the number of divisions on the head scale.
(A) $ 50 $
(B) $ 100 $
(C) $ 200 $
(D) $ 500 $

Answer 1

Hint : To solve this question, we need to obtain the formula for the least count of a screw gauge in terms of its pitch and the number of divisions. Then putting the values given in the question, we will get the final answer.

Complete step by step answer
The pitch of a screw gauge is defined as the distance moved by its screw on the main scale in its one complete rotation. Also the least count of the screw gauge is defined as the distance moved by the screw when it is rotated through one head scale or the circular scale division.
Let the number of divisions on the head scale be $ N $ .
For one rotation, the screw moves a distance equal to the pitch on the main scale. So in one rotation, the screw is rotated through $ N $ divisions. Therefore, when the screw is rotated through one division, the distance moved by the screw, known as the least count, is given by
 $ LC = \dfrac{{Pitch}}{N} $
 $ \Rightarrow N = \dfrac{{Pitch}}{{LC}} $
According to the question, the pitch is equal to $ 0.5mm $ , and the least count is equal to $ 0.01mm $ . Substituting these above, we have the number of divisions as
 $ N = \dfrac{{0.5}}{{0.01}} $
On solving, we finally get
 $ N = 50 $
Thus, the number of divisions on the head scale is equal to $ 50 $ .
Hence, the correct answer is option A.

Note
The screw gauge is popularly used in the measurement of the thickness of a metal sheet or wire in millimetres. From this calculation, the gauge number of the wire is determined. As the screw gauge is used for measuring the gauge number of a metal wire or sheet, so it is named so.