Question

The circular head of a screw gauge is divided into 50 divisions and the
screw moves 1 mm ahead in two revolutions of the circular head. Find
its pitch and least count.
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Answer 1

Explanation:

According to the question, the screw gauge has 50 divisions on its circular scale.  

The Pitch is calculated as the screw completes one rotation completely, the distance traveled by the screw defines the pitch.

Thereby, from given data in question we have movement of 1 mm by 2 rotations  

• Pitch = Distance / Rotation

$=>Pitch=\frac{1}{2}\:mm\\\\=Pitch=0.5\:mm$  

Least count is given by the ratio of pitch and the divisions present over the circular scale of screw gauge.

• Least Count = Pitch / Number of division over circular scale

$=>Least\:Count=\frac{0.5}{50}\:mm\\\\=>Least\:Count=0.01\:mm$

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Answer 2

Answer:

least count is given by the ratio of the pitch and the division present over the circular scale of screw gauge