Please answer fast
The circular head of a screw gauge is divided into 50 divisions and the screw moves
1mm ahead in two revolutions of the circular head. Find its
(i) pitch
(ii) least count
Answers
Answer:
answer is least count hope it's helpful to you
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 = 1/2 mm
= pitch = o. 5mm
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
\begin{gathered}= > Least\:Count=\frac{0.5}{50}\:mm\\\\= > Least\:Count=0.01\:mm\end{gathered}
=>LeastCount=
50
0.5
mm
=>LeastCount=0.01mm
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