A screw gauge has 50 divisions on its circular scale and its screw moves by 1 mm on turning it by two revolutions. When the flat end of the screw is in contact with the stud, the zero of circular scale lies below the baseline and 4th division of circular scale is in line with the baseline. Find the least count.

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Solution

Number of divisions on circular head $= 50$
Distance moved in two revolution $= 1 m m$

Pitch $=$ distance moved in 1 revolution $= \frac{1}{2} = 0.5 m m = 0.5 c m$
Least count $= \frac{p i t c h}{n u m b e r o f d i v i s i o n s o n c i r c u l a r h e a d} = \frac{0.5}{50} = 0.01 m m$

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Number of divisions on circular head =50Distance moved in two revolution =1 mmPitch = distance moved in 1 revolution =12=0.5 mm=0.5 cmLeast count =pitchnumber of divisions on circular head=0.550=0.01 mm

Number of divisions on circular head $= 50$
Distance moved in two revolution $= 1 m m$

Pitch $=$ distance moved in 1 revolution $= \frac{1}{2} = 0.5 m m = 0.5 c m$
Least count $= \frac{p i t c h}{n u m b e r o f d i v i s i o n s o n c i r c u l a r h e a d} = \frac{0.5}{50} = 0.01 m m$