The head scale of a screw gauge has 200 divisions and it thimble moves by 0.5 mm along the main scale for one complete rotation of the screw. Find its least count.
Answers
Answered by
1
Answer:
Pitch of the screw gauge P=0.5 mm
Number of divisions on circular scale N=200
So, least count L.C.= N P
= 200
0.5 mm
=0.0025 mm=0.00025 cm
Explanation:
Answered by
1
Answer:
lc=pitch/no of circular scale division.
Explanation:
LC=0.5/200
LC=0.0025mm
I hope it will help you.bro main class 10 ka hi hokar are Sara padh liya hai mujhe brainleast mark Karo..
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