A screw gauge has 50 divisions on its circular sale. The pitch of the screw gauge is 0.5 mm. What is its least count? (0.001 cm)
Difficulty: Easy
Solution:
Number Of divisions on the circular scale = 50
Pitch of screw gauge = 0.5 mm
Least count Of screw gauge L.C. =?
Least count = Pitch / Number Of division on the circular scale
Least count = $\frac{0.5mm}{50}$
= 0.01 mm = $0.01\times \frac{1}{10}$cm
Least count = 0.001 cm
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