A) 1mm and 100
B) 0.5 mm and 50
C) 1 mm and 50
D) 0.5 mm and 100
Correct Answer: C
Solution :
[c] Least count of a screw gauge \[=\frac{Pitch}{Number\,of\,circular\,scale\,divisions}\] \[=\frac{1\,mm}{50}=0.02\,mm\] Therefore the pitch and no. of circular scale divisions are 1mm and 50 respectively.
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