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A screw gauge has least count of 0.01 mm and there are 50 divisions in its circular scale. The pitch of the screw gauge is : |
0.25 mm 0.5 mm 1.0 mm 0.01 mm |
0.5 mm |
Least count of screw gauge = $\frac{Pitch}{(Number of division on circular scale)}$ ⇒ 0.01 mm = $\frac{Pitch}{50}$ ⇒ Pitch = 0.5 mm |
