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The circular head of a screw gauge is divided into 50 divisions and the screw moves 1 mm ahead in two revolutions of the circular head. Find its least count. |
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Answer» SOLUTION :Given, number of DIVISIONS on circular head = 50, and DISTANCE moved in two revolutions = 1 mm LEAST count = `("Pitch")/("Number of divisions on circular head")` `therefore L.C. =(0.05)/(50) = 0.001 cm` |
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