The circular scale (head scale) of a screw gauge is divided into $100$ equal parts and it moves $0.5 m m$ ahead in one rotation. The pitch and the least count are:

0.5 m m, 0.005 m m

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0.6 m m, 0.006 m m

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0.2 m m, 0.002 m m

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None of the above

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Solution

The correct option is A $0.5 m m, 0.005 m m$
Pitch of screw is distance moved by screw in one rotation = $0.5 m m$

Least count = pitch / number of divisions in circular scale $= 0.5 / 100 = 0.005 m m$

Answer. A) $0.5 m m, 0.005 m m$

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The circular scale (head scale) of a screw gauge is divided into 100 equal parts and it moves 0.5 mm ahead in one rotation. The pitch and the least count are:

The correct option is A 0.5 mm,0.005 mmPitch of screw is distance moved by screw in one rotation = 0.5 mmLeast count = pitch / number of divisions in circular scale =0.5/100=0.005 mm Answer. A) 0.5 mm,0.005 mm

The circular scale (head scale) of a screw gauge is divided into $100$ equal parts and it moves $0.5 m m$ ahead in one rotation. The pitch and the least count are:

The correct option is A $0.5 m m, 0.005 m m$
Pitch of screw is distance moved by screw in one rotation = $0.5 m m$

Least count = pitch / number of divisions in circular scale $= 0.5 / 100 = 0.005 m m$

Answer. A) $0.5 m m, 0.005 m m$