A screw gauge has $500$ scale on its head scale and the head scale moves $5$ mm when rotated $10$ times, then determine the pitch and least count of the screw gauge.

0.001

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0.001

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Solution

The correct option is C $0.001$ mm
Pitch of screw gauge $P = \frac{5 m m}{10} = 0.5 m m$
Number of divisions $N = 500$
Thus, Least count $L . C = \frac{P}{N} = \frac{0.5 m m}{500} = 0.001 m m$

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A screw gauge has 500 scale on its head scale and the head scale moves 5 mm when rotated 10 times, then determine the pitch and least count of the screw gauge.

The correct option is C 0.001 mmPitch of screw gauge P=5 mm10=0.5 mmNumber of divisions N=500Thus, Least count L.C=PN=0.5 mm500=0.001 mm

A screw gauge has $500$ scale on its head scale and the head scale moves $5$ mm when rotated $10$ times, then determine the pitch and least count of the screw gauge.

The correct option is C $0.001$ mm
Pitch of screw gauge $P = \frac{5 m m}{10} = 0.5 m m$
Number of divisions $N = 500$
Thus, Least count $L . C = \frac{P}{N} = \frac{0.5 m m}{500} = 0.001 m m$