The thimble of a screw gauge has 50 divisions. The spindle advances 1 mm when the screw is turned through two revolutions. What is the least count of the screw gauge?

0.001 m m

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0.1 m m

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0.01 m m

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0.0001 m m

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Solution

The correct option is C $0.01 m m$
Pitch of screw gauge $= \frac{D i s t a n c e m o v e d}{N o . o f c o m p l e t e r e v o l u t i o n s} = \frac{1}{2} = 0.5 m m$

Least count of screw gauge $= \frac{P i t c h}{N o . o f d i v i s i o n s} = \frac{0.5}{50} = 0.01 m m$

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The thimble of a screw gauge has 50 divisions. The spindle advances 1 mm when the screw is turned through two revolutions. What is the least count of screw gauge?

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The thimble of a screw gauge has 50 divisions. The spindle advances 1 mm when the screw is turned through two revolutions. What is the least count of the screw gauge?

The correct option is C 0.01mmPitch of screw gauge =Distance movedNo.of complete revolutions=12=0.5 mmLeast count of screw gauge =PitchNo.of divisions=0.550=0.01 mm

The correct option is C $0.01 m m$
Pitch of screw gauge $= \frac{D i s t a n c e m o v e d}{N o . o f c o m p l e t e r e v o l u t i o n s} = \frac{1}{2} = 0.5 m m$

Least count of screw gauge $= \frac{P i t c h}{N o . o f d i v i s i o n s} = \frac{0.5}{50} = 0.01 m m$