A screw gauge has 50 divisions on its circular scale. The circular scale is 4 units ahead of the pitch scale making, prior to use. Upon one complete rotation of the circular scale, a displacement of 0.5 mm is noticed on the pitch scale. The nature of zero error involved, and the least count of the screw gauge, are respectively :

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Negative, 2 $μ m$

Positive, 10 $μ m$

Positive, 0.1 mm

Positive, 0.1 $μ m$

answer is B.

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Detailed Solution

No. of division on circular scale of screw gauge = 50

Pitch = 0.5 mm
Least count of screw gauge
$= \frac{p i t c h}{N o . o f d i v i s i o n o n c i r c u l a r s c a l e}$ $= \frac{0.5}{50} m m = 10 μ m$

And nature of zero error is positive because the circular scale is 4 units ahead of the pitch scale making, prior to use, when studs are in contact.

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