What is the least count of a screw gauge if the head scale of a screw gauge contains 100 divisions and its pitch is 1 mm?

0.1 mm

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0.01 mm

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0.001 mm

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0.0001 mm

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Solution

The correct option is B 0.01 mm
Least count of screq gauge is given by:
$L = \frac{Pitch}{Number of head scale divisions}$
$L = \frac{1 m m}{100}$
$L = 0.01 m m$

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What is the least count of a screw gauge if the head scale of a screw gauge contains 100 divisions and its pitch is 1 mm?

The correct option is B 0.01 mmLeast count of screq gauge is given by: L=PitchNumber of head scale divisions L=1 mm100 L=0.01 mm

The correct option is B 0.01 mm
Least count of screq gauge is given by:
$L = \frac{Pitch}{Number of head scale divisions}$
$L = \frac{1 m m}{100}$
$L = 0.01 m m$