The distance between the two consecutive threads of the screw of a screw gauge is 2 mm and the number of head scale divisions is $50$ . What is the least count (in mm) ?

$0.04 c m$

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$0.4 m m$

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$0.04 m m$

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$4 m m$

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Solution

The correct option is C
$0.04 m m$

Step 1: Given data and formula used
$P i t c h (P) = 2 m m$
Number of head scale divisions(N) $= 50$
$(L . C)$ stands for least count
$L . C = \frac{P}{N}$
Step 2:Calculation of $L . C$
On putting given value in the equation
$L . C = \frac{2 m m}{50} = 0.04 m m$
Hence the correct option is C.

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The distance between the two consecutive threads of the screw of a screw gauge is 2 mm and the number of head scale divisions is 50. What is the least count (in mm) ?

The correct option is C 0.04mmStep 1: Given data and formula usedPitch(P)=2mmNumber of head scale divisions(N)=50 (L.C)stands for least countL.C=PNStep 2:Calculation of L.COn putting given value in the equationL.C=2mm50=0.04mmHence the correct option is C.

The distance between the two consecutive threads of the screw of a screw gauge is 2 mm and the number of head scale divisions is $50$ . What is the least count (in mm) ?