Calculate the least count of the screw gauge.

10^{- 2} c m

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10^{- 3} m m

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10^{- 3} c m

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10^{- 3} m

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Solution

The correct option is C $10^{- 3} c m$
As the number of divisions on main scale are 10 to a centimeter, so the smallest division on main scale will be $1 m m$ and it will be the pitch (p) of the screw gauge. The number of circular divisions is 100.

$L C = \frac{P}{N o . o f C S D}$

$= \frac{1 m m}{100} = 0.01 m m$

$= 10^{- 2} m m o r 10^{- 3} c m$

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Calculate the least count of the screw gauge.

The correct option is C 10−3cmAs the number of divisions on main scale are 10 to a centimeter, so the smallest division on main scale will be 1mm and it will be the pitch (p) of the screw gauge. The number of circular divisions is 100.LC=PNo.ofCSD=1mm100=0.01mm=10−2mm or 10−3cm

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As the number of divisions on main scale are 10 to a centimeter, so the smallest division on main scale will be $1 m m$ and it will be the pitch (p) of the screw gauge. The number of circular divisions is 100.