A vernier calipers has its main scale of 10 cm equally divided into 200 equal parts. Its verner scale of 25 divisions coincides with 12 mm on the main scale. The least count of the instrument is

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0.020 cm

0.010 cm

0.001 cm

0.002 cm

answer is A.

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Detailed Solution

$\begin{array}{l} 1 MSD = \frac{10 cm}{200} = \frac{1}{20} cm \\ IVSD = \frac{1.2 cm}{2.5} \end{array}$
LC = 1 MSD – 1 VSD
$LC = \frac{1}{20} - \frac{1.2}{25} = 0.002 cm$

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