Errors in measurementMedium
Question
In a vernier callipers, 50 vernier scale divisions are equal to 48 main scale divisions. If one main scale division $= 0.05\text{ }mm$, then the least count of the vernier callipers is $\_\_\_\_$ mm .
Options
A.0.002
B.0.05
C.0.02
D.0.005
Solution
$LC = 1MSD - 1MSD = 1MSD - \frac{48}{50}MSD$
$$= \frac{2}{50}MSD = \frac{2}{50} \times .05\text{ }mm = 0.002\text{ }mm$$
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