A vernier calipers having 1 main scale division = 0.1 cm is designed to have a least count of 0.02 cm. If n be the number of divisions on vernier scale and m be the length of vernier scale, then

A vernier calipers having 1 main scale division = $0.1 cm$ is designed to have a least count of $0.02 cm$ . If n be the number of divisions on vernier scale and m be the length of vernier scale, then

A
$n = 10, m = 0.5 cm$
B
$n = 9, m = 0.4 cm$
C
$n = 10, m = 0.8 cm$
D
$n = 10, m = 0.2 cm$

Solution:

$1 MSD = 0.1 cm$

$L C = 1 M S D - 1 V S D$

$0.02 cm = 0.1 cm - 1 VSD$

$1 VSD = 0.1 - 0.02 = 0.08 cm$

$1 V S D = \frac{l e n g t h o f v e r e n i e r s c a l e}{n o . o f d i v i s i o n s} = 0.08 cm \frac{m}{n} = 0.08 c m$

only option with the values n=10 and m=0.8cm will satisfy it.

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