Q. A vernier callipers having 1 main scale division = 0.1 cm is designed to have a least count of 0.02 cm. Let n be the number of divisions on vernier scale and m be the length of vernier scale, then which of the following is correct?
NTA AbhyasNTA Abhyas 2020
Solution:
Least count of vernier calliper is,
$LC=\frac{1 M S D}{n o . o f d i v i s i o n s o n v e r n i e r}$
So $0.02=\frac{0 . 1}{n}$
We get n = 5
∴ $LC=1MSD-1VSD$
So, 0.02 = 0.1 - 1 VSD
1 VSD = 0.08 cm
so length of vernier scale is $5\times 0.08=0.4cm$
$LC=\frac{1 M S D}{n o . o f d i v i s i o n s o n v e r n i e r}$
So $0.02=\frac{0 . 1}{n}$
We get n = 5
∴ $LC=1MSD-1VSD$
So, 0.02 = 0.1 - 1 VSD
1 VSD = 0.08 cm
so length of vernier scale is $5\times 0.08=0.4cm$