n=5
vernier const.=1/n * smallest MSD
A vernier calliper having 1 main scale div. .1cm is designed to have a least count of .02cm. If n be the no. of divisions on vernier scale and m be the length of vernier scale. Find n&m
n * L.C = MSR
n * 0.02 = 0.1
n = 10/2
n = 5
n * lc = m
5* 0.02 = 0.1 cm
abhirup sop editing the answer after me .
If you edit pls write editted .
@virang
i answered 5 before u...ok i delete my second ans
some one posted wrong ans in post #2...and deleted...not i
i never delete wrong ans..i always hide them