The correct option is C 1
We know that least count of the vernier callipers =1M.S.D−1V.S.D.
But since we are given that N divisions of the main scale coincide with N+m divisions of the vernier scale, we have:
N divisions of M.S.D=N+m divisions of V.S.D and,
1V.S.D=NN+mM.S.D
Now, L.C=1M.S.D−(NN+m)M.S.D or
L.C=mN+mM.S.D
Thus, for a minimum least count that is 0.001cm or 0.01mm, the value of m should be 1.