A Vernier calipers has $$1 \,mm$$ marks on the main scale. It has $$20$$ equal divisions on the Vernier scale which match with $$16$$ main scale divisions. For this Vernier calipers, the least count is-
A.
$$0.02 \,mm$$
B.
$$0.05 \,mm$$
C.
$$0.1 \,mm$$
D.
$$0.2 \,mm$$
Answer :
$$0.2 \,mm$$
Solution :
20 divisions on the Vernier scale = 16 divisions of main scale
$$\therefore $$ 1 division on the Vernier scale $$ = \frac{{16}}{{20}}$$ divisions of main scale $$ = \frac{{16}}{{20}} \times 1\,mm = 0.8\,mm$$
We know that least count $$= 1\, MSD - 1\,VSD = 1\, mm - 0.8\, mm = 0.2\ mm$$
Releted MCQ Question on Basic Physics >> Unit and Measurement
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