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Consider a Vernier callipers in which each $1 \ cm$ on the main scale is divided into $8$ equal divisions and a screw gauge with $100$ divisions on its circular scale. In the Vernier callipers, $5$ divisions of the Vernier scale coincide with $4$ divisions on the main scale and in the screw gauge, one complete rotation of the circular scale moves it by two divisions on the linear scale. Then:
$(A)$ If the pitch of the screw gauge is twice the least count of the Vernier callipers, the least count of the screw gauge is $0.01 \ mm$.
$(B)$ If the pitch of the screw gauge is twice the least count of the Vernier callipers, the least count of the screw gauge is $0.005 \ mm$.
$(C)$ If the least count of the linear scale of the screw gauge is twice the least count of the Vernier callipers, the least count of the screw gauge is $0.01 \ mm$.
$(D)$ If the least count of the linear scale of the screw gauge is twice the least count of the Vernier callipers, the least count of the screw gauge is $0.005 \ mm$.
$(A,D)$
$(B,D)$
$(B,C)$
$(C,D)$
Solution
For vernier callipers, $1$ main scale division $=\frac{1}{8} cm$ $1$ vernier scale division $=\frac{1}{10} cm$
So least count $=\frac{1}{40} cm$
For screw gauge, pitch $(p)=2$ main scale division
So least count $=\frac{ p }{100}$
So option $(B)$ and $(C)$ are correct.
