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 $0.005 \ mm$.

  • [IIT 2015]
  • A

    $(A,D)$

  • B

    $(B,D)$

  • C

    $(B,C)$

  • D

    $(C,D)$

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