Consider a Vernier callipers in which each \(1 \mathrm{~cm}\) 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 \(0.01 \mathrm{~mm}\).
(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 \(0.005 \mathrm{~mm}\).
(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
\(0.01 \mathrm{~mm}\).
(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
\(0.005 \mathrm{~mm}\).