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 division 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

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If the pitch of the screw gauge is twice the least count of the Vernier callipers, the least count of the screw gauge is 0.01mm.

If the pitch of the screw gauge is twice the least count of the Vernier callipers, the least count of the screw gauge is 0.005mm.

If the least count of the linear scale of the screw gauge is twice the least count of the Vernier calipers, the least count of the screw gauge is 0.01 mm.

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.

answer is B, C.

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Detailed Solution

FOR VERNIER CALIPERS
$1 MSD = \frac{1}{8} cm$
$1 VSD = \frac{4}{5} MSD = \frac{4}{5} \times \frac{1}{8} = \frac{1}{10} cm$
$LC of Vernier calliper = \frac{1}{8} - \frac{1}{10} cm = 0.025 cm$
(A) & (B) pitch of screw gauge $= 2 \times (0.025) = 0.05 c m$
Least count of screw gauge $= \frac{0.05}{100} c m = 0.005 m m$
(C) & (D) Least count of liner scale of screw gauge = 0.05 cm
Pitch = 0.05 x 2 cm = 0.1 cm
Least count screw gauge $= \frac{0.1}{100} c m = 0.01 m m$