A Vernier calipers in which 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 calipers, 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

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

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 calipers, the least count of the screw gauge is 0.005 mm

answer is B, C.

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Detailed Solution

Smallest division on main scale
$= \frac{1}{8} cm=0.125 cm$
5 division of Vernier scale = 4 divisions of main scale
$∵$ 1 division of Vernier scale
$= \frac{4×0.125}{5} cm=0.1 cm$
Least count of Vernier calipers = 0.125 – 0.1 = 0.025 cm
Least count of screw gauge $= \frac{pitch}{100}$
If pitch of screw gauge is twice the least count of Vernier calipers:
$∵$ Least count screw gauge $= \frac{2×0.025cm}{100}$ = 0.005 mm
Also, least count of linear scale of screw gauge = 2 × 0.025 cm = 0.05 cm
$∵ pitch=2 \times 0.05 cm=0.1 cm$
Least count of screw gauge
$= \frac{0.1cm}{100} =0.001 cm$