Question

Consider a Vernier calipers 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 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 :

• A. 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.
• B. If the pitch of the screw gauge is twice the least count of the Vernier calipers, the least count of the screw gauge is 0.005 mm.
• C. If the least count of 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.
• D. 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 1

Answer: (B), (C)

Vernier scale least count calculations
M.S.D. = Main scale division $= \frac{1}{8} \,cm$
V.S.D. = Vernier scale division
L.C. = M.S.D. - V.S.D.
5 × V.S.D. = 4 × M.S.D.
V.S.D. $= \frac{4}{5} M.S.D. = \frac{4}{5} \times \frac{1}{8} = \frac{1}{10}\,cm$
L.C. $= \frac{1}{8} - \frac{1}{10}\quad\quad\Rightarrow \frac{2}{10}\quad\quad\Rightarrow \frac{1}{40}\,cm\quad\quad\Rightarrow \,0.25 \, cm$
If pitch of screw gauge $= 2 × 0.25 = 0.5\, mm$
then L.C. of screw gauge $= \frac{0.5}{100} = 0.005\, mm$
option B is ans.
If L.C. of linear scale of screw gauge $= 2 × 0.25 \quad\quad⇒ 0.5 \,mm$
then L.C. of screw gauge $= \frac{2\times0.5}{100} = 0.01\, mm$