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:

Step-by-step solution

1 main scale division (M.S.D)   = 1 8 c m 5 Vernier scale division (V.S.D) = 4   M . S . D 1   V . S . D . = 4 5 M . S . D Least count of Vernier scale L . C . = 1   M . S . D . - 1   V . S . D . = 1   M . S . D . - 4 5 M . S . D L . C =   1   M . S . D 5 = 1 40 c m For option A and B If the pitch of the screw gauge is twice the least count of the Vernier callipers then pitch = 2 × L . C . of Vernier scale = 1 20 c m Hence least count of screw gauge = P i t c h 100 = 0.5