Consider a vernier calliper in which each 1cm on the main scale is divided into 8 equal divisions and a screw gauge with 100 divisions on its circular scale. In the vernier calliper, 5 divisions of the vernier scale coincides with 4 divisions of main scale and in the screw gauge, one complete rotation of the circular scale moves it by two divisions on the linear scale.

1MSD=18cm;  5VSD=4MSD1VSD=45MSD=45×18=110 cmLeast count of vernier caliper =1MSD−1VSD=18 cm−110 cm=0.025  cmPitch of scres gauge P=L⋅C 100If L⋅C=0.005  mm, then  P=0.5  mmNow  PL⋅C  of  vernier  callipers=0.5 mm0.025 cm=2So option A is correctIf L.C of screw gauge = 0.01 mm, then P = 1 mmNow L.C of linear scale of screw gauge  =P2=0.5  mm P/2L⋅C  of  V⋅C=0.5 mm0.025 cm=2So option C is correct