Discussion Forum

Consider 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

Answer:

Explanation:

 For Vernier Callipers

     $1MSD=\frac{1}{8}cm$

       5  VSD=4 MSD

  $\therefore$                  $1VSD=\frac{4}{5}MSD$

                      = $\frac{4}{5}\times\frac{1}{8}=\frac{1}{10}cm$

 Least count of Vernier callipers

  = 1MSD-1VSD

      =  $\frac{1}{8}cm-\frac{1}{10}cm=0.025 cm$

 (a) and (b)

   Pitch of screw gauge

            =   $2\times0.025 =0.05 cm$

 Least count of screw gauge

                     $=\frac{0.05}{100}cm=0.005mm$

  (c) and (d)

 Least count of linear  scale of screw gauge=0.05

  Pitch= $0.05\times 2=0.1cm$

      Least count of screw gauge

                                         $=\frac{0.1}{100}cm=0.01 mm$