Discussion Forum



1)

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


A) If the pitch of the screw gauge is twice the least count of the Vernier callipers, the least count of the screw gauge is 0.1 mm

B) If the pitch of the screw gauge is twice the least count of the Vernier callipers, the least count of the screw gauge is 0.05mm

C) If the least count of the linear scale of the screw gauge is twice the least count of the Vernier callipers, the least count of the screw gauge is 0.01mm

D) If the least count of the linear scale of the screw gauge is twice the least count of the Vernier callipers, the least count of the screw gauge is 0.005mm

Answer:

Option B,C

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

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