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PhysicsMultiple correct+4 / -2Question 9 of 60

JEE Advanced 2015 · Paper 1, Physics question 9

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:

Choose one or more options
Answer and worked solution

Official answer: B, C

Vernier callipers. 1 MSD =18=\frac18 cm. Since 5 VSD = 4 MSD, 1 VSD = 0.8 MSD, and the least count is 0.20.2 MSD =0.025=0.025 cm =0.25=0.25 mm.

Screw gauge. The pitch equals two linear-scale divisions, and the least count is pitch/100.

  • (A)/(B) Pitch =2×0.25=0.5=2\times0.25=0.5 mm, so LC =0.005=0.005 mm. (B) is true.
  • (C)/(D) One linear division =0.5=0.5 mm, so the pitch is 1 mm and LC =0.01=0.01 mm. (C) is true.

(B), (C)

Topics:Vernier callipers and screw gaugeLeast count
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