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JEE Advanced Physics Experimental Physics 2021 JEE Advanced 2021 (Paper 1)

JEE Advanced Physics Question (2021) — Solution

Question

The smallest division on the main scale of a Vernier calipers is  0 . 1   cm . Ten divisions of the Vernier scale correspond to nine divisions of the main scale. The figure below on the left shows the reading of this calipers with no gap between its two jaws. The figure on the right shows the reading with a solid sphere held between the jaws. The correct diameter of the sphere is

Options

  1. A. 3.07 cm
  2. B. 3 . 11   cm
  3. C. 3 . 15   cm
  4. D. 3 . 17   cm

Answer

C. 3 . 15   cm

Step-by-step solution

Given,  10 V S D = 9 M S D Least count of the vernier caliper,  L C = M S D - V S D ⇒ L C = M S D - 9 M S D 10 = M S D 10 = 0 . 1   cm 10 = 0 . 1   mm ⇒ L C = 0 . 01   cm Case I: When sphere is not present between jaws then, main value  =  Zero error  = M . S . R + L . C . × V . S . R Zero error  = - 1 + 0.1 × 10 - 6   mm = − 0.04   cm Case II: When sphere is present between jaws then, main value  = 3.1 + 0.01 × 1 = 3.11   cm So, True value  =  Diameter of sphere  = 3.11 − − 0.4 = 3.15   cm

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