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
- A. 3.07 cm
- B. 3 . 11   cm
- C. 3 . 15   cm
- 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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