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JEE Advanced Physics Experimental Physics 2022 JEE Advanced 2022 (Paper 2)

JEE Advanced Physics Question (2022) — Solution

Question

Area of the cross-section of a wire is measured using a screw gauge. The pitch of the main scale is 0 . 5   mm . The circular scale has  100 divisions and for one full rotation of the circular scale, the main scale shifts by two divisions. The measured readings are listed below. Measurement condition Main scale reading Circular scale reading Two arms of gauge touching each other without wire 0  division 4  divisions Attempt- 1 : With wire 4  divisions 20 divisions Attempt- 2 : With wire 4  divisions 16 divisions What are the diameter and cross-sectional area of the wire measured using the screw gauge?

Options

  1. A. 2 . 22 ± 0 . 02 mm ,   π 1 . 23   ±   0 . 02 mm 2
  2. B. 2 . 22 ± 0 . 01 mm , π 1 . 23 ± 0 . 01 mm 2
  3. C. 2 . 14 ± 0 . 02 mm , π 1 . 14 ± 0 . 02 mm 2
  4. D. 2 . 14 ± 0 . 01 mm , π 1 . 14 ± 0 . 01 mm 2

Answer

C. 2 . 14 ± 0 . 02 mm , π 1 . 14 ± 0 . 02 mm 2

Step-by-step solution

Note: This question is given bonus by JEE council. In one rotation 2 divisions of the main scale are crossed. Therefore, the least count of the screw gauge is L C = 2 × 0 . 5 100 = 0 . 01   mm And the zero error is  4 × L C = 0 . 04   mm Reading-1 R 1 = MSR + LC × CSR - Zero   error = 2 + 0 . 20 - 0 . 04 mm = 2 . 16 mm Reading-2 R 2 = 2 + 0 . 16 - 0 . 04 mm = 2 . 12 mm Therefore, average reading  R m = R 1 + R 2 2 = 2 . 14   mm Average mean error  = R m - R 1 + R m - R 2 2 = 0 . 02   mm ⇒  Diameter  d = 2 . 14 ± 0 . 02   mm Area  = π d 2 4 ⇒   ∆ Area = 2 π d 4 ∆ d ⇒   ∆ Area ≅ π 0 . 02 ⇒   Area = π d 2 4 ± ∆ Area = π 1 . 14 ± 0 . 02   mm 2

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