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Match List-I with List-II regarding the least count of various measuring instruments. List-I (Instrument Configuration) List-II (Least Count) (A) Screw gauge: Pitch 1 mm , 100 circular divisions (I) 0.02 mm (B) Screw gauge: Pitch 0.5 mm , 100 circular divisions (II) 0.1 mm (C) Vernier caliper: 1 MSD = 1 mm , 10 VSD = 9 MSD (III) 0.01 mm (D) Vernier caliper: 1 MSD = 1 mm , 50 VSD = 49 MSD (IV) 0.005 mm Choose the corr

Options

  1. A(A)-(II), (B)-(III), (C)-(I), (D)-(IV)
  2. B(A)-(III), (B)-(IV), (C)-(I), (D)-(II)
  3. C(A)-(IV), (B)-(III), (C)-(II), (D)-(I)
  4. D(A)-(III), (B)-(IV), (C)-(II), (D)-(I)

Correct answer

D. (A)-(III), (B)-(IV), (C)-(II), (D)-(I)

Step-by-step solution

For a screw gauge, Least Count (LC) = Pitch Number of circular divisions (A) LC = 1 mm 100 = 0.01 mm . This matches (III). (B) LC = 0.5 mm 100 = 0.005 mm . This matches (IV). For a Vernier caliper, Least Count (LC) = 1 MSD - 1 VSD (C) 10 VSD = 9 MSD 1 VSD = 0.9 MSD . LC = 1 MSD - 0.9 MSD = 0.1 MSD . Since 1 MSD = 1 mm , LC = 0.1 mm . This matches (II). (D) 50 VSD = 49 MSD 1 VSD = 49 50 MSD . LC = 1 MSD - 49 50 MSD = 1 50 MSD = 0.02 mm . This matches (I). Therefore, the correct matching is (A)-(III), (B)-(IV), (C)-(

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