JEE Main 2022 · 26 July · morning · Single correct · +4 / −1

A screw gauge of pitch 0.5 mm is used to measure the diameter of uniform wire of length 6.8 cm, the main scale reading is 1.5 mm and circular scale reading is 7 . The calculated curved surface area of wire to appropriate significant figures is :
[Screw gauge has 50 divisions on its circular scale]

  1. (a) 6.8 cm²
  2. (b) 3.4 cm²
  3. (c) 3.9 cm²
  4. (d) 2.4 cm²
Answer and solution

Answer (b) 3.4 cm²

Solution

Physics sir · 4 steps

Read the diameter off the gauge, use π d L, then trim the answer to the 2 sig figs of the length.

  1. Least count. Least count = pitch ÷ circular divisions = (0.5 mm)/50 = 0.01 mm.
  2. Diameter. Diameter = main scale reading + (circular reading × least count) = 1.5 + 7 × 0.01 = 1.57 mm = 0.157 cm.

    S = 2π r L = π d L = 3.1416 × 0.157 cm × 6.8 cm ≈ 3.354 cm²

  3. Significant figures. In a product, the result keeps as many significant figures as the least precise factor. d = 1.57 has 3 and L = 6.8 has only 2, so the area is rounded to 2 significant figures.
  4. So. 3.354 cm² rounds to 3.4 cm², which is option (b).

The idea to keep. Screw gauge reading = MSR + CSR × (pitch/divisions). When you multiply measured values, round the final answer to the fewest significant figures among them.

Answer from the NTA final answer key. Solution by PrepWiser.

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Asked about 60 times in JEE Main since 2020. This question is from 2022.