Physics Circle home Student Login

Home›Chapters›Units, Dimensions and Measurement PYQs›Least Count, Vernier Callipers & Screw Gauge›Question 6

JEE Main 2026 (4 Apr, Shift 1) · Least Count, Vernier Callipers & Screw Gauge

Least Count, Vernier Callipers & Screw Gauge question from JEE Main 2026 (4 Apr, Shift 1)

Q6JEE Main 2026 · 4 APR · SHIFT 1Medium

Least Count, Vernier Callipers & Screw Gauge

In a screw gauge when the circular scale is given five complete rotations it moves linearly by 2.5 mm. If the circular scale has 100 divisions, the least count of screw gauge is ______ mm.

Show answer and solution

Answer: (d)

  1. Pitch =2.5 mm5=0.5 mm= \dfrac{2.5\text{ mm}}{5} = 0.5\text{ mm}
  2. LC =0.5 mm100=0.005 mm= \dfrac{0.5\text{ mm}}{100} = 0.005\text{ mm}
  3. So the answer is (d) 5×10−3 mm5\times 10^{-3}\text{ mm}.
Report a mistake

More questions on Least Count, Vernier Callipers & Screw Gauge