Physics Circle home Student Login

Home›Chapters›Units, Dimensions and Measurement PYQs›Least Count, Vernier Callipers & Screw Gauge›Question 7

JEE Main 2026 (2 Apr, Shift 2) · Least Count, Vernier Callipers & Screw Gauge

Least Count, Vernier Callipers & Screw Gauge question from JEE Main 2026 (2 Apr, Shift 2)

Q7JEE Main 2026 · 2 APR · SHIFT 2Medium

Least Count, Vernier Callipers & Screw Gauge

In a screw gauge the zero of main scale reference line coincides with the fifth division of the circular scale when two studs are in contact. There are 100 divisions in circular scale and pitch of screw gauge is 0.1 mm. When diameter of a sphere is measured, the reading of main scale is 5 mm and 50th division of circular scale coincides with the reference line of main scale. The diameter of sphere is ______ mm.

Show answer and solution

Answer: (a)

  1. LC =0.1 mm100=0.001 mm= \dfrac{0.1\text{ mm}}{100} = 0.001\text{ mm}
  2. With studs in contact the 5th division is on the reference line: positive zero error =+5×0.001=+0.005 mm= +5\times 0.001 = +0.005\text{ mm}
  3. Observed reading =5+50×0.001=5.050 mm= 5 + 50\times 0.001 = 5.050\text{ mm}
  4. Diameter =5.050−0.005=5.045 mm= 5.050 - 0.005 = 5.045\text{ mm}. So the answer is (a).
Report a mistake

More questions on Least Count, Vernier Callipers & Screw Gauge