Physics Circle home Student Login

Home›Chapters›Units, Dimensions and Measurement PYQs›Least Count, Vernier Callipers & Screw Gauge›Question 49

JEE Main 2020 (3 Sep, Shift 1) · Least Count, Vernier Callipers & Screw Gauge

Least Count, Vernier Callipers & Screw Gauge question from JEE Main 2020 (3 Sep, Shift 1)

Q49JEE Main 2020 · 3 SEP · SHIFT 1Easy

Least Count, Vernier Callipers & Screw Gauge

Using screw gauge of pitch 0.1 cm and 50 divisions on its circular scale, the thickness of an object is measured. It should correctly be recorded as

Show answer and solution

Answer: (b)

  1. LC =0.1 cm50=0.002= \dfrac{0.1\text{ cm}}{50} = 0.002 cm
  2. A correct reading must be a whole multiple of 0.0020.002 cm (last digit even).
  3. Only 2.1242.124 cm =1062×0.002= 1062\times 0.002 cm fits.
  4. So the answer is (b) 2.1242.124 cm.
Report a mistake

More questions on Least Count, Vernier Callipers & Screw Gauge