Physics Circle home Student Login

Home›Chapters›Units, Dimensions and Measurement PYQs›Least Count, Vernier Callipers & Screw Gauge›Question 13

JEE Main 2025 (24 Jan, Shift 1) · Least Count, Vernier Callipers & Screw Gauge

Least Count, Vernier Callipers & Screw Gauge question from JEE Main 2025 (24 Jan, Shift 1)

Q13JEE Main 2025 · 24 JAN · SHIFT 1Medium

Least Count, Vernier Callipers & Screw Gauge

The least count of a screw guage is 0.01 mm. If the pitch is increased by 75%75\% and number of divisions on the circular scale is reduced by 50%50\%, the new least count will be ________ ×10−3 mm\times 10^{-3}\text{ mm}

Numerical answer type. Work it out, then open the answer.

Show answer and solution

Answer: 35

  1. LC =pN=0.01 mm= \dfrac{p}{N} = 0.01\text{ mm}
  2. New LC =1.75p0.5N=3.5×0.01=0.035 mm= \dfrac{1.75p}{0.5N} = 3.5\times 0.01 = 0.035\text{ mm}
  3. =35×10−3 mm= 35\times 10^{-3}\text{ mm}, so the answer is 35.
Report a mistake

More questions on Least Count, Vernier Callipers & Screw Gauge