Physics Circle home Student Login

Home›Chapters›Units, Dimensions and Measurement PYQs›Least Count, Vernier Callipers & Screw Gauge›Question 12

JEE Main 2025 (28 Jan, Shift 1) · Least Count, Vernier Callipers & Screw Gauge

Least Count, Vernier Callipers & Screw Gauge question from JEE Main 2025 (28 Jan, Shift 1)

Q12JEE Main 2025 · 28 JAN · SHIFT 1Medium

Least Count, Vernier Callipers & Screw Gauge

A tiny metallic rectangular sheet has length and breadth of 5 mm and 2.5 mm, respectively. Using a specially designed screw gauge which has pitch of 0.75 mm and 15 divisions in the circular scale, you are asked to find the area of the sheet. In this measurement, the maximum fractional error will be x100\frac{x}{100} where xx is _______.

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

Show answer and solution

Answer: 3

  1. LC =0.7515=0.05 mm= \dfrac{0.75}{15} = 0.05\text{ mm}, so Δl=Δb=0.05 mm\Delta l = \Delta b = 0.05\text{ mm}
  2. A=lb⇒ΔAA=Δll+ΔbbA = lb \Rightarrow \dfrac{\Delta A}{A} = \dfrac{\Delta l}{l} + \dfrac{\Delta b}{b}
  3. =0.055+0.052.5=0.01+0.02=3100= \dfrac{0.05}{5} + \dfrac{0.05}{2.5} = 0.01 + 0.02 = \dfrac{3}{100}
  4. So x=3x = 3.
Report a mistake

More questions on Least Count, Vernier Callipers & Screw Gauge