Physics Circle home Student Login

Home›Chapters›Units, Dimensions and Measurement PYQs›Least Count, Vernier Callipers & Screw Gauge›Question 42

NEET 2021 · Least Count, Vernier Callipers & Screw Gauge

Least Count, Vernier Callipers & Screw Gauge question from NEET 2021

Q42NEET 2021Medium

Least Count, Vernier Callipers & Screw Gauge

A screw gauge gives the following readings when used to measure the diameter of a wire Main scale reading: 0 mm

Circular scale reading: 52 divisions

Given that 1 mm on main scale corresponds to 100 divisions on the circular scale. The diameter of the wire from the above data is:

Show answer and solution

Answer: (a)

  1. Least count =pitchN=1 mm100=0.01 mm= \dfrac{\text{pitch}}{N} = \dfrac{1\text{ mm}}{100} = 0.01\text{ mm}
  2. Reading =MSR+CSR×LC=0+52×0.01=0.52 mm= \text{MSR} + \text{CSR}\times \text{LC} = 0 + 52\times 0.01 = 0.52\text{ mm}
  3. So the answer is (a) 0.052 cm0.052\text{ cm}.
Report a mistake

More questions on Least Count, Vernier Callipers & Screw Gauge