Quantrex Quantrex AcademyJEE · NEET · NDA PYQs with solutions Open app
NEET2026PhysicsExperimental PhysicsActual

In a vernier callipers, 20 VSD coincide with 16 MSD (each division of length 1 mm). The least count of the vernier callipers is :

Options

  1. A0.2 cm
  2. B0.01 cm
  3. C0.02 cm
  4. D0.1 cm

Correct answer

C. 0.02 cm

Step-by-step solution

Given 20 VSD = 16 MSD. 1 VSD = 16 20 MSD = 0.8 MSD. Least count is given by 1 MSD - 1 VSD. Least count = 1 MSD - 0.8 MSD = 0.2 MSD. Since 1 MSD = 1 mm, we get Least count = 0.2 mm. Converting to cm, Least count = 0.02 cm. Answer: 0.02 cm

Practice Experimental Physics on Quantrex Academy →

More from Experimental Physics

One main scale division of a Vernier calliper is equal to 1 mm and the number of divisions on the Vernier scale is 10 . When both the jaws touch each other, the Vernier scale shift 2026Which of the following measurements require ‘index correction’ ? 2026Consider the diameter of a spherical object being measured with the help of a Vernier callipers. Suppose its 10 Vernier Scale Divisions (V.S.D.) are equal to its 9 Main Scale Divis 2025The pitch of an error free screw gauge is 1 mm and there are 100 divisions on the circular scale. While measuring the diameter of a thick wire, the pitch scale reads 1 mm and 63^ r 2024In a vernier callipers, (N+1) divisions of vernier scale coincide with N divisions of main scale. If 1 MSD represents 0.1 mm , the vernier constant (in cm ) is: 2024A 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 &# 2021The angle of 1 ' (minute of arc) in radian is nearly equal to 2020A plano-convex lens of unknown material and unknown focal length is given. With the help of a spherometer we can measure the 2020 Full Experimental Physics list All NEET PYQs