Two full turns of the circular scale of a screw gauge cover a… | JEE Main 2008 (AIEEE)

Q66JEE Main 2008 · AIEEEDifficult

Least Count, Vernier Callipers & Screw Gauge

Two full turns of the circular scale of a screw gauge cover a distance of 1 mm on its main scale. The total number of divisions on the circular scale is 50. Further, it is found that the screw gauge has a zero error of − 0.03 mm. While measuring the diameter of a thin wire, a student notes the main scale reading of 3 mm and the number of circular scale divisions in line with the main scale as 35. The diameter of the wire is

Show answer and solution

Answer: (d)

Pitch $= \dfrac{1}{2} = 0.5\text{ mm}$ ; LC $= \dfrac{0.5}{50} = 0.01\text{ mm}$
Observed reading $= 3 + 35\times 0.01 = 3.35\text{ mm}$
Diameter $= 3.35 - (-0.03) = 3.38\text{ mm}$ . So the answer is (d).

Report a mistake

More questions on Least Count, Vernier Callipers & Screw Gauge

In a vernier calliper, 20 VSD coincide with 16 MSD (each division of length 1 mm). The…NEET 2026
One main scale division of a Vernier calliper is equal to 1 mm and the number of…NEET 2026 (Re-neet)
In a Vernier calipers, when both jaws touch each other, zero of the Vernier scale is…JEE Main 2026 (5 Apr, Shift 1)
In an experiment, a set of readings are obtained as follows: 1.24 mm, 1.25 mm, 1.23 mm…JEE Main 2026 (28 Jan, Shift 2)
In a vernier callipers, 50 vernier scale divisions are equal to 48 main scale divisions…JEE Main 2026 (24 Jan, Shift 2)
In a screw gauge when the circular scale is given five complete rotations it moves…JEE Main 2026 (4 Apr, Shift 1)
In a screw gauge the zero of main scale reference line coincides with the fifth division…JEE Main 2026 (2 Apr, Shift 2)
In a screw gauge, the zero of the circular scale lies 3 divisions above the horizontal…JEE Main 2026 (23 Jan, Shift 1)

Least Count, Vernier Callipers & Screw Gauge question from JEE Main 2008 (AIEEE)

More questions on Least Count, Vernier Callipers & Screw Gauge