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Screw Gauge And Micrometer

Easyphysics

A screw gauge has a pitch of 1 mm and a circular scale divided into 100 equal divisions, so what is the least count of this instrument?

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About This Question

Subject
physics
Chapter
experimental skills
Topic
screw gauge and micrometer
Difficulty
Easy
Year
2025
Tags
screw gauge least countpitchcircular scale divisionsmicrometer resolutionspindle advance

Solution

Correct Answer:

0.01 mm

NCERT Class 11, Chapter 2 defines the least count of a screw gauge as the pitch divided by the number of divisions on the circular (head) scale. The pitch is the linear distance the spindle advances along the main scale in one complete rotation of the circular scale. Given a pitch of 1 mm and 100 circular divisions, the least count is mm. This is why a screw gauge measures far finer lengths than a vernier calipers of comparable size. The value 0.1 mm is wrong because it divides by 10 instead of 100, confusing the head scale with a vernier scale. The value 0.001 mm is wrong because it divides by 1000, over-counting the circular divisions by a factor of ten. The value 1 mm is wrong because it quotes the pitch itself, forgetting to divide by the circular divisions. It is worth noting that the pitch itself is defined by measuring the total distance travelled over several complete rotations and dividing by the number of rotations, which reduces the effect of any single reading error. This screw-based magnification of small displacements is what makes the instrument suitable for measuring wire diameters and thin sheets. A magnitude check confirms that 100 divisions covering a 1 mm advance sensibly resolve to one-hundredth of a millimetre, matching 0.01 mm, which is finer than a typical vernier calipers.

This easy difficulty physics question is from the chapter experimental skills, covering the topic of screw gauge and micrometer. It appeared in the 2025 exam.

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