In a screw gauge, two complete rotations of the circular scale advance the spindle by 1 mm on the main scale, and the circular scale has 50 equal divisions. The instrument is known to have a zero error of -0.03 mm. When the diameter of a fine wire is measured, the main scale reading is 3 m

Question

In a screw gauge, two complete rotations of the circular scale advance the spindle by 1 mm on the main scale, and the circular scale has 50 equal divisions. The instrument is known to have a zero error of -0.03 mm. When the diameter of a fine wire is measured, the main scale reading is 3 mm and the circular scale division coinciding with the reference line is 35. What is the wire’s diameter?

Accepted Answer

3.38 mm. Two rotations advance 1 mm so pitch = 0.5 mm, and with 50 divisions least count = 0.5/50 = 0.01 mm. Observed reading = 3 + 35*0.01 = 3.35 mm; correcting for zero error of -0.03 mm: 3.35 - (-0.03) = 3.38 mm.

Solution

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