In a screw gauge, five complete rotations of the circular scale produce a linear displacement of 2.5 mm along the main scale. The circular scale has 100 equal divisions. What is the least count of the screw gauge?
Options

Question

In a screw gauge, five complete rotations of the circular scale produce a linear displacement of 2.5 mm along the main scale. The circular scale has 100 equal divisions. What is the least count of the screw gauge?

Options

G1 Physics · Accepted Answer

For five complete rotations, the screw advances by 2.5 mm. Therefore the pitch is Pitch = 2.5 ÷ 5 = 0.5 mm. The least count is LC = Pitch ÷ number of circular-scale divisions = 0.5 ÷ 100 = 0.005 mm = 5×10⁻³ mm. Correct option: D.