Least count = Number of parts on vernier scale Value of 1 part on main scale =5×1000.25 cm=5×10−4 cm Reading =4×0.05 cm+30×5×10−4 cm =(0.2+0.0150)cm=0.2150 cm (Thickness of wire)
In a screw gauge, 5 complete rotations of the screw cause it to move a linear distance of 0.25 cm. There are 100 circular scale divisions. The thickness of a wire measured by this screw gauge gives a reading of 4 main scale divisions and 30 circular scale divisions. Assuming negligible zero error, the thickness of the wire is:
Held on 15 Apr 2018 · Verified 6 Jul 2026.
0.0430 cm
0.3150 cm
0.4300 cm
0.2150 cm
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